Class 9 Chemistry - BeTOPPERS IIT / NEET Foundation Series - 2022 Edition

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IIT FOUNDATION Class IX

CHEMISTRY

© USN Edutech Private Limited The moral rights of the author’s have been asserted. This Workbook is for personal and non-commercial use only and must not be sold, lent, hired or given to anyone else.

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Published by

:

USN Eductech Private Limited Hyderabad, India.

PREFACE Our sincere endeavour in preparing this Book is to enable students effectively grasp & understand the Concepts of Chemistry and help them build a strong foundation in this Subject. From among hundreds of questions being made available in this Book, the Student would be able to extensively practice in each concept exclusively, throughout that Chapter. At the end of each Chapter, two or three Worksheets are provided with questions which shall cover the entire Chapter, helping each Student consolidate his / her learning. This Book help students prepare for their respective Examinations including but not limited to i.e. CBSE, ICSE, various State Boards and Competitive Examinations like IIT, NEET, NTSE, Science Olympiads etc. It is compiled by our inhouse team of experts who have a collective experience of more than 40 years in their respective subject matter / academic backgrounds. This books help students understand concepts and their retention through constant practice. It enables them solve question which are ‘fundamental / foundational’ as well questions which needs ‘higher order thinking’. Students gain the ability to concentrate, to be self-reliant, and hopefully become confident in the subject matter as they traverse through this Book. The important features of this books are: 1.

Lucidly presented Concepts: For ease of understanding, the ‘Concepts’ are briefly presented in simple, easy and comprehensible language.

2.

Learning Outcomes: Each chapter starts with ‘Learning Outcomes’ grid conveying what the student is going to learn / gain from this chapter.

3.

Bold-faced Key Terms: The key words, concepts, definitions, formulae, statements, etc., are presented in ‘bold face’, indicating their importance.

4.

Tables and Charts: Numerous strategically placed tables & charts, list out etc. summarizes the important information, making it readily accessible for effective study.

5.

Box Items: Are ‘highlighted special topics’ that helps students explore / investigate the subject matter thoroughly.

6.

Photographs, Illustrations: A wide array of visually appealing and informative photographs are used to help the students understand various phenomena and inculcate interest, enhance learning in the subject matter.

7.

Flow Diagrams: To help students understand the steps in problem-solving, flow diagrams have been included as needed for various important concepts. These diagrams allow the students visualize the workflow to solve such problems.

8.

Summary Charts: At the end of few important concepts or the chapter, a summary / blueprint is presented which includes a complete overview of that concept / chapter. It shall help students review the learning in a snapshot.

9.

Formative Worksheets: After every concept / few concepts, a ‘Formative Worksheet’ / ‘Classroom Worksheet’ with appropriate questions are provided from such concept/s. The solutions for these problems shall ideally be discussed by the Teacher in the classroom.

10. Conceptive Worksheets: These questions are in addition the above questions and are from that respective concept/s. They are advised to be solved beyond classroom as a ‘Homework’. This rigor, shall help students consolidate their learning as they are exposed to new type of questions related to those concept/s.

11. Summative Worksheets: At the end of each chapter, this worksheet is presented and shall contain questions based on all the concepts of that chapter. Unlike Formative Worksheet and Conceptive Worksheet questions, the questions in this worksheet encourage the students to apply their learnings acquired from that entire chapter and solve the problems analytically. 12. HOTS Worksheets: Most of the times, Summative Worksheet is followed by an HOTS (Higher Order Thinking Skills) worksheet containing advanced type of questions. The concepts can be from the same chapter or as many chapters from the Book. By solving these problems, the students are prepared to face challenging questions that appear in actual competitive entrance examinations. However, strengthening the foundation of students in academics is the main objective of this worksheet. 13. IIT JEE Worksheets: Finally, every chapters end with a IIT JEE worksheet. This worksheet contains the questions which have appeared in various competitive examinations like IIT JEE, AIEEE, EAMCET, KCET, TCET, BHU, CBSE, ICSE, State Boards, CET etc. related to this chapter. This gives real-time experience to students and helps them face various competitive examinations. 14. Different Types of Questions: These type of questions do appear in various competitive examinations. They include:

• Objective Type with Single Answer Correct

• Non-Objective Type

• Objective Type with > one Answer Correct

• True or False Type

• Statement Type - I (Two Statements)

• Statement Type - II (Two Statements)

• MatchingType - I (Two Columns)

• MatchingType - II (Three Columns)

• Assertion and Reasoning Type

• Statement and Explanation Type

• Roadmap Type

• FigurativeType

• Comprehension Type

• And many more...

We would like to thank all members of different departments at BeTOPPERS who played a key role in bringing out this student-friendly Book. We sincerely hope that this Book will prove useful to the students who wish to build a strong Foundation in Chemistry and aim to achieve success in various boards / competitive examinations. Further, we believe that as there is always scope for improvement, we value constructive criticism of the subject matter, as well as suggestions for improving this Book. All suggestions hopefully, shall be duly incorporated in the next edition. Wish you all the best!!!

Team BeTOPPERS

CONTENTS 1.

Language of Chemistry

..........

01 - 28

2.

Elements, Compounds & Mixtures

..........

29 - 56

3.

Atomic Structure

..........

57 - 82

4.

Periodic Classification

..........

83 - 108

5.

Chemical Bonding

..........

109 -140

6.

Gaseous State

..........

141 - 158

7.

Mole Concept

..........

159 - 182

8.

Solutions

..........

183 - 198

9.

Compounds of Carbon

..........

199 - 220

10.

Compounds of Nitrogen

..........

221 - 244

11.

Compounds of Sulphur

..........

245 - 266

12.

Key and Answers

..........

267 - 377

By the end of this chapter, you will understand • Atom and its basic structure • Molecular weight • Atomic number • Ions • Valency • Mass number Valency chart • Complete representation of an element • • Atomic weight • Chemical formula • Atomic mass unit • Chemical equations • Molecules • Methods to balance a chemical equation i)

1. Introduction Chemistry is a branch of science which deals with the study of the composition of matter and the chemical changes involved in it. In other words, chemistry is the science of matter and its transformation. Like all other branches of science, chemistry has its own technical terms. In chemistry, all chemical reactions are expressed in terms of chemical equations which involve symbols and formulae. All chemical substances are expressed in symbols and formulae which form the language of chemistry. For example, in the language of chemistry; sodium is expressed as ‘Na’, potassium as ‘K’, water as ‘H2O’, calcium carbonate as ‘CaCO3’ and so on. When sodium chloride is heated with conc. sulphuric acid, we get sodium sulphate and hydrogen chloride. In the language of chemistry, the above reaction is expressed as below:

a)

NaCl  H 2SO4  NaHSO 4  HCl Sodium. chloride

Sulphuric acid

Sod. bisulphate

Hydrogen chloride

2. Atom and its basic structure An atom is the smallest part of an element which takes part in chemical reactions and may or may not exist independently. An atom essentially consists of two parts: i) Nucleus, and ii) Electrons, which revolve round the nucleus in fixed paths, much the same way as the planets revolve round the sun. The path along which electrons revolve is called orbit or shell.

b)

Chapter -1

LANGUAGE OF CHEMISTRY

Learning Outcomes

Nucleus It is a very small region situated in the centre of the atom. The diameter of the nucleus of an atom is 10, 000 times smaller than the diameter of an atom. The nucleus of hydrogen consists of a single proton. However, the nuclei of all other atoms consists of protons and neutrons. The protons and neutrons collectively present within the nucleus are called nucleons. The nucleus in an atom is positively charged, because of the presence of protons. Protons The mass of one proton is taken as 1 a.m.u. (atomic mass unit). The charge on one proton is taken as unit positive charge. The proton is held within the nucleus. It is the number of protons present within the nucleus, which distinguishes atoms of one element from the other elements. For example, if the number of protons within the nucleus is only one, the atom of an element is hydrogen. However, if nucleus has two protons, the atom is of an element is helium. Neutrons Neutrons have no electrical charge on them. The mass of a neutron is almost equal to the mass of a proton. For the sake of simplicity, we regard the mass of one neutron equal to 1 a.m.u.

9th Class Chemistry

2 ii)

Electrons

1 times the 1837 mass of 1 hydrogen atom or the mass of one proton. For all practical purposes, an electron is regarded weightless. The electric charge on an electron is negative, i.e., each electron has a unit negative charge. The mass of an electron is

How are electrons arranged around the nucleus? The paths of electron around the nucleus are called orbits or shells. Outermost shell Nucleus Inner shells

The orbit closest to the nucleus is called first orbit. The next orbit is called second orbit, third orbit and so on. It has been found that maximum number of electrons in the first orbit can be 2, in the second orbit 8 and in the third orbit 18. However, the outermost orbit cannot have more than 8 electrons.

3. Atomic number Though every human being is apparently similar to fellow human beings, yet certain unique features are exclusively associated with an individual. No two individuals can have these features in common. Do you know, what these unique features are? They are the DNA and fingerprints. Similarly, elements too have a unique feature. No two elements have this unique feature. Can you name this unique feature? This unique feature is atomic number. What DNA and fingerprints are to human beings, the atomic number is to elements. The atomic number of an element is defined as the number of electrons present in the neutral atom of the element. It is represented by the symbol ‘Z’. If the atomic number of an element is equal to ‘X’, it means that the number of electrons in a neutral atom of that element are equal to ‘X’ each (or) the number of protons in the nucleus of an atom of that element are equal to ‘X’. www.betoppers.com

Eg: The atomic number of ‘Phosphorus’ is 15. This means that the number of protons in the nucleus of the Phosphorus atom are equal to 15 and also the number of electrons in the neutral atom of Phosphorus are equal to 15.

4. Mass number We know that an atom consists of protons, neutrons and electrons and the mass of the electrons is negligible. Therefore, the real mass of an atom is determined by the total number of protons and neutrons it contains. The total number of protons and neutrons present in the atom of an element is known as its mass number. It is represented by ‘A’.  Mass number (A) = Number of protons + Number of neutrons (n)  Mass number (A) = Atomic number (Z) + Number of neutrons (n)  A = Z + n Number of Neutrons in Nucleus: We know, A = Z + n  n = A – Z  Number of neutrons = Mass number – Atomic number Atomic Mass no. of Name number number neutrons (Z) (A) (A-Z) Hydrogen 1 1 0 Helium 2 4 2 Lithium 3 7 4 Beryllium 4 9 5 Boron 5 11 6 Carbon 6 12 6 Nitrogen 7 14 7 Oxygen 8 16 8 Fluorine 9 19 10 Neon 10 20 10 Sodium 11 23 12 Magnesium 12 24 12 Aluminium 13 27 14 Silicon 14 28 14 Phosphorus 15 31 16 Sulphur 16 32 16 Chlorine 17 35 18 Argon 18 40 22 Potassium 19 39 20 Calcium 20 40 20

Language of Chemistry

3

5. Complete representation of an element We know that symbol is that shorthand form of representing an element. All the details of an element cannot be conveyed by a symbol. Hence, there is a need for representation that conveys all the possible details of an element. One such representation of an element is as follows:

X or Z X A The atomic number (Z) is written on the lower left side of the symbol. The mass number (A) is written on either the upper left side or upper right side of the symbol (x).

Conceptive Worksheet 1.

A Z

For example: Oxygen is represented as 8 O16 . This means that: i) The element is oxygen ii) Its atomic number is 8. iii) Its mass number is 16. iv) Number of protons in it are 8. v) Number of electrons in it are 8. vi) Number of neutrons in it are 16 – 8 = 8 Hence, we can draw the maximum details of an element, if its complete representation is known.

Formative Worksheet 1.

Atoms consists of electrons, protons and neutrons. If the mass attributed to neutron were halved and that attributed to the electrons were doubled, atomic mass of

2.

4. 5. 6. 7. 8.

C would be

3.

4.

5. 6.

7.

approximately 1) same 2) doubled 3) halved 4) reduced by 25% If three neutrons are added to the nuclei of 235 92

3.

12 6

2.

U the new particles have an atomic number

of 1) 89 2) 9 5 3) 90 4) 92 A tripositively charged ion of an element 'X' has the same number of electrons as in trinegatively charged N–3. Then identify 'X'. Atomic number is always equal to the number of electrons present in an atom. True / False. Explain. The atoms of all the elements contain electrons, protons and neutrons. True / False. Explain. What is the ratio of number of neutrons in Silicon to phosphorus? Why does an atomic number not change during a chemical reaction? Find the number of electrons, protons in a) Nitrate ion b) Sulphate ion

8.

9.

The atomic number of sodium is 11. i) How many protons are there in the nucleus of a sodium atom? ii) How many electrons does an atom of sodium contain? iii) How many electrons and protons are present in Na+ ? State the number of protons, electrons and neutrons in each of the following atoms: 1) Atom 'A' [Z = 4, A = 9] 2) Atom 'B' [Z = 11, A = 23] 3) Atom 'C' [Z = 15, A = 31] 4) Atom 'D' [Z = 18, A = 40] An atom 'Y' contains ten electrons and ten neutrons. State a) The number of protons it contains b) The mass number of 'Y' An atom 'Z' has a mass number twelve and contains six electrons. State a) The number of neutrons it contains b) The atomic number of 'Z' Mass number of chlorine is 35.5. True / False. The number of neutrons present in bivalent +vely charged zinc ion with mass number 70 1) 32 2) 30 3) 40 4) 28 Which one of the following statements about atomic number is false? 1) It is equal to the number of protons present in the nucleus. 2) It is a more fundamental property of the atom than the atomic weight. 3) No two elements can have the same atomic number. 4) The atomic number of an element decides its stability. Which of the following species has more electrons than protons and more protons than neutron? 1) D– 2) H+ 3) D3O+ 4) OH– The atomic weight of an element is 52 and its atomic number is 24. The number of electrons, protons and neutrons in an atom of this element will be respectively 1) 24, 24, 28 2) 24, 28, 24 3) 28, 24, 24 4) None www.betoppers.com

9th Class Chemistry

4 10. The atomic mass of lead is 208 and its atomic number is 82. The atomic mass of bismuth is 209 and its atomic number is 83. The ratio of neutrons/protons in the atom 1) higher of Pb 2) higher of Bi 3) same of both

4) None

6. Atomic weight Do you know the mass of a single atom of hydrogen? The mass of a single atom of hydrogen is 0.000000000000000000000017 gms. Do you know the mass of single atom of helium.? The mass of a single atom of helium is 0.00000000000000000000000664gm. Learning and remembering the above values of mass of atoms is very difficult or impossible . How to remember the masses of different atoms? A simple system to represent atomic mass has been designed. In this system, atomic weight or atomic mass is obtained by comparing the mass of one atom of an element with a standard mass of atom. The latest universally accepted reference is the atom C – 12 atom. 1/12th of mass of C–12 atom is taken as standard. Let’s try to find the weight of an oxygen atom. If one atom of oxygen is on left pan of the balance, then we require 16 standard units to balance the pan. Then atomic weight of oxygen is 16. Atomic weight is defined as the number of times an atom is heavier than 1/12th of the mass of C-12 atom. Mathematically, Atomic Weight 

Weight of one atom of the element Weight of 121 th of C  12 atom

Example: The atomic weight of calcium is 40. It implies that one calcium atom is 40 times heavier than the mass of 1/12th of C–12 atom. Note: Atomic weights do not have any units since they are relative atomic masses of substances.

Table showing atomic weights of different elements

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Element Hydrogen Helium Lithium Beryllium Boran Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon Potassium Calcium

Symbol H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Atomic weight 1.0079 = 1 4.0026 = 4 6.941 = 7 9.0122 = 9 10.81 = 11 12.011 = 12 14.0067 = 14 15.9994 = 16 18.9984 = 19 20.179 = 20 22.98977 = 23 24.305 = 24 26.9815 = 27 28.0855 = 28 30.9738 = 31 32.06 = 32 35.453 = 35.5 39.948 = 40 39.0983 = 39 40.08 = 40

Easy way to remember atomic weights The atomic weights can be remembered easily by correcting them with the atomic numbers, even though there is no standard relationship between the atomic weights and the atomic numbers. On observing the above chart, it is seen that, for most of the even atomic numbered elements, the atomic weight is twice the atomic number i.e., Atomic weight = 2Z. The exceptions are Beryllium (9) and Argon (40). For odd atomic numbered elements, the atomic weights can be expressed as 2Z + 1. The exceptions here are hydrogen (1) nitrogen (14) and chlorine (35.5). Note: A naturally occurring element with maximum atomic weight is Uranium. One of its isotopes has an atomic number weight of 238. The element with minimum atomic weight is Hydrogen, with an atomic weight of 1.

7. Atomic mass unit For day - to - day measurement of weight, the units used are kilograms and grams. For measuring bigger weights, we use quintals and tones. Can we use these units for measuring the weight of atoms, molecules and elementary particles? Let us check. The mass of hydrogen atom in terms of grams is 1 × 1. 66 × 10–24 grams and the mass of carbon is 12 × 1.66 × 10–24 grams.

Language of Chemistry

5

As such a method of expression is very complex, we get decimals and negative exponents in the weights. To avoid these complications, we have a special unit to measure the weight of atoms, molecules and elementary particles. It is called a.m.u (atomic mass unit). The mass of the lightest atom, hydrogen, is taken as a standard and is used as a unit - amu. One amu is the mass of the hydrogen atom. In terms of grams, one a.m.u = 1.66 × 10– 24 grams. In modern times, the reference is changed from hydrogen to C – 12 isotope and a.m.u is defined as follows: One atomic mass unit (a.m.u) is equal to

Note: Atomic mass unit is also called Avogram (or) Dalton. Let us now observe the following table, showing the mass of single atoms, of the elements.

Mass in a.m.u

Electron

0.000549

Proton

1.0072651

Neutron

1.0086651

Relation between the atomic weight and the mass of one atom of that elements

1th of the 12

Consider an element with atomic weight ‘X’. It means that one atom of the element is ‘X’ times heavier than 1/12th of C–12 isotope’s atom. Mathematically, Mass of one atom of element = X (mass of 1/12th of C-12 isotope’s atom = X (1 a.m.u) = X a.m.u. ( 1 a.m.u = Mass of 1/12th C-12 isotope’s atom) For example, the atomic weight of Carbon is 12. So, the mass of an atom of Carbon is 12 a.m.u.

mass of one carbon – 12 atom.

1 1 (mass of one C – 12)  × mass 12 12 of one C – 12 atom 1 a.m.u 



Atomic particle

1  12  1.66  1024  1.66 1024 g . 12

H

Atomic weight (approximately) 1

Atomic weight (in a.m.u) 1a.m.u=1×1.66×10-24g

Helium

He

2

2 a.m.u=2×1.66×10-24g

3.

Lithium

Li

7

7 a.m.u=7×1.66×10-24g

4.

Beryllium

Be

9

9 a.m.u=9×1.66×10-24g

5.

Boron

B

11

11 a.m.u=11×1.66×10-24g

6.

Carbon

C

12

12 a.m.u=12×1.66×10-24g

7.

Nitrogen

N

14

14 a.m.u=14×1.66×10-24g

8.

Oxygen

O

16

16 a.m.u=16×1.66×10-24g

9.

Fluorine

F

19

19 a.m.u=19×1.66×10-24g

S.No.

Element

Symbol

1.

Hydrogen

2.

10. Neon Ne 20 20 a.m.u=20×1.66×10-24g _____________________________________________________________________________ 12. The weight of Helium atom in grams is: ormative orksheet 1) 2 2) 4 –24 3) 6.64 × 10 4) 1.66 × 10–24 9. Atomic weight of an element is ‘x’. Find the weight 13. The mass of an atom of an element ‘x’ is 39. The of one atom of that element number of atoms of it present in Gram atomic 10. The ratio of weight of one atom of an element to its weight is: atomic weight is equal to: – 1) 1 2) 1.66 × 1024 11. The mass of one atom of an element is 40×1.66×10 24 3) 6.023 × 1023 4) 96500 g. Find the number of protons present in its 14. The total mass of 100 atoms of silicon is nucleus 1) 2800 2) 2800 amu –22 3) 28 × 1.66 × 10 g 4) 280 kg

F

W

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9th Class Chemistry

6

Conceptive Worksheet 11. The modern atomic weight scale is based on: 1) C12 2) O16 3) H1 4) C13 12. 1 amu is equal to: 1 1 1) th of C - 12 2) th of O-16 12 14 3) 1g of H2 4) 1.66 × 10–23 kg 13. 1 atomic mass unit = 1 th 1) mass of a carbon - 12 atom 12 2) 1.66 × 10–24g 3) 6.023 × 10–23g 4) 6.023 × 1023g

8. Molecules It was the Italian chemist, Amedo Avogadro (1776-1856 A.D), who introduced the word ‘molecule’ The smallest particle of an element or a compound that can exist independently is called a molecule. Examples: O2 is an oxygen molecule, O3 is an ozone molecule, NH3 is an ammonia molecule, CH4 is a methane molecule, etc. Molecules are made up of atoms; atoms in turn are made up of sub-atomic particles. Its size is generally expressed in Angstrom (A°) Different molecules have different shapes. A few examples are as follows: i)

The shape of water molecule (H2O) is angular.

ii)

The shape of ammonia molecule (NH3) is pyramidal.

Types of molecules Molecules like, H2 , Cl2, F 2 are formed by the combination of atoms of the same element and molecules like H2O, N2O, NO2 are formed by the combination of atoms of different elements. Thus, molecules can be categorised as follows: Homogeneous molecules: Molecules that are formed by the combination of atoms of the same element are called homogeneous molecules. Eg: H2, Cl2, F2, O3, S6, S8

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Heterogeneous molecules: Molecules that are formed by the combination of atoms of different elements are called heterogeneous molecules. Example: H2O, NO2, N2O.

Atomicity The number of atoms present in a homogeneous molecule is known as Atomicity. No. of Named as atoms He, Ne, Ar, Kr, Xe 1 Monoatomic H2, Cl2, O2, F2, l2 2 Diatomic O3 3 Triatomic P4 4 Tetra atomic More S6, S8, C60 Poly atomic than 4 Molecule

Highest atomicity: The element with the highest atomicity is Carbon. Carbon has the property of self-linkage, known as Catenation. A group of Carbon atoms can interlink in a network, resulting in the formation of a spherical football-like structure. The most famous one is C–60 molecule called Buckminster fullerene. This C–60 molecule looks like a soccer ball and is called “Bucky ball”. Similar fullerenes like C32, C50, C70, C76 and C84 may also be formed.

9. Molecular weight Relative molecular mass or molecular weight is defined as the number of times a molecule is heavier 1 th the mass of C-12 isotope’s atom. than 12 Relative molecular mass RMM

Average mass of one molecule = Weight of 1 12 th of C -12 atom . Relative molecular mass or molecular weight has no units. If the relative molecular mass or molecular weight of any compound is M, then its molecular 1 mass is ‘M’ a.m.u.= Mol.wt  th the mass of C12 12 atom.

Language of Chemistry

7

Table showing molecular weights of some compounds

For example, the molecular weight of calcium carbonate is 100. So, the mass of one molecule of calcium carbonate is 100 a.m.u.

S .No.

Molecule

Symbol

Molec ul ar weight

1.

Hydrogen

H2

2

2. 3.

Nitrogen Oxygen

N2 O2

28 32

4. 5.

Carbon dioxide Carbon monoxide

CO2 CO

44 28

6. 7. 8. 9.

Ammonia Methane Nitrogen dioxide Water

NH3 CH4 NO2 H 2O

17 16 46 18

Relation between molecular weight and weight of one molecule

10. 11. 12.

Hydrochloric acid Sulphuric acid Nitric acid

HCl H 2SO4 HNO3

36.5 98 63

Consider a compound whose molecular weight is ‘Y’. It means that its molecule is ‘Y’ times heavier than 1/12th of C12 isotope’s atoms.

13. 14.

Phosphoric acid Sodium hydroxide

H 3PO4 NaOH

98 40

 Mass of one molecule of the compound

15. 16.

Magnesium hydroxide Aluminium hydroxide

Mg(OH) 2 Al (OH) 3

58 78

= Y (mass of 1/12th of C12 isotope).

17. 18.

Sodium chloride Magnesium oxide

NaCl MgO

58.5 40

19. 20.

Sodium sulphate Sodium carbonate

Na2SO4 Na2CO3

142 106

Note: Mass of one molecule of any compound in grams = (molecular weight) × 1.66 ×10–24 grams ( a.m.u. = 1.66 × 10–24 grams) For example, mass of one molecule of Calcium carbonate = 100 a.m.u. = 100 × 1.66 × 10–24 grams = 1.66 × 10–22 grams

= ‘Y’ times 1/12th of C12 isotope

= Y × 1 a.m.u (since mass of 1/12th of C12 isotope’s atom = 1 a.m.u.) = Y a.m.u.

Relation between molecular weight and weight of one molecule

So, we may conclude that if the molecular weight of an element is Y, then the mass of an atom of that element is Y a.m.u.

Consider a compound whose molecular weight is ‘Y’. It means that its molecule is ‘Y’ times heavier than 1/12th of C12 isotope’s atoms.

For example, the molecular weight of calcium carbonate is 100. So, the mass of one molecule of calcium carbonate is 100 a.m.u.

 Mass of one molecule of the compound = ‘Y’ times 1/12 of C isotope

Note: Mass of one molecule of any compound in grams = (molecular weight) × 1.66 ×10–24 grams

= Y (mass of 1/12th of C12 isotope).

( a.m.u. = 1.66 × 10–24 grams)

= Y × 1 a.m.u

For example, mass of one molecule of Calcium carbonate = 100 a.m.u.

th

12

(since mass of 1/12th of C12 isotope’s atom = 1 a.m.u.) = Y a.m.u.

= 100 × 1.66 × 10–24 grams = 1.66 × 10–22 grams

So, we may conclude that if the molecular weight of an element is Y, then the mass of an atom of that element is Y a.m.u.

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9th Class Chemistry

8

Table showing molecular weights of some compounds in a.m.u S.No.

Molecule

Symbol

1.

Hydrogen

H2

Molecular weight 2

Molecular weight (in a.m.u) 2 a.m.u = 2×1.66×10-24g

2.

Nitrogen

N2

28

28 a.m.u = 28×1.66×10-24g

3.

Oxygen

O2

32

32 a.m.u = 32×1.66×10-24g

4.

Carbon dioxide

CO2

44

44 a.m.u = 44×1.66×10-24g

5.

Carbon monoxide

CO

28

28 a.m.u = 28×1.66×10-24g

6.

Ammonia

NH3

17

17 a.m.u = 17×1.66×10-24g

7.

Methane

CH4

16

16 a.m.u = 16×1.66×10-24g

8.

Nitrogen dioxide

NO2

46

46 a.m.u = 46×1.66×10-24g

9.

Water

H2O

18

18 a.m.u = 18×1.66×10-24g

10.

Hydrochloric acid

HCl

36.5

36.5 a.m.u = 2×1.66×10-24g

11.

Sulphuric acid

H2SO4

98

98 a.m.u = 98×1.66×10-24g

12.

Nitric acid

HNO3

63

63 a.m.u = 63×1.66×10-24g

13.

Phosphoric acid

H3PO4

98

98 a.m.u = 98×1.66×10-24g

14.

Sodium hydroxide

NaOH

40

40 a.m.u = 40×1.66×10-24g

15.

Magnesium hydroxide

Mg(OH)2

58

58 a.m.u = 58×1.66×10-24g

16.

Aluminium hydroxide

Al (OH)3

78

78 a.m.u = 78×1.66×10-24g

17.

Sodium chloride

NaCl

58.5

18.

Magnesium oxide

MgO

40

40 a.m.u = 40×1.66×10-24g

19.

Sodium sulphate

Na2SO4

142

142 a.m.u =142×1.66×10-24g

20.

Sodium carbonate

Na2CO3

106

106 a.m.u = 106×1.66×10-24g

58.5 a.m.u = 58.5×1.66×10-24g

Summary

Molecule

Definition

Types of molecules

Homogeneous

Formed by same kind of atoms The smallest particle of matter that exists independently.

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Molecular Weight

Heterogeneous Formed by different kind of atoms

Gram Molecular Weight It is the molecular weight expressed in grams.

It is the number of times a molecule is heavier than 1/12th of C–12 atom

Language of Chemistry

Formative Worksheet 15. Match the following: Column - I Column - II i) Sodium p) Monoatomic ii) Helium q) Diatomic iii) Oxygen r) Triatomic iv) Ozone s) Poly atomic v) Sulphur 1) i, ii - p; iii - q; iv - r; v - s 2) iii, ii - p; i - q; iv - r; v - s 3) i,iii - p; iv - q; ii - r; v - s 4) i,iv - p; iii - q; ii - r; v - s 16. The units of molecular mass (or) molecular weight is 1) amu 2) grams 3) Both ‘1’ and ‘2’ 4) None 17.

weight of oneatomof an element = x. Its atomic weight weight of onemoleculeof a compound = y. Its molecular weight Then, x : y is

1 2) 2 : 1 3) 1 : 2 4) 1 : 1 12 18. Calculate the molecular weights of the following compounds. (i) H2SO4 (ii) HCl (iii) NaCl (iv) HNO3 (v) MgCl2 (vi) Na2CO3 (vii) Na2SO4 (viii) SO2 (ix) CO2 (x) O3. 19. The mass of one molecule of a compound is 1.66 × 10–22 g. The molecular weight of the compound is: 1) 166 2) 1.66 3) 100 4) 1000 20. The weight of ammonia molecule in grams is: 1) 17g 2) 17 × 10–3 –24 3) 17 × 1.66 × 10 4) 17 × 1.66 × 10–27 1) 1:

Conceptive Worksheet 14. Atoms of which of the following elements exist independently? 1) Helium 2) Sodium 3) Argon 4) Magnesium 15. Which of the following exist independently? 1) Atoms 2) Molecules 3) Ions 4) All

9 16. Which of the following is the smallest particle of matter that exist independently? 1) Atom 2) Molecule 3) element 4) compound 17. A: H2O, CH4, NH3 ; B: H2, N2, O2, F2 1) ‘A’ contains homogeneous molecules. 2) ‘B’ contains hetereogeneous molecules. 3) ‘A’ contains hetereogeneous molecules. 4) ‘B’ contains homogeneous molecules. 18. In which of the following do the smallest particles exist in diatomic form? 1) Hydrogen gas 2) Nitrogen gas 3) Oxygen gas 4) Fluorine gas 19. The number of atoms present in a molecule is called its atomicity. Then, the atomicity of phosphorus and sulphur is_______and _______respectively. 20. Which of the following exists in mono-atomic form? 1) He 2) Ne 3) Ar 4) All 21. Which among the following takes part actively in a chemical reaction? 1) He 2) Ne 3) Ar 4) None

10. Ions An ion is a positively or negatively charged atom (or group of atoms). Removal of electron from an atom or addition of electron to the atom leads to the formation of a charged particle called ion.

Classification of ions i)

ii)

Based on the kind of atom that form an ion, they are classified into simple ion and compound ion. Simple ions: Ions which are formed from single atoms are called simple ions. Simple ions are also known as monoatomic ions. Example: Na+, K+ ,Cl–, Br–, I–..... Compound ions: Ions which are formed from groups of joined atoms are called compound ions (or polyatomic ions). Example: NH4+,NO3–, NO2– , Based on the mode of formation of ion, they are classified into electron negative and electropositive ions. There are two types of ions: Electronegative ion (Anions) and Electropositive (cations). Electronegative ion(Anion): The atom that gains electron(s) and forms a negative ion is called an anion. Anion bears the same charge as the number of electrons gained. www.betoppers.com

9th Class Chemistry

10 Cl + e–  Cl– O + 2e–  O2– N + 3e–  N3– – Anions: Cl , Br–, I–, OH–, NO3–, NO2–, N3. Electropositive(Cation): The atom that loses electron(s) and forms a positive ion is called a cation. Cation bears the same charge as the number of electrons lost. Examples: Na – e–  Na+ Ca – 2e–  Ca2+ Al – 3e–  Al3+ Cations: Na+, K+, NH4+, Mg2+, Ca2+, Zn2+, Hg2+, Cu2+, Sn2+, Pb2+, Fe2+, Fe3+, Al3+. Examples:

11. Valency Old concept: Valency of an element may be defined as the number of hydrogen atoms or the number of chlorine atoms or double the number of oxygen atoms with which one atom of that element combines. Examples: i) One atom of nitrogen combines with three atoms of hydrogen to form NH3. Thus, the valency of nitrogen is 3. ii) AlCl3 is formed when one atom of aluminium combines with three atoms of chlorine. Thus, the valency of Al is 3. iii) ZnO is formed when one atom of Zn combines with one atom of oxygen. Thus double the number of oxygen atoms is the valency of Zn. The valency of zinc is 2. Now let us consider a few compounds of C and H, viz: i) Methane (CH4 ) ii) Ethane (C2H6 ) iii) Ethene (C2H4) and iv) Ethyne (C2H2). i) In methane (CH4), the valency of carbon is 4. ii) In C2H6, the valency of carbon is 3, as three hydrogen atoms combine with one carbon atom. iii) In C2H4, the valency of carbon is 2. iii) In C2H2, the valency of carbon is 1. In the above compounds, we see that the valency of carbon is 4, 3, 2 and 1. But we know that the valency of carbon is always 4. Thus, the above definition is not perfect. Modern concept : According to the modern concept, the valency may be defined as the number of electrons lost, gained or shared with one atom of the element in order to acquire the stable configuration of the nearest inert gas element. www.betoppers.com

Variable Valency: Observe the valency of the elements in the following compounds:

Element Compound formed Valency Cu2O 1 CuO 2 FeO 2 Iron Fe2O 3 3 Hg2O 1 Mercury HgO 2 From the table, we can see that the atom of same element combine differently with atoms of other element, to form different compounds. By applying modern concept of valency, different values of valencies of same element are obtained. Each of the above elements are exhibiting more than one valency. In such case, we say that, an element exhibits variable valency. Reason for variable valency: Let us understand it from the behaviour of copper during the formation of Cu2O and CuO. Copper in Cu2O: The copper in Cu2O is in Cu+ form with valency of 1.The Cu+ is formed by the loss of one electron from the outermost shell of copper as shown below. Copper in CuO: The copper in CuO is in Cu++ form with valency of 2. The Cu++ is formed due to the lose of two electrons from copper. One electron is lost from the outer most shell and the other from inner shell. Hence, bivalency of copper is due to the loss of electron from its inner shell. Therefore, we can say that the participation of electrons present in the inner shells (penultimate shell) during reaction, is responsible for variable valency. Naming an element with variable valencies: If an element exhibits two different valencies, the suffix ‘–ous’ is attached to the lowest valency ion/ radical and the suffix ‘ic’ is attached to the highest valency ion/radical Examples: The lowest valency exhibited by iron is 2 and highest valency is 3. Therefore, i) Fe+2 is named as ferrous and FeCl2 is called Ferrous chloride. ii) Fe+3 is named as ferric and FeCl3 is called Ferric chloride. Copper

Language of Chemistry

11

Element Copper i) Cuprous

Valency Cu+

Compound formed

1

Cuprous oxide

or Copper [I] oxide

++

2 2

Cupric oxide Ferrous oxide

or Copper [II] oxide CuO or Iron [II] oxide FeO

ii) Ferric Fe+++ Mercury i) Mercurous Hg+

3. 1

Ferric oxide or Iron [III] oxide Fe2O3 Mercurous oxide or Mercury [I] oxide Hg2 O

ii) Mercuric Hg++

2

Mercuric oxide

Iron

ii) Cupric i) Ferrous

Cu Fe++

or Mercury [II] oxide HgO

Conceptive Worksheet

Formative Worksheet 21. All ions are radicals, but all radicals are not ions. True or false. 22. i)

Cations are called __________

ii)

Anions are called __________

23. Iron and chlorine combine to form FeCl2, FeCl3.

22. The number of electrons, lost, gained or shared with one atom of the element in order to acquine stable configuration of nearest noble gas element is called _______. 1) Valency 2) Atomicity

The valency of iron in these compounds are :

3) Molecularity

1) 2, 3

4) None of these

2) 3, 4

3) 2, 4

4) 3, 4

24. i)

The name of the element with lower valency in a compound ends with suffix ______.

ii)

The name of the element with higher valency ends with suffix ________.

1) Ion

(i)

(ii)

3) Compound

a)

ous

ous

4) Substance

b)

ic

ous

c)

ous

ic

d)

ic

ic

25. Statement A : Copper combines with chlorine to form Cu2Cl2 and CuCl2. Statement B : The valency of copper in these compounds is 1 and 2. 1) ‘A’ is true, ‘B’ is false.

Cu2O

23. An atom (or) a group of atoms which can exist independently with charge(s) is called : 2) Molecule

24. Radicals are formed by the loss of electron (or) electrons. Such radicals are called ________. 1) Anions 2) Cations 3) Compounds 4) Molecules 25. Radicals are formed by the gain of electrons. Such radicals are called _______.

2) ‘A’ is false, ‘B’ is true.

1) Anions

3) Both ‘A’ and ‘B’ are true.

2) Cations

4) Both ‘A’ and ‘B’ are false.

3) Compounds 4) Molecules

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12 Valency chart I.

Monovalent Electropositive ions Ion

Symbol

H+ Li+ Na+ K+ Rb+ Cu+ Cuprous or Copper (I) +1 Ag+ Silver Au+ Aurous or gold (I) Gold Hg+ Mercurous or Mercury Mercury (I) NH4+ Ammonium + Phosphonium PH4

II. Monovalent Electronegative ions Symbol

Acetate

CH3COO– or C2H3O2–

Formate

Charge

HCOO– or CHO2–

Bicarbonate

HCO3

(or) Hydrogen carbonate

HSO4

Bisulphate Bisulphite (or) Hydrogen

HSO3

Symbol

Chloride

Cl–

Bromide

Br–

Iodide

I–

Hypochlorite

ClO–

Hypobromite

BrO–

Hypoiodite

IO–

Chlorite

ClO2–

Bromite

BrO2–

Iodite

IO2–

Chlorate

ClO3

Bromate

BrO3

–1

Iodate

IO3

Perchlorate

ClO 4

Perbromate

BrO4

Periodate

IO4

Nitrite

NO 2

Nitrate Hypophophite or Dihydrogen phosphate Dihydrogen phosphate or (biphosphate) Cyanide

NO 3

Cyanate

H 2PO4 CN 

Sulphide (or)

Permanganate

Bisulphide

Hydride

H

Hydroxide

OH

Superoxide Hydrogen peroxide

O 2

Fluoride

HS–

F–

Such radicals are called _______.

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Thiocyanate

–1

H 2PO2

CNO  SCN  Sulphocyanide MnO 4

sulphite Hydrogen

Charge

Charge

Hydrogen Lithium Sodium Potassium Rubidium Copper

Ion

Ion

HO2

-1

Language of Chemistry

13

III. Divalent electropositive ions

IV. Divalent electronegative ions Ion Carbonate

Symbol CO32

Chromate

CrO 24

Mg2+

Dichromate

Cr2O72

Calcium

Ca2+

Manganate

MnO24

Sulphide

Strontium

Sr2+

S-2 SO32

Barium

Ba2+

Radium

Ra2+

Copper

Ion

Symbol

Beryllium

Be2+

Magnesium

Mercury

Charge

Sulphite Thiosulphate

Iron

S4O62

Cu2+

Sulphate

SO 24

Cupric (or) Copper (II)

Perdisulphate

S2O82

Hg2+

Oxide

Mercuric (or) Mercury

Peroxide

O2O22

(II)

Stannite

SnO 22

Stannate

SnO32

Silicate

SiO32

Ferrous (or)

COO 

Iron (II) Cr2+ (or) Chromium

Cobalt

Nickel Manganese

Zinc Lead

Oxalate

COO 

Molybdate

Co2+

Plumbite

PbO 22

(or) Cobaltous or

Plumbate

PbO32

Cobalt (II)

Pyroantimonate

H 2Sb2O72

Ni2+

B4O72

Cd2+

Tetraborate Mono hydrogen phosphite Hydrophosphate (monohydrogen phosphate) Zincate

Zn2+

Tartarate

C4 H 4O 62

Tungstate

WO 24

Titanate

TiO32

Fluorosilicate

SiF62

Chromous

Mn2+ Manganous or

Pb2+ Plumbous or lead (II) Sn2+

Tin

+2

-2

C2 O42 or |

or  COO 2  MoO24

Manganese (II)

Cadmium

S2O32

Tetrathionate

Fe2+

Charge

2

HPO23 HPO24 ZnO 22

Stannous or Tin

(II)

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9th Class Chemistry

14

VI. Trivalent electronegative ions

Formative Worksheet 26. Write the symbol and valency of the following ions. i) Potassium ii) Cuprous ii) Aurous iv) Stannous v) Plumbous vi) Zinc vii) Charmous viii) Mercuric ix) Ferrous x) Radium 27. Write the symbol and valency of the following ions I) Acetate ii) Cyanide iii) Nitrate iv) Carbonate v) Hydrogen peroxide vi) Chromate vii) Manganate viii) Sulphite ix) Oxalate x) Zincate 28. x1 , x 2 , x 3 , x 4 , x 5 are the valencies of hydrogen sulphide, dichromate, sulphate, superoxide, hypophosphite ions represents y1, y2, y3, y4, y5 are the valencies of foimate, bisulphite fluoride, ammonium and phosponium ions respectively. If a = x1 + x2 + x3 + x4 + x5 and b = y1 + y2 + y3 + y4 + y5, find a/b.

Conceptive Worksheet 26. Ferrous ion is : 1) Monovalent 2) Bivalent 3) Trivalent 4) Both (2) and (3) 27. ____ may be defined as an atom or group of atoms which behaves as a single unit in chemical changes. 1) Compound 2) Molecule 3) Ion 4) None 28. Write the formula of Chromate and dichromate ions 29. How is Cyanide ion represented 30. The monovalent ion/radical among the following is : 1) Calcium 2) Carbonate 3) Chromate 4) Bicarbonate

V. Trivalent electropositive ions Ion

Symbol Fe3+ Iron Ferric or Iron (III) 3 Manganese Mn Manganic or Manganese (III) Aluminium Al3  Au 3  Gold Auric or gold (III) Sb3+ Antimony Antimonous or Antimony (III) As 3+ Arsenic Arsenous or Aresenic (III) Chromium Cr 3 Co3+ Cobalt Cobaltic or Cobalt (III) Boron B3 

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Charge

Ion Aluminate

Symbol AlO33

Arsenite

AsO33

Arsenate

AsO34

Arsenide Boride Borate

As 3 B3  BO33

Nitride Phosphide Phosphite

N 3 P 3PO 33

Phosphate

PO34

Ferricyanide

 Fe  CN 6 

–3

3

Co  NO 2 6  OCl3 OBr 3 Ol3

Cobaltnitrite Oxychloride Oxybromide Oxyiodide

Iron  III  3

Cobalt  III 

VII. Tetravalent electropositive ions Ion Platinum Lead Tin

Symbol Pt 4+ Platinic or Platinum (IV) Pb 4 Plumbic or Lead (IV) Sn 4 Stannic or Tin (IV)

Charge

+4

VIII. Tetravalent electronegative ions Ion

Symbol

Ferrocyanide Pyrophosphate

 Fe  CN 6  Iron  II  P2O74 

Pyroarsenate

As 2 O74

Pyroantimonate

Sb 2 O 74 

Carbide

C4

Charge 4-

–4

IX. Pentavalent electropositive ions Ion Arsenic Antimony

+3

Charge

Symbol As5 Arsenic (V) Sb5+ Antimonic or Antimony(V)

Formative Worksheet 29. Match the following Column - I Column - II a) Aluminate i) 1 b) Phosphate ii) 2 c) Ferrocynide iii) 3 d) Carbide iv) 4

Charge +5

Language of Chemistry 30. Match the following i) Antimonic a) 1 ii) Stannic b) 2 iii) Boron c) 3 iv) Osmium d) 4 v) Manganese e) 5 vi) Boride f) 6 vii) Phosphate g) 7 viii) Arseatc i) 8 31. Ion Valency Ion Valency i) Aurous a1 Iodide b1 ii) Auric a2 Iodate b2 iii) Stannous a 3 Sulphate b3 iv) Stannic a4 Sulphite b4 v) Chromous a 5 Manganate b5 vi) Chromic a 6 Permanganate b6 If x0 = a1 + a2 + a3 + a4 + a5 + a6 y0 = b1 + b2 + b3 + b4 + b5 + b6 x Find : y

Conceptive Worksheet 31. Arrange the following into monovalent, bivalent trivalent, tetravalent cations. A) Phosphonium B) Stannic C) Cobaltous D) Antimonous 1) A,B,D,C 2) A,C,D,B 3) A,B,C,D 4) D,B,C,A 32. Which of the following form tetravalent ions? 1) Platinum 2) Lead 3) Tin 4) All 33. Among the following trivalent ion/radical is : 1) Zinc 2) Boride 3) Barium 4) Oxide 34. Choose the trivalent anions from the following: i) Aluminate ii) Dichromate iii) Bromide iv) Boride 1) i, ii, iii 2) (i), (iv) 3) i, iii 4) i, ii, iii, iv

13. Chemical formula The representation of a molecule of a substance (element or compound) in terms of symbols and subscript numbers is known as a formula. The symbol of an element represents one individual atom of the element. Some of the elements no doubt exist independently. Examples : He, Ne, Ar, Kr, Xe, Fe, Hg, Co etc. However, many elements occur in combination with one or more atoms of its own kind or with one or more atoms of other elements as molecules. According to the molecular concept of matter, a

15 molecule is the smallest unit of matter capable of independent existence. Molecules containing 1, 2 or 3 atoms are called monatomic, diatomic or triatomic molecules respectively. If they contain more than 3 atoms they may be described as polyatomic. Thus the representation of a molecule of an element or a compound in terms of symbols and figures is defined as chemical formula. Examples : H2, H2O, SO2, NH3, O3, H2SO4. H2 stands for a molecule of hydrogen consisting of two hydrogen atoms. H2O stands for a molecule of water consisting of 2 atoms of hydrogen and one atom of oxygen. Thus the number of atoms of each present in the molecule is indicated by the number on its right hand corner as subscript. Thus H2SO4 stands for one molecule of sulphuric acid which contains 2 atoms of hydrogen, one atom of sulphur and 4 atoms of oxygen.

Significance of a formula Like the symbols, a formula has also qualitative as well as quantitative significance. Qualitative significance: Qualitatively, it represents : i) The name of the substance. ii) The names of various elements present in the substance. Quantitative significance: Quantitatively, it represents : i) One molecule of the substance. ii) The actual number of atoms of each element present in one molecule of the substance. iii) The number of parts by weight of the substance (molecular weight). iv) The number of parts by weight of each element. Examples: The formula of CaCO 3 has two significances. Qualitatively, it represents. i) Calcium carbonate. ii) It contains calcium, carbon and oxygen as the elements. Quantitatively it represents. i) One molecule of calcium carbonate ii) One molecule of calcium carbonate is made up of one atom of calcium, one atom of carbon and three atoms of oxygens. iii) One mole of calcium carbonate. iv) 100 parts by weight of calcium carbonate (atomic weights : Ca - 40, C - 12,O - 16)

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9th Class Chemistry

16 v)

40 parts by weight of calcium, 12 parts by weight of carbon and 48 parts by weight of oxygen. The following are the formulae of some of the molecules of elements and compounds generally used in the laboratory.

Element/Compound Hydrogen Nitrogen Oxygen Ozone Water Sulphur dioxide Sulphuric acid Nitric acid Hydrochloric acid Sodium hydroxide (caustic soda) Calcium hydroxide Sodium Chloride (common salt) Sodium carbonate (washing soda) Sodium bicarbonate (cooking soda, baking powder) Ammonium chloride Silver nitrate Magnesium sulphate (epsom salt) Sodium sulphate (glauber’s salt)

Formula H2 N2 O2 O3 H2O SO2 H2SO4 HNO3 HCl NaOH Ca(OH)2 NaCl Na2CO3.10H2O NaHCO3 NH4Cl AgNO3 MgSO4.7H2O Na2SO4.10H2O

Criss cross method One of the most important points to remember while writing the formula of a chemical compound is that it is always electrically neutral. In other words, the positive and negative valencies of the ions or radicals present in the chemical compound add up to zero. To write a formula follow the steps given below. This method of writing a formula is called the crisscross method. Step I : Write the symbol of the positive ion or the radical to the left and the negative ion or the radical to the right. Step II: Put the valency number of each of the radical or ion on its top right. Divide the valency number by the highest common factor, if any, to get simple ratio. Now ignore the (+) and (-) symbols. Interchange the valency numbers of the radicals. Step III: Shift the valency number to the lower right of the ion or the radical. If radical receives a number more than 1, enclose it within brackets. Do not enclose single atom within brackets. Examples-1: Writing the formula of alumunium sulphate

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Step I : Write the symbols of the combining units side by side. Cation on the L.H.S. and the anion on R.H.S Al SO4 Step.II: Enclose the compound Ion, if any, in a bracket, eg. keep SO4 as (SO4 ) Step.III: Write the numerical value of the charge on top Al3 (SO4) 2 Step IV: If the charges of the two ions are divisible by a common factor, then reduce the numerals. Step V: Finally, criss-cross the numerical values of their charges to obtain the formula.

Al 3

(SO4)2

Formative Worksheet 32. Correct formula of a trivalent metal nitride is: 1) M3 N2 2) MN2 3) MN 4) M2 N3 33. Chemical formula for calcium sulphate is CaSO4. The formula for ferric sulphate is: 1) Fe2(P 2O 7) 3 2) Fe4 P 3O 14 3) Fe2 (SO4 )3 4) Fe3 PO 4 34. Formula of phosphoric acid is H3PO4. A metal M forms chloride of the formula MCl2. The formula of its phosphate is: 1) M3 PO4 2) M2 PO4 3) M3 (PO4 )2 4) M2 (PO4 )3 35. Formula of chromic acid is H2CrO4. Formula of divalent metal chromate is: 1) MCrO4 2) M2CrO4 3) M2 (CrO4 )3 4) M3CrO4 36. A metal ‘M’ is divalent in its ‘ous’ state and trivalent in its ‘ic’ state. Write the formulae of the following compounds. i) Metal nitrate ii) Metal ferricyanide iii) Metal hypophosphite iv) Metal nitride v) Metal phosphide 37. Write the formula of the compound formed by each of the following pairs of ions. a) Ca 2+ and PO3-4

b) Na+ and S2 O2-3

c) Mg2+ and N3–

d) Mn2+ and O2–

e) Ba 2+ and S 2–

f) NO3 and Mg2+

Language of Chemistry

17

Conceptive Worksheet 35. What is the valency of inert gases? 36. A metal M forms a compound M2HPO4. What will be the formula of the metal sulphate? 37. The phosphate of a metal has the formula MPO4 . The formula of its nitrate will be: 38. What is the formula of sodium phosphate? 39. Which of the following is not the correct formula? 1) H2 S 2) NaHSO4 3) SiO2 4) NaCl2

40. Match the following in list A with those in List B: List A List B i) Oil of vitriol a) Na3AlF6 or AlF33NaF ii) Cinnabar b) CuSO4.5H2O iii) Cryolite c) FeS2 iv) Blue vitriol d) NaNH4.HPO4.4H2 O v) Green vitriol e) FeSO4.7H2O vi) White vitriol f) H2SO4 vii) Iron pyrites g) ZnSO4.7H2O viii) Red lead h) AgNO3 ix) Microcosmic salt i) Pb3O4 x) Lunar caustic j) HgS

41. Complete the following table :

Cations

Sodium Copper(II) Aluminium Zinc Silver Calcium Cobalt Mercurous Magnesium Tin (IV)

Anions

Chloride Carbonate Na2CO3

Nitrate

Sulphate

+ Phosphate

Oxalate

Cu(NO3)2 AlPO4 ZnCl2 Ag2C2O4 CaCO3 CoSO4 Hg2SO4 MgSO4 SnCl4

14. Chemical equations There are a large variety of substances in nature, and we come across innumerable transformations of substances in our day-to-day life. For example, rusting of iron, curdling of milk, transformation of raw vegetables/spices into delicious dishes etc. All these transormations are due to the change in the chemical composition of the original substances. Such a chemical change, which a substance undergoes, is called a chemical reaction. To understand a chemical reaction, it is important to know, which compound has transformed into what compound. Note: The compounds which are involved in the chemical reaction are called reactants and the compounds obtained after the reaction are called products. (Detailed study of chemical reactions can be learnt in the chapter ‘Types of chemical reactions

and changes’.) We know that elements and compounds are represented by symbols and formulae, respectively. In the same way, a chemical reaction can also be represented symbolically. Such a symbolic representation of a chemical reaction in terms of a chemical formula is called a chemical equation. Example: Magnesium reacts with Oxygen to form Magnesium Oxide. This can be symbolically conveyed with a chemical equation as follows: Mg  O2  MgO

How to write a chemical equation Following are the steps for writing a chemical equation. Step – 1: Write down the symbols and the formulae of the reactants, on the left hand side. Note: The different reactants are separated by plus (+) sign. www.betoppers.com

9th Class Chemistry

18 Step – 2: Write down the symbols and the formulae of the products, on the right hand side. Note: The different reactants are separated by plus (+) sign. Step – 3: The products and reactants are separated, by an arrow    . The equation obtained by the above steps is called a skeleton equation. The skeleton equation has to be supported with some additional information to make it a complete equation. What does a chemical equation convey ? A chemical equation conveys the following information: 1. The reactants that enter into a reaction. 2. The products which are formed by the reaction. 3. The number of atoms and molecules of the reactants and the products involved. 4. The state of matter, in which the substance is present, or formed. State of substance Symbol

5.

Solid s Liquid l Gas g Aqueous solution aq The direction of the reaction.

    6. 7.

8.

Is this equation correct in all aspects ? We observe that the number of atoms of each element on reactants side are not equal to the number of atoms of the same element, on the products side. Hence, the above equation is not balanced and should be balanced. What is a balanced equation ? A balanced equation is one, in which the number of atoms of each element, are the same on the side of the reactants (i.e., on the left hand side of the arrow) and also on the side of the products (i.e., on the right hand side of the arrow). Why should an equation be balanced ? All equations must be balanced in order to comply with the Law of conservation of matter, which states that ‘matter is neither created nor destroyed.’ In the course of a chemical reaction, an unbalanced equation would imply that atoms have been created or destroyed, which is not possible, and hence, an equation should be balanced. Limitations of a balanced chemical equation 1. It does not give information about the physical state of reactants and products. For example, the equation given below does not tell whether the substances involved in chemical reaction are in solid, liquid or gaseous states. CaCO3 + 2HCl  CaCl2 + H2O + CO2 How is the above limitation overcome in a balanced equation ? This difficulty is overcome by putting symbols like (s) for solids, (l) for liquids and (g) for gases. For chemicals, which react in solution form, a symbol (aq) is used. Following examples will illustrate the point : i) CaCO3(s) + 2HCl(aq)

Irreversible Re versible

The sign means “yields”, and shows the direction of the action. A ‘  ’ above the arrow shows that the reaction takes place in the presence of heat. i) Light or sun light above the arrows shows that the reaction is photochemical reaction. ii) Catalyst mentioned above the arrow shows that the reaction is catalytic reaction.

 CaCl2(aq) + H2O(l) + CO2(g) ii) AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) iii) 2Pb(NO3)2 (s)

2. 3.

Balancing a chemical equation Let us consider the formation of hydrogen chloride. Hydrogen  Chlorine  Hydrogen chloride (or) H 2  Cl2  HCl

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4.

 2PbO(s) + 4NO2(g) + O2(g) A balanced chemical equation does not tell whether a chemical reaction will come to completion or not. A balanced chemical equation does not tell anything about the speed of a chemical reaction. For example, the reaction between silver nitrate solution and sodium chloride solution completes in a few seconds. However, decomposition of lead nitrate crystals takes place in a few minutes. A balanced chemical equation does not tell about the physical conditions which bring about the

Language of Chemistry

19

chemical reaction, e.g., it does not tell whether heat energy, light energy, pressure, catalyst, etc., are required for a chemical reaction or not. The problem is partly solved by writing the conditions of reaction on the arrow head as illustrated by a balanced equation. Fe  Mo   2NH 3  g  N 2  g   3H 2  g    900 atm,450  C

From the above equation we can say that 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gas, under a pressure of 900 atmospheres, at a temperature of 450°C in the presence of catalyst iron containing molybdenum, when a reversible reaction takes place with the formation of 2 volumes of ammonia gas. A balanced chemical equation does not tell about changes such as precipitation, change in colour, evolution of heat, light and sound energy during the chemical change.

5.

Formative Worksheet 38. Statement A : A balanced chemical equation tells all about the physical conditions of a reaction. Statement B : A balanced chemical equation tells about only some physical conditions like light, heat energy and pressure but not of catalyst of a reaction. 1) ‘A’ is true, ‘B’ is false 2) ‘A’ is false, ‘B’ is true 3) Both ‘A’ and ‘B’ are true 4) Both ‘A’ and ‘B’ are false 39.

Fe-Mo   2NH 3  g  N 2  g  + 3H 2  g   900 atm - 450°C 

The above reaction is a balanced one with corrected limitations. Identify the corrected limitations. 1) Physical states of reactants 2) Number of atoms and molecules 3) Symbols and formulae of all the substances. 4) Physical conditions of a reaction on the arrow. 40.

Fe-Mo   2NH3 N 2 + 3H 2  900 atm - 450°C 

Which of the following statements is not true? 1) One litre of nitrogen and three litres of hydrogen combine to form two litres of ammonia insame conditions of temperature and pressure. 2) One molecule of nitrogen and three molecules of hydrogen combine to form two molecules of ammonia in same conditions of temperature and pressure. 3) One mole of nitrogen and three moles of hydrogen combine to form two moles of ammonia.

4)

41.

One gram of nitrogen and three grams of hydrogen combine to form two grams of ammonia.

K 2 Cr2 O7 + H 2SO4  K 2SO 4 + Cr2 SO 4 3 + H 2 O + O 2

In the above reaction, to get the balanced equation, how many moles of sulphuric acid and how many number of water are required respectively?. 1) 2 and 4 2) 3 and 5 3) 4 and 4 4) 3 and 3 42. Which of the following is true regarding the amount of substances involved and formed in the following reaction?  CaCO3   CaO + CO2 CaCO3 CaO 1) 50g 28g 2) 100g 28g 3) 100g 56g 4) 100g 56g

CO 2 22g 22g 22g 44g

Conceptive Worksheet 42. The chemical equation is a statement that describes a chemical change in terms of _____________ and _______________. 43. In a chemical reaction the atoms are neither __________ nor ___________. 44. The substances which take part in a chemical reaction are called ______. 45. The substances formed as a result of chemical change during a chemical reaction are called________. 46. The sign of an arrow    is read “ _________” 47.

48. 49. 50.

or “_________”. A chemical equation in which number of atoms of each element is same on the side of reactants and products is called _________. _________ decomposes on heating to form potassium nitrite and oxygen An unbalanced equation is also called as _______. Which of the following is not true of a balanced chemical equation? 1) A balanced chemical equation gives information about physical states of all reacting substances. 2) A balanced equation gives information about the number of atoms of all substances involved in the reaction. 3) Both 4) None

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9th Class Chemistry

20 51. Statement A : The completion and speed of a reaction can be known from a balanced chemical equation. Statement B : Formulae and number of atoms of all reacting substances in a reaction can be known by a balanced chemical equation. 1) ‘A’ is true, ‘B’ is false 2) ‘A’ is false, ‘B’ is true 3) Both ‘A’ and ‘B’ are true 4) Both ‘A’ and ‘B’ are false 52. Which of the following is true about the following reaction? Fe-Co   2NH3  g  N 2  g  + 3H 2  g   900atm-450°C 

1) 2) 3) 4)

8 atoms of reactants react to give 8 atoms of products. 4 molecules of reactants react to give two molecules of product. 34 grams of reactants react to give 34 grams of products. All the above.

c)

In case the elementary gases like hydrogen, oxygen, etc., appear, the equation is balanced by keeping these gases in the atomic state. d) Atoms of the elementary gases are balanced last of all. e) The balanced equation in th atomic state is changed in the molecular form. (If the number of atoms of the elementary gas is odd, multiply the whole equation by 2 to make it molecular) Drawbacks of Hit and Trial method Following are the drawbacks of the hit and trial method. i) It is tedious and takes a long time. ii) It is quite difficult to balance a chemical equation in which different atoms of the same element occur at a number of places on both sides of the arrow head. Try to balance the following equation. Mg  HNO 3  Mg  NO 3 2  NH 4 NO 3  H 2 O

iii)

15. Methods to balance a chemical equation Things to remember, before balancing a chemical equation 1. Before beginning to balance an equation, check each formula, to see if it is correct, and NEVER change a formula while balancing an equation. 2. Balancing is done by placing coefficients in front of the formula to ensure the same number of atoms of each element on both sides of the arrow.

Hit and Trial Method This method is generally employed to balance simple equations. It only involves hit and trial process till the number of atoms of each kind are the same on both sides of the equation. Following steps may be helpful in balancing a chemical equation by this method: a) Select the biggest formula and balance the same kinds of atoms in it on both sides of the arrow. b) If the above step fails, then select the element which occurs at the minimum number of places and this element is balanced first. The element which occurs at the maximum number of places is balanced last of all.

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It does not given any information regarding the mechanism of reaction.

Formative Worksheet 43. Balance the following skeleton equation: Mg + CO2   MgO + C 44. Balance the following equation: KClO3   KCl + O 45. Rewrite the following equation in the balanced form and indicate in this equation that the reaction is exothermic. C6H6 + O2   CO2 + H2O 46. Balance the following skeleton equation: Fe3O4 + H2   Fe + H2O 47. Balance the following skeleton equation: FeS2 + O   Fe2O3 + SO2

f-number Method Frequency of number of places at which an element occurs in a chemical equation. If an atom of an element is present at one place towards the reactants side it must be present at one place towards the products side. For example in the equation : KNO3  KNO2 + O2 Potassium atom is present at one place towards the side of reactants and at one place towards the side of products. Thus, frequency of occurrence in the whole equation is 2. Similarly, nitrogen atom

Language of Chemistry

21

occurs at one place towards the side of reactants and at one place towards the side of products. Thus, frequency of occurrence of nitrogen in the whole equation is 2. However, oxygen atom occurs at one place in the reactants side and two places in the products side; thus the frequency of occurrence of oxygen is 3. The frequency of occurrence of various elements in an equation , in short is called f-number. Thus, in the above equation f-number for potassium is 2, nitrogen is 2 and oxygen is 3. 1.

Note : While calculating f-numbers of various elements just count the number of places where the given elements occur. Do not count the actual number atom as they do not represent places. In the above equation f-number of oxygen is 3, because it is at three places in the equation. Do not say fnumber of oxygen is 7, as it has 7 atoms in the equation.

Examples for finding f-number : Ca (HCO3)2 + Ca (OH)2

Ca CO3 + H2O (unbalanced equation)

f – number for Ca = 3 f = number for hydrogen = 3

f – number for C = 2

f = number for oxygen = 4

Element f-number

Ca 3

H 3

C 2

O 4

2. Al2(CO3)3 +

H2SO4 

Al2 (SO4)3 + H2O + CO2 [Skelton equation]

Al = f = 2 C:f=2

O:f=5

H:f=2 S:f=2

Element f-number 3.

Al 2

C 2

O 5

H 2

S 2

K 2 Cr2O7  HC  KC  CrC 3  H2O  C 2

Element K Cr O H Cl f-number 2 2 2 2 4 As stated earlier, the balancing of chemical equation is related to the atomic number of element. List of common metals in the increasing order of their atomic number Na 11 Mg 12 Al 13 K 19 Ca 20 Cr 24 Mn 25 Fe 26 Ni 28 Cu 29 Zn 30 Ag 47 Ba 56 Au 79 Hg 80 Pb 82

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9th Class Chemistry

22 List of common non-metals in the increasing order of their atomic number H 1 C 6 N 7 O 8 F 9 Si 14 P 15 S 16 Cl 17 Br 35 I 53

A list of common metallic elements and non-metallic elements in the increasing order of their atomic numbers is given above.

Rules for Balancing a Chemical Equation 1. 2. 3. 4. 5.

6.

Write the frequency numbers of all the elements in a given chemical equation. Start balancing equation from that element which has least frequency number. Other elements should be balanced in the order of increasing frequency numbers. If two or more elements have same frequency number, then balance the metallic element first. If there are two or more metallic elements with same frequency number, first balance the metal with highest atomic number. Then balance the next metallic elements with lower atomic number and so on. If there are two or more non-metallic elements, with same frequency number, first balance the nonmetallic element with highest atomic number. Other non-metallic elements should be balanced in the decreasing order of atomic numbers. Note: Key to balance an equation is its f-number. However, if f-numbers are same, then first metallic elements should be balanced in the order of decreasing atomic number, followed by non-metals in the order of decreasing atomic numbers.

Solved examples for writing chemical equations Example1 : Calcium hydrogen carbonate solution reacts with calcium hydroxide solution to form calcium carbonate and water. This reaction can be written in equation form as under: Ca(HCO3)2 + Ca(OH)2  CaCO3 + H2O Writing f-numbers and order of balancing each element. www.betoppers.com

Element Ca H C O f-number 3 3 2 4 nd rd st Order of balancing 2 3 1 4th each element Explanation : i) As the frequency number of carbon is 2, which is least f-number, we must start balancing with carbon. The carbon atoms on reactants side are two, whereas towards products side is 1. Therefore, we will multiply CaCO3 by numeral 2 as under : Ca(HCO3)2 + Ca(OH)2  2CaCO3 +H2O –(a) ii) The elements calcium and hydrogen have f-number 3. However, calcium, being a metal should be balanced first and then the hydrogen. In equation (a), counting calcium atoms towards reactants side, there are 2 calcium atoms. On the products side there are 2 calcium atoms. Thus, calcium atoms are balanced. In equation (a) counting hydrogen atoms towards reactants side, there are 2 atoms in Ca(HCO3)2 and 2 atoms in Ca(OH)2. Thus, there are four hydrogen atoms. On the products side there are only 2 hydrogen atoms in H2O. Thus, if we multiply H2O by 2, there will be 4 hydrogen atoms as shown in equation (b). Ca(HCO3)2 + Ca(OH)2  2CaCO3 +2H2O–(b) iii) The element oxygen has f-number 4. Thus, it is balanced last of all. In equation (b) there are 6 atoms of oxygen in Ca(HCO3)2 and 2 atoms of oxygen in Ca(OH)2. Thus, total number of atoms of oxygen towards the reactants side is 6 + 2 = 8 atoms. Towards the products side there are 6 atoms of oxygen in 2 molecules of calcium carbonate and 2 atoms of oxygen in 2 molecules of water. Thus, total number of oxygen atoms towards the products side is 6 + 2 = 8 atoms. As number of atoms of all elements towards the reactants and the products side is same, the above equation (b) is fully balanced.

Formative Worksheet 48. The correct set of f-numbers of different elements in the equation: Al2  CO 3 3 + H 2SO 4

1) 2) 3) 4)

Al 2 2 2 2

C 2 2 5 5

O 5 5 2 2

 Al2  SO 4 3 + H2O + CO2 H 2 2 2 5

S 2 1 2 2

Language of Chemistry 49. The correct order of balancing of each element in the equation: Pb(NO)3  PbO + NO2 + O2 1) Pb O N 2) N O Pb 3) Pb N O 4) O N Pb 50. Identify the elements that are balanced in the equation: Al2(SO4)3 + HNO3  2Al(NO3)3 + 3SO2 + H2O 1) Al 2) H 3) N 4) S 51. Statement-I : The element having highest frequency should be balanced first. Statement-II: When two non-metals have same fnumber, the element with higher atomic number is balanced first. 1) Both are true 2) Both are false 3) I is true, II is false 4) I is false, II is true 52. Give the balancing order of each element for the following equation: K2Cr 2O7 + H2SO4  K2SO4 + Cr2(SO4)3 +H2O + O2 i) balancing chromium ii) balancing hydrogen iii) balancing oxygen iv) balancing sulphur v) balancing potassium 1) i, ii, iii, iv, v 2) i, iii, iv, v, ii 3) v, iv, iii, ii, I 4) i, v, ii, iv, iii 53. Elements Atomic number a. H p. 9 b. C q. 1 c. N r. 6 d. F s. 8 e. O t. 7 1) a-q, b-p, c-t, d-r, e-s 2) a-q, b-r, c-t, d-p, e-s 3) a-q, b-r, c-t, d-s, e-p 4) a-q, b-r, c-p, d-t, e-s 54. Aluminium carbonate reacts with dilute nitric acid to form aluminium nitrate, water and carbon dioxide. This reaction can be written in equation form as under: AL 2(CO3)3 +HNO3  2AL(NO3)3 +CO2 + H2 O 55. Copper reacts with hot and concentrated sulphuric acid to form copper sulphate, sulphur dioxide gas and water. This reaction can be written in the form of chemical equation as under: Cu + H2SO4  CuSO4 + SO2 + H2O 56. Lead nitrate crystals on strong heating decompose to form lead monoxide, nitrogen dioxide gas and oxygen gas. The reaction can be written in the form of chemical equation as under : Pb(NO3)2  PbO + NO2 + O2

23 57. Potassium dichromate on heating with cone, sulphuric acid forms potassium sulphate, chromium sulphate, water and oxygen. The reaction can be written in the form of chemical equation as under: K2Cr2 O7 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O + O2 58. Mg + HClO3  Mg(ClO3)2 + H2O + HCl 59. FeS2 + O2  Fe2O3 + SO2

Conceptive Worksheet 53. When a metal and a non-metal have same frequency, then the element that is balanced first is metal true/ false 54. Balance the following equations i) K2Cr2O7 + H2SO4  K2SO4 + Cr2 (SO4)3 + H2 O + F O2 ii) CaOCl2 + NH3  CaCl2 + H2O + N2 iii) FeSO4 + H2SO4 + HNO3  Fe2(SO4)3 + NO + H2O 55. Identify the balanced equations from the following: 1) Hg 2 (NO3)2  2Hg + 2NO2 + O2 2) CuSO4 + 4KI  Cu2I2 + 2K2SO4 + I2 3) SO2 + Cl2 + 2H2O  H2SO4 + HCl 4) 2NaOH + Cl2  2NaCl + NaClO + H2O

Summative Worksheet 1.

2.

3.

Statement (A) : For some of the elements, the first letter of its English name is used as symbol to represent that element in short form. Only capital letters are used. Statement (B) : When two letters are used, the first letter is in capital form and the second letter is always a small one. 1) Statement A is true, statement B is false 2) Statement A is false, statement B is true 3) Both statements, ‘A’ and ‘B’ are true 4) Both statements, ‘A’ and ‘B’ are false Identify the correct statement/s: 1) The symbol for an element represents the element either in pure state or in combined state. 2) A symbol represents one atom of that element. 3) An atom is the smallest particle of an element. 4) None are correct. Find the number of protons, neutrons and electrons present in HD+ ion 1) 1 proton, 1 neutron, 1 electron 2) 2 protons, 1 neutron, 2 electrons 3) 2 protons, 2 neutrons, 0 electron 4) 2 protons, 1 neutron, 1 electron www.betoppers.com

9th Class Chemistry

24 4. 5. 6.

7. 8.

The atomic number and mass number of an element is same. Identify the element. Write the names of a few elements, whose mass number is double their atomic numbers. An atom 'X' contains twelve protons. State a) the number of electrons it contains b) the atomic number of 'X'. An atom has net charge of –1.It has 18 electrons and 20 neutrons. Its mass number is Which of the following is true for O2 ?

1) 8 protons, 8 neutrons and 6 electrons 2) 8 protons, 8 neutrons and 10 electrons 3) 6 protons, 8 neutrons and 10 electrons 4) 6 protons, 8 neutrons and 8 electrons 9. The nitride ion in lithium nitride consists of: 1) 7 protons + 7 electrons 2) 10 protons + 7 electrons 3) 7 protons + 10 electrons 4) 10 protons + 10 electrons 10. Atomic weight of an element is x. It means that weight of one atom of that element is:

1  xg 12 3) 12 × xg 4) 1.66x × 10–24g The approximate number of electrons that are required to make 1 smallest unit of mass is: 1) 6.023 × 1023 2) 1.66 × 1024 3) 1852 4) 2500 The mass of one atom of an unknown element is 4 × 1.66 × 10–24g. Identify the element Calculate the molecular weights of the following: (i) Glucose (ii) Sucrose (iii) Laughing gas (iv) Table salt (v) Marsh gas (vi) Caustic soda (vii) Baking soda (viii) Blue vitriol Which of the following is true of H2SO4? 1) It represents 1 molecule of sulphuric acid 2) It represents that 1 molecule of H 2 SO 4 contains 2 atoms of H, 1 atom of S & 4 atoms of O. 3) It represents that sulphuric acid is made of hydrogen, oxygen and sulphur. 4) All. What is the weight of one carbon dioxide molecule? 1) ‘x’ g

11.

12. 13.

14.

15.

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2)

16. Statement A: The number of atoms present in gram atomic weight of different elements are equal. Statement B: The number of molecules present in gram molecular weight of different substances are equal. 1) ‘A’ is true, ‘B’ is false 2) ‘A’ is false, ‘B’ is true 3) Both ‘A’ and ‘B’ are true 4) Both ‘A’ and ‘B’ are false 17. Gold can exhibit: 1) Monovalent 2) Bivalent 3) Trivalent 4) Both (1) and (3) 18. Match the following: Level - I Level - II i) Antimony A) Bi and trivalent ii) Iron B) Penta and trivalent iii) Ammonium C) Trivalent iv) Boron D) Monovalent 1) (i)  (D), (ii)  (B),(iii)  (A),(iv)  (C) 2) (i)  (B), (ii)  (A),(iii)  (D),(iv)  (C) 3) (i)  (A), (ii)  (B),(iii)  (C),(iv)  (D) 4) None 19. Ferric ion is : 20. Arrange the following into monovalent, bivalent trivalent, tetravalent cations. A)Phosphonium B)Stannic C)Cobaltous D)Antimonous 1) A,B,D,C 2) A,C,D,B 3) A,B,C,D 4) D,B,C,A 21. Which of the following forms tetravalent ions? 1) Platinum 2) Lead 3) Tin 4) All 22. Super oxide ion is : 23. 24. 25.

26.

1) O22  2) O–2 3) O2 4) O2 Sulphite and sulphate ions are represented as Phosphide and phosphate ions are : The bivalent ion/radical among the following is : 1) Nitride 2) Phosphide 3) Antimony 4) Sulphate Identify the correct statement/s : 1) The representation of a molecule of a substance (element or compound) in terms of symbols and subscript numbers is known as the formula. 2) Atoms of different elements combine in certain fixed ratio to form a compound. 3) All chemical compounds are represented by their respective formulae. 4) None of the above.

Language of Chemistry

25

27. A formula has : 1) Qualitative significance only 2) Quantitative significance only 3) Both qualitative and quantitative significance. 4) None of these 28. Quantitatively, formula represents : 1) One molecule of the substance. 2) The actual number of atoms of each element present in one molecule of the substance. 3) The number of parts by weight of the substance (molecular weight). 4) The number of parts by weight of each element. 29. The formula, CaCO3, quantitatively signifies : 1) One molecule of calcium carbonate. 2) One molecule of calcium carbonate is made up of one atom of calcium, one atom of carbon and three atoms of oxygen. 3) 100 parts by weight of calcium carbonate [At. wts. Ca = 40, C = 12, O = 16]. 4) 40 parts by weight of calcium, 12 parts by weight of carbon and 48 parts by weight of oxygen.

30. The formula of potassium arsenate is K3AsO4. The formula of potassium ferrocyanide is K4Fe(CN)6. Write the formulas of a) Iron (III) arsenate b) Barium ferrocyanide c) Aluminium ferrocyanide d) Zinc arsenate e) Magnesium Arsenate f) Nickel Arsenate 31. Write the formula of the compound formed by each of the following pair of ions. a) Ca2+ and PO3-4 c) Mg2+ and N3–

b) Na+ and S2 O2-3

e) Ba2+ and S2–

f) NO3 and Mg2+

d) Mn2+ and O2–

32. Complete the following table :

Cations

Anions

Bicarbonate Nitrite

Borate

Iodide

Cadmium

CdS2O3

Chromium Nickel Potassium

Thiosulphate

CrI3 Ni3(BO3)2 KHCO3

Lead (II)

PbI2

Barium Ammonium

Ba3(BO3)2 NH4NO 2

Manganese Iron (III)

MnS2O3 Fe(HCO3)3

33. Give the balanced symbolic representation for the reaction: Arsenic + sodium hydroxide  sodium arsenate + hydrogen 1) As + 6NaOH  2Na3AsO3 + 3H2 2) 2As + 6NaOH  2Na3AsO3 + 3H2 3) 2As + 6KOH  2K3AsO3 + 3H2 4) As + 3NaOH  Na3AsO3 + H2 34. Statement-I : The frequency of oxygen in the reaction CH4 + O2  CO2 + H2O is 5. Statement-II: Frequency of an element is the no.of atoms on both reactants and products side. 1) Both are true 2) Both are false 3) I is true, II is false 4) I is false, II is true www.betoppers.com

9th Class Chemistry

26 35. In an unbalanced chemical equation the f-numbers of various elements are as follows: Al

C

O

H

N

2

2

5

2

2

HOTS Worksheet I)

Following the various steps for writing a formula, write formulae of the following compounds :

Arrange the elements in the order of balancing.

1) Zinc chloride

1) Al, C, N, H, O

2) Al, N, C, H, O

2) Copper bromide

3) Al, N, C, O, H

4) Al, N, H, C, O

3) Potassium chloride 4) Bismuth iodide

Balance the following skeleton equations.

5) Calcium Ferricyanide

i)

Sn + HCl + NO  SnCl2 + NH2OH

ii)

Ca3(PO4)2 + SiO2  P2O5 + CaSiO3

7) Potassium ferrocyanide

iii)

Al2O3 + C  Al4C3 + CO

8) Sodium meta-aluminate

iv)

C2 H 4

+

O2

 CO 2 + H2O

9) Sodium cobaltnitrite

v)

C2 H 2

+

O2

 CO 2 + H2O

10) Sodium peroxide

vi)

CH 4

+

O2

 CO 2 + H2O

11) Bismuth oxychloride

vii) NH 3

+

Na  NaNH2 + H 2

6) Cadmium phosphate

12) Zinc Ferrocyanide

viii) Cr(OH)3 + Na2O2  Na2CrO4 + H2O + NaOH

13) Ammonium sulphocyanide

ix)

Mn(OH)2 +Na2O2  Na2MnO4 +NaOH

15) Nickel bromide

x)

Al2(SO4) 3 +NaOH  Al(OH)3 + Na2SO4

16) Chromium sulphate

xi)

KI + H2SO4  KHSO4+ H2O+ SO2+ I2

17) Lead acetate

14) Calcium nitrate

xii) CuFeS2 + O2  Cu2S + FeS + SO2

18) Ammonium nitrite

xiii) FeS + O2  FeO + SO2

19) Cuprous chloride

xiv) Cu2S + O2  Cu2O + SO2

20) Mercurous nitrate

xv) Cu2S + CuSO4

 Cu +

SO2

21) Manganese phosphate

xvi) Cu2O + Cu2S

 Cu +

SO2

22) Manganese sulphate

xvii) CuSO4 + KI

 CuI +I2+ K2SO4

xviii) CuCl2+H2O+SO2  CuCl + HCl + H2SO4 xix) NaBrO + NH2CONH2  NaBr + CO2 + H2O + N2 xx) Fe + N2O  N2 + Fe3O4

23) Sodium dichromate 24) Barium silicate 25) Stannic chloride 26) Potassium zincate 27) Sodium cyanide 28) Potassium metaborate 29) Zinc perchlorate 30) Strontium oxalate

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Language of Chemistry

27

II.

Balance the following skeleton equations.

1.

CuCl CuCl2

+ +

SO2 + HCl H2O + S



 Zn Zn(CN)2 + Ag

2.

NaAg(CN)2 NaCN

+ +

3.

FeCl3 Fe(CNS)3 +

+ NH4CNS  NH4Cl

Ca 3(PO 4 ) 2 Ca(H2 PO 4 ) 2

+ +

H2 SO 4 CaSO4



CaCN2 +

+

H2 O



4. 5. 6.

6.

+ +

7.

CaCO3

8.

Hg2 (NO3 ) 2



9.

AlN Al(OH) 3

+ +

H2 O NH 3



+ +

Al Mn



4) K2CO3

The chemical formula of potassium superoxide is: 2) K2O

3) KO

4) K2O2

The chemical formula of Epsom’s salt is:

4) Na2SO4 . 10H2O

Plaster of Paris is a compound of 1) ZnSO4

2) CaSO4

3) BaSO4

4) MgSO4

An example of alum is:

2) FeSO4 . (NH4)2SO4 . 6H2O

Hg + NO2 +

O2

3) K2SO4.Al2 (SO4)3 .24H2 O 4) Fe2(SO4)3.Al2 (SO4)3.24H2 O 10. The compound which is commonly known as inorganic benzene is: 1) B3N3H6

2) C3N3H6

3) C6H6

4) C5H5B

11. Chemically Borax is:

Washing soda has a formula

1) Sodium metaborate

1) Na2CO3 . 7H2O

2) Na2CO3 . H2O

2) Sodium orthoborate

3) Na2CO3 . 10H2O

4) Na2CO3

3) Sodium tetraborate decahydrate

Molecular formula of Glauber’s salt is 1) MgSO4 . 7H2O

2) FeSO4 . 7H2O

3) CuSO4 . 5H2O

4) Na2SO4 . 10H2O

The chemical formula of prussian blue is 1) Na4 [Fe(CN)5 NOS] 2) Na2 [Fe(CN)5 NO] 3) Na4 [Fe(CN)5 NO]

4.

3) KHCO3

1) Al2(SO4)3 . Na2SO4 . 12H2O

IIT JEE Worksheet

3.

8.

9.



S + H2SO4 + NO2 +

2.

2) Na2CO3

3) 2CaSO4 . H2O

7.

1.

1) NaHCO3

1) MgSO4 . 10H2O 2) MgSO4 . 7H2O

 HNO 3 NH4NO3 + H2 O HNO 3 H2 O

Baking powder has one of the following constituents.

1) KO2

NH 3

Mg Mg(NO3) 2

10. Mn3O 4 Al 2 O 3

5.

4) Na2 [Fe(CN)5 NO2]

4) Sodium tetraborate pentahydrate 12. Chemical composition of white lead is: 1) Pb3O4

2) Pb(OH)2 . 2PbCO3

3) PbO

4) PbO2

13. Dry powder extinguisher’s contains: 1) Sand

Microcosmic salt is

2) Sand and Na2CO3

1) Na 2HPO4H2 O

3) Sand and Baking soda

2) NaNH4HPO4 . 4H2O

4) Sand and K2CO3

3) Na(NH4)2HPO4 . 2H2 O 4) None of these www.betoppers.com

9th Class Chemistry

28 True or False

14. An alum is represented by:

23. Carborundum is SiC

1 III 1) M1M 2  SO 4 3 .24H 2O 1 1

III 1

2) M M

24. Phosgene is the common name given to Carbonyl chloride (COCl2)

 SO 4 3 .12H 2O

25. A dry ice piece is composed of solid CO2.

1 III 3) M 2SO 4 M  SO 4 2 .24H 2O

26. Pb3O4 is called litharge.

2 III 4) M1 SO 4 M 2  SO 4 3 .24H 2 O

27. Nitrolim is mixture of Ca(CN)2 and C. 28. The formula of Nitrosyl chloride is NOCl2.

15. Semi water gas is a mixture of

29. Peroxidisulphuric acid’s commercial name is Oleum.

1) Water gas and CO2 2) Water gas and producer gas

30. Calcium pyrophosphate is represented by the formula Ca2P2O7. The molecular formula of ferric pyrophosphate is Fe4(P2O7)3.

3) Producer gas and oil gas 4) Producer gas and CO2 16. Which of the following formula represents fuming sulphuric acid (oleum) 1) H2S2O6

2) H2SO5

3) H2S2O8

4) H2S2O7

17. K2CS3 can be called potassium: 1) Thiocarbide

2) Sulpho cyanide

3) Thiocarbonate

4) Thiocyanate

2 18. The species that do not contain peroxide  O 2 

ions is: 1) PbO2 2) BaO2 3) H2O2 4) Na2O2 19. Plaster of Paris is 1) CaSO4 . H2O

2) CaSO4 . 2H2O

3) CaSO4 . 5H2O

4) 2CaSO4 . H2O

20. Dolomite has the composition 1) KCl . MgCl2 . 6H2O 2) Na3AlF6 3) CaCO3 . MgCO3 4) CaCl2 . MgCl2 . 6H2O 21. Carnallite is the mineral of 1) Mg

2) Na

3) Zn

4) Ca

22. The chief ore of Aluminium is: 1) Felspar

2) Bauxite

3) Kaolin

4) Carnallite

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31. The correct formula of Haematite is Fe3O4. Fill in the blanks 32. Calgon is the trade name of ________ and its chemical formula is ________. 33. Cu, Ag, Au are called coinage ________ 34. Uni negative inorganic groups resembling the halide ions are called ________ 35. Balanced Equation of (NH 4 ) 3 PO 4  NH 3 + H 2 O + HPO 3 is ____________.



By the • • • • • •

end of this chapter, you will understand Matter and its classification • Types of mixtures Pure and Impure substances • Separation of Solid – Solid mixtures Element, Compounds & Mixtures • Separation of Solid – Liquid mixtures Metals, Nonmetals and metalloids • Separation of Liquid – Liquid Mixtures Properties of elements, compounds • Separation of Liquid – Gas mixtures and mixtures • Separation of Gas – gas mixtures Characteristics of elements, compounds and mixtures

1. Introduction In the previous chapter, we have learnt in detail about the classification based on physical properties. In this chapter, let us explore the details of the classification based on chemical composition. Based on chemical composition, we can classify matter on the basis of pure substances and mixtures. However, before we start classification of matter on the above basis, we must make ourselves very clear about what are pure substances and mixtures.

Chapter -2

ELEMENTS, COMPOUNDS & MIXTURES

Learning Outcomes

For a layman, pure substances are pure honey, pure milk, pure cheese, pure water, etc. However, to a chemist none of the above mentioned substances are pure. For example, pure milk is made of a number of substances like proteins, carbohydrates, mineral salts, vitamins, water etc., present in variable amounts in the milk of different animals of same breed. Thus, milk can be called a mixture in which amount of various substances are not present in same fixed ratio. Before we answer the question : What is a pure substance, let us see the broad classification of matter.

Classification of matter Matter Pure substances

Elements

Metals

Mixtures

Compounds

Non-Metals

Metalloids

Homogeneous Mixtures

Heterogeneous Mixtures

Inert gases

2. Pure substances To answer this question, let us see, with what, gold and water are made of. If you consider gold, it is made up of only one type of particles called gold atoms. Water is also made up of only one type of particles called water molecules. Such substances are called pure substances. Thus, a homogeneous material which contains particles of only one kind and has a definite set of properties is called pure substance. Iron, silver, oxygen, sulphur are pure substances, because each one has only one kind of particles. However, if a substance is composed of two or more different kinds of particles combined together in fixed proportion by weight, then the substance is also regarded as a pure substance. Sodium chloride is a pure substance because it has a fixed number of sodium and chlorine particles combined

9th Class Chemistry

30 together in fixed proportion by weight. Similarly, magnesium oxide and carbon dioxide are pure substances. Note : It does not imply that all homogeneous substances are pure. For example, common salt solution in water is a homogeneous solution, yet it cannot be called a pure substance, as it is made of two different substances, e.g. salt and water.

Characteristics of a pure substance: i) ii)

iii)

A pure substance is homogeneous in nature. A pure substance has definite set of properties. These properties are different from the properties of other substances. The composition of a pure substance cannot be altered by any physical means.

3. Element To understand it, let us consider silver. What happens if you break it into tiny pieces? Do you get any new substances ? You get tinier particles of silver but you will not end up with gold or copper. Thus, silver remains as silver. Such substances are called element. Thus, an element is a pure substance that cannot be broken into two or more simpler substances by any known physical or chemical means. An element is made of only one kind of atoms. Chemists have discovered 115 elements so far. Amongst 115 elements, 82 are normal elements and 33 are radioactive elements. Normal elements are those which do not give harmful radiation. Radioactive elements are those which give harmful radiation. Thus, we can say, Eighty-two elements are nonradioactive and Thirty-three elements are radioactive elements.

Characteristics of Elements i) ii)

iii) iv) v) vi)

An element is a pure homogeneous substance, made up of only one kind of atoms. Except during nuclear reactions, an element cannot be broken into two or more smaller parts. An atom is the smallest unit of an element. It shows all the properties of that element. Elements may occur in free state in nature or are found in the form of their compounds. Some elements (radioactive elements) can be prepared artificially by the nuclear reactions. Any element may chemically react with other element(s) to form compound (s).

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Types of elements I.

The elements are further classified into : Metals, Non-metals, Metalloids and Noble gases. Metals The solid state of matter, in which the atoms are very closely packed together and have a special type of bond known as metallic bond is called a metal. Because of very tight, or close packing, the metals are quite hard. Out of 115 elements, nearly 70 elements are found to be metals. Metals have the following characteristics i) It has a lustre, i.e., it has a metallic glow. ii) It is a good conductor of heat and electricity. iii) It is ductile, i.e., it can be drawn into wires. iv) It is malleable, i.e., it can be beaten into sheets. v) It is solid at room temperature. vi) It has a high melting point and high boiling point. vii) It produces a sonorous sound on being hit. Exceptions i) Mercury and Gallium are liquid metals at 30 °C. ii) Zinc is not malleable and ductile at room temperature. iii) Sodium, potassium, calcium and lead do not have high melting points. List of common metals Name in English 1. Lithium 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

Sodium Magnesium Aluminium Potassium Calcium Va nadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Silver Tin Barium Platinum

Symbol Li Na Mg Al K Ca V Cr Mn Fe Co Ni Cu Zn Ga Ag Sn Ba Pt

Elements, Compounds & Mixtures Name in English 20. Gold 21. Mercury 22. Lead 23. Radium 24. Uranium 25. Tungsten 26. Thorium

31

Conceptive Worksheet

Symbol Au Hg Pb Ra U W

1.

Th

2.

Formative Worksheet 1.

2.

3.

4. 5.

Assertion : Sodium chloride is a pure substance. Reason : It has a fixed number of sodium and chlorine particles. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of as sertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. Statement A: All homogeneous substances are pure. Statement B: A pure substance is homogeneous in nature. Statement C: Common salt solution in water is a homogeneous solution. a) Statement (A) is true but statements (B) and (C) are false. b) Statement (A) is false but statements (B) and (C) are true. c) Statement(s) (A) and (C) are true but statement(B) is false. d) Statement(s) A, B, C are all true. Which of the following is a true statement ? a) Magnesium oxide, Carbon dioxide are pure substances. b) Milk, Honey, Cheese are pure substances. c) Iron, Silver, Oxygen are impure substances. d) Sodium chloride is an impure substance. Name two elements that which are liquid metals at 30°C. Statement A: Metals are ductile and malleable. Statement B: Zinc is the most ductile and malleable metal. a) Statement ‘A’ is false. b) Statement ‘B’ is false. c) Statements ‘A’ and ‘B’ are both true. d) Statement ‘A’ and ‘B’ are both false.

3.

Which of the following statement(s) is/are false? a) A pure substance is heterogeneous in nature. b) The composition of a pure substance can be altered by any physical means. c) A pure substance has definite set of properties. d) A pure substance contains a fixed number of particles. Which of the following elements can be prepared artificially by the nuclear reactions? a) Rn b) Fe c) Mg d) Ca Identify the metals present in the following compounds (i) Potassium dichromate (ii) Calcium nitrate (iii) Aluminium chloride a) (i)  K, Cr (ii)  Ca, (iii)  Al b) (i)  N2, (ii)  Cl2, (iii)  O2 c) (i)  Po, (ii)  C, (iii)  Ag d) (i)  O2, (ii)  N2, (iii)  Cl2

II. Non-metals As the name suggests, non-metals are opposite to metals, which means that their properties are quite different, from the metals. They are comparatively less in number. Out of 115 elements, only about 14 to 15 elements are found to be non-metals. A non-metal has the following characteristics: i)

It has no lustre, i.e., it cannot be polished.

ii)

It is a bad conductor of heat and electricity.

iii)

It is not ductile, i.e., it cannot be drawn into wires.

iv)

It is not malleable, i.e., it cannot be beaten into sheets.

v)

It is a gas or a br ittle solid at room temperature.

vi)

It has low melting point and low boiling point.

vii) It does not produce a sonorous sound on being hit. Exceptions 1.

Graphite (an allotrope of carbon) has a lustre and is a good conductor of heat and electricity.

2.

Bromine is a liquid non-metal.

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9th Class Chemistry

32 List of common non-metals Non-metal

State

Colour

Hydrogen

Gas

Colourless

Nitrogen

Gas

Colourless

Oxygen

Gas

Colourless

Fluorine

Gas

Colourless

Chlorine

Gas

Greenish yellow

Bromine

Liquid

Reddish brown

Iodine

Solid

Greyish brown

Carbon

Solid

Grey

Phosphorus

Solid

Waxy yellow

Sulphur

Solid

Yellow

V. Abundance of elements in earth’s crust

Symbol

Helium He Neon Ne Argon Ar Krypton Kr Xenon Xe Radon Rn Note : Helium is the second lightest element after hydrogen. Radon is given out by the radioactive emission from earth.

Percentage by weight

Oxygen

49.85

Silicon

26.03

Aluminium

7.28

Iron

4.12

Calcium

3.18

Sodium

2.33

Potassium

2.33

Magnesium

2.11

Hydrogen Titanium

0.97 0.41

Other elements

1.39

VII. Atomicity of an Element Generally, elements exist as single atoms. However, sometimes two or more atoms of an element combine with one another to form a compound atom or molecule. Depending upon the number of atoms present in its molecule, the elements can be classified as under: i) The molecule of a monoatomic element contains only one atom. Ex: He, Ne, Ar, Kr, Xe ii). The molecule of a diatomic element contains two atoms, e.g., hydrogen (H 2 ), oxygen (O2), nitrogen (N2),etc. iii) The molecule of a polyatomic element contains more than two atoms, e.g., ozone (O3 ), phosphorus (P 4), sulphur (S 8 ), boron (B12) and carbon (C60)

Silicon Solid Grey III. Metalloids: Elements that exhibit some properties of metals and some properties of non-metals are called metalloids. Examples : Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te) and Polonium (Po) IV. Noble Gases : These elements are found in air in the form of gas in very small amounts. Therefore, they are sometimes called rare gases. They are also called noble gases, because they do not react chemically with any known element. List of noble gases

Noble gases

Element

4. Compound A compound is a pure substance, which is composed of two or more elements, combined chemically in a fixed proportion by weight, and can be broken into smaller parts or elements by chemical methods only. Most of the elements do not exist in elementary form in nature. Two or more different elements combine in a fixed proportion by weight to form a compound. The combination of elements takes place due to chemical reaction.

Characteristics of a Compound i)

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Nature: The nature of elements constituting a chemical compound remains same.

Elements, Compounds & Mixtures Examples : i) In case of water, the constituting elements are hydrogen and oxygen only. ii) In case of mercury (II) oxide, the constituting elements are mercury and oxygen only. ii) Behaviour: A pure chemical compound is homogeneous in nature. Example : If one molecule of water contains hydrogen and oxygen in some fixed pattern, then all other molecules of water will also contain hydrogen and oxygen combined with each other in the same pattern. iii) Separation: A chemical compound can be broken into two or more different elements. It can be synthesized from these elements by chemical means. Examples : Water can be broken into hydrogen and oxygen, and the gases so formed can be made to form water by chemical means. Similarly, mercuric oxide can be broken into mercury and oxygen. Mercury and oxygen can be made to combine chemically to form mercury (II) oxide. iv) Composition: A chemical compound has a fixed composition, i.e., its constituent elements combine together in a fixed ratio by weight. Examples : Hydrogen and oxygen combine in the fixed ratio of 1 : 8 by weight to form water. Similarly, in magnesium oxide, magnesium and oxygen combine in the fixed ratio of 3:2 by weight. v) Properties: A chemical compound has distinct set of properties which do not resemble the properties of its constituent elements. Examples : Water is a colourless liquid which extinguishes fire, whereas hydrogen is a combustible gas and oxygen is a supporter of combustion.

33 Similarly, common salt (sodium chloride) is harmless when we use it as table salt, whereas its constituents sodium and chlorine are highly dangerous. Sodium is a waxy white metal, which catches fire when placed in air or water. Chlorine is a greenish yellow gas with a suffocating smell and is highly poisonous in nature. vi) Mechanical separation : During the formation of a chemical compound, its constituent elements completely lose their individual properties. Thus, it is not possible to separate the constituent elements by simple means like nitrate ion; evaporation sublimation, etc. However, we can separate the constituent elements by applying suitable chemical means. Examples : It is not possible to separate hydrogen and oxygen gas from water by simple processes like heating, cooling filtration, etc., unless we break molecules of water chemically with the help of electric current. Similarly, sodium and chlorine cannot be separated from sodium chloride unless electric current is passed through its saturated solution in water. vii) Energy exchange: Formation of a compound involves energy changes : It is impossible to form a chemical compound, unless the constituent elements either absorb energy or give out energy. The energy is generally in the form of heat energy. However, in certain cases it could be electrical energy, light energy or sound energy. Examples : Sodium and chlorine react with the liberation of heat and light energy to form sodium chloride. Similarly, carbon dioxide gas and water react in the presence of chlorophyll to form carbohydrates and oxygen only, when sunlight is absorbed by the leaves of the plant.

Comparison between elements and compounds

Nature Types of atoms Separation of constituents

Nature of properties

Element An element is a pure and homogeneous substance. An element contains only one type of atoms An element cannot be broken down into any other simpler substances by physical or chemical means. An element has characteristic physical and chemical properties.

Compound A compound is also a pure and homogeneous substance. A Compound contains different types of atoms. A compound may be broken down into simpler substances by chemical means.

A compound has characteristic physical and chemical properties, but these are entirely constituent elements. www.betoppers.com

9th Class Chemistry

34

Formative Worksheet 6.

Statement A : An element is a substance that can be split into simpler substances by a chemical means or a physical means such as filtration, sublimation and distillation. Statement B : A compound is a substance that can be split into two or more simpler substances by a chemical means. a) Statement ‘A’ is false. b) Statement ‘B’ is false. c) Statements ‘A’ and ‘B’ are both true. d) Statements ‘A’ and ‘B’ are both false. 7. Match the following: Column ‘A’ Column ‘B’ (a) Sulphur (i) Greyish brown solid (b) Fluorine (ii) Greenish yellow gas (c) Bromine (iii) Yellow solid (d) Iodine (iv) Red liquid (e) Chlorine (v) Colourless gas 8. Statement A :During the formation of a chemical compound, its constituent elements completely lose their individual properties. Statement B : We can separate the constituent elements in a chemical compound by simple means like filtration; evaporation; sublimation etc. a) Statement ‘A’ is true but statement ‘B’ is false. b) Statement ‘A’ is false but statement ‘B’ is true. c) Statement(s) ‘A’ and ‘B’ are both true. d) Statement(s) ‘A’ and ‘B’ are both false. 9. Match the following: Column ‘A’ Column ‘B’ (i) Metal (p) Kr (ii) Non-metal (q) Ag (iii) Rare gas (r) Si (s) Xe 10. Assertion : The atoms of pure substances like Helium and Neon are regarded as molecules. Reason : The atoms of Helium and Neon can exist independently. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but rea son is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct.

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Conceptive Worksheet 4.

Which of the following are the characteristics of non-metals? a) Non-lustrous b) Bad conductor c) Non-ductile d) All 5. Select the metalloid group from the following: a) He, Ne, Ar b) Ge, Te, Po c) Sn, Ba, Pt d) H2, N2, O2 6. Which of the gases are colourless? a) Hydrogen b) Nitrogen c) Oxygen d) All 7. Which of the following non-metals is a good conductor of electricity? a) Carbon b) Nitrogen c) Fluorine d) All 8. The most abundant element in ear th’s crust is : 9. Which type of elements are chemically inert? 10. Find the atomicity of Carbon in fullerene_____

5. Mixtures If two or more substances (elements, compounds or both) mixed together in any proportion, do not undergo any chemical change, but retain their characteristics, the resulting mass is called a mixture.

Kinds of Mixtures a)

b)

Heterogeneous mixture : A mixture in which various constituents are not mixed uniformly is called heterogeneous mixture. Examples : A mixture of sand, salt and sulphur is a heterogeneous mixture. Similarly, a handful of soil is a heterogeneous mixture. Homogeneous mixture : A mixture in which different constituents are mixed uniformly is called homogeneous mixture. Examples : Brass is an alloy of copper and zinc and is a homogeneous mixture. Similarly, all solutions are homogeneous mixtures.

Characteristics of a Mixture i.

ii.

Variable composition : The constituents of a mixture are present in any ratio. Example : A mixture of sand and salt can be in a ratio of 1:2 or 5:6, etc, by weight. Physical change: The mixture is a result of physical change. The constituents of a mixture do not bind each other by chemical bonds. Example: In air the main constituents, i.e., oxygen, nitrogen and carbon dioxide, do not bind each other with chemical bonds.

Elements, Compounds & Mixtures iii. No specific properties : The properties of a mixture are the average of the properties of its constituents. Example: The properties of air are midway between properties of nitrogen and oxygen. iv. Homogeneity : Most of the mixtures are heterogeneous, i.e., their constituents are not spread evenly throughout. However, some mixture are homogeneous. Example: In the mixture of iron and sulphur, at some places iron is more and at some places sulphur is more. v. Separation : Generally, the constituents of mixture can be separated by employing suitable physical means.

35 Example: Iron can be separated from the mixture of iron and sulphur with the help of a magnet. vi. Energy changes: The formation of heterogeneous mixture, does not involve any energy (heat) change, but, that of a homogeneous mixture generally does. For example during the formation of heterogeneous mixtures like : Sand + common salt ,Sand + water and Oil + water no heat is exchanged. But during the formation of homogeneous mixtures like magesium sulphate + water, concentrated sulphuric acid + water and hydrogen chloride + water heat is evolved and during the formation of glucose + water, heat is absorbed. Exception: During the formation of homogeneous mixture like nitrogen + oxygen , no heat is evolved.

Differences between a mixture and a compound Mixture

Compound 1. Nature When two or more elements or compounds or When two or more elements unite chemically, a both are mixed together, such that they do not compound is formed. combine chemically, a mixture is formed. 2. Structure Mixtures are generally heterogeneous. However, Compounds are always homogeneous. some mixtures can be homogeneous. 3. Composition In case of mixtures their constituents can be In case of compounds, the constituents are present in any ratio, i.e., mixtures have variable present in a fixed ratio by weight. composition. 4. Properties The constituents of a mixture retain their The properties of a compound are entirely individual chemical and physical properties. different from the properties of its constituents 5. Separation of constituents The constituents of a mixture can be separated The constituents of a compound cannot be by applying physical methods like solubility, separated by applying physical methods. filtration, evaporation, distillation, use of However, constituents of a compound can be magnet, etc. separated by chemical means. 6. Energy change There may or may not be energy change during During the formation of a compound either the the formation of mixture. energy is absorbed or given out.

Is air a mixture or a compound? We know that, matter that can be separated into different kinds of substances, by any physical process is called a mixture and substance which can be broken further by chemical means is called a compound. Air is made up of different constituents like oxygen, nitrogen, carbon dioxide, water vapour etc., and these constituents may be separated, by means of water vapour etc., and these constituents may be separated, by means of physical process like fractional distillation. Moreover, air has a variable composition, because, it contains different amounts of variable gases at different places and hence, it has no definite formula. Whereas, the constituents in a compound are present in a fixed ratio and hence, the compound has a definite formula. Therefore, due to the above reasons air is considered a mixture and not a compound. www.betoppers.com

9th Class Chemistry

36

Formative Worksheet 11. Bronze is an example of a) Compound b) Element c) Homogeneous mixture d) Heterogeneous mixture. 12. When two substances A and B are powdered together in a Pestle and Mortar, a large amount of heat is evolved and a new substance C is formed. What is the nature of the new substance? a) Mixture

b) Compound

c) Element

d) None of these

13. Assertion:When we mix sugar and sand together, there is no change in energy i.e., energy is neither evolved nor absorbed. Reason : In the preparation of a mixture, energy is neither given out nor absorbed. a)

Both assertion and reason are correct and reason is the correct explanation of assertion.

b)

Both assertion and reason are correct but rea son is not the correct explanation of assertion.

c)

Assertion is correct and reason is incorrect.

d)

Assertion is incorrect and reason is correct.

14. Which of the following tests confirms that brass is a mixture? a)

The constituents cannot be separated by physical means.

b)

The constituents lose their identities when mixed together.

c)

When we try to melt brass, it does not have a sharp melting point.

d)

None of these.

15. Match the following: Column ‘A’

Column ‘B’

(l)

(i) Element

Diamond

(m) Brass

(ii) Compound

(n) Marble

(iii) Mixture

(o) Washing soda (p) Ink (q) Coke www.betoppers.com

16. Classify the following into elements, compounds and mixtures. Column ‘A’ Column ‘B’ (p) Air (i) Element (q) Silver (ii) Compound (r) Carbon dioxide (iii)Mixture (s) Zinc (t) Sugar solution 17. Statement A: When a magnet is rolled in the mixture of iron and sulphur, iron is attracted to the magnet. Statement B: When a magnet is rolled in iron sulphide powder, iron is attracted to the magnet. a) Statement ‘A’ is true but statement ‘B’ is false. b) Statement ‘A’ is false but statement ‘B’ is true. c) Statement(s) ‘A’ and ‘B’ are both true. d) Statement(s) ‘A’ and ‘B’ are both false.

Conceptive Worksheet 11. Which of the following represents a mixture? a) Ammonia b) Calcium carbonate c) Gun powder d) Hydrogen chloride 12. If two or more substances are mixed together in any proportion and do not undergo any chemical change, but retain their characteristics, the resulting mass is called 13. Which of the following is not a compound? a) Marble b) Washing soda c) Quick lime d) Coal 14. Carbon burns in oxygen to form carbon dioxide. The properties of carbon dioxide are: a) Similar to carbon b) Similar to oxygen c) Totally different from both carbon and oxygen. d) Much similar to both carbon and oxygen. 15. Which of the following statements are false? a) The particles of iron and sulphur mixture can be seen. b) The particles of iron sulphide cannot be seen. c) Iron sulphide is homogeneous mixture. d) None of the above.

Elements, Compounds & Mixtures

37

Kinds of mixtures A mixture can exist in a solid, liquid or gaseous state. Furthermore, a mixture can be homogeneous or heterogeneous. Depending on the physical state of mixtures, they can be classified as under:

Nature of mixture Constituents of mixture Homogeneous Heterogeneous Homogeneous Heterogeneous Homogeneous

Solid-Solid mixture Solid-Solid mixture Solid-Liquid mixture Solid-Liquid mixture Liquid-Liquid mixture

Heterogeneous

Liquid-Liquid mixture

Examples Alloys (brass, fusible alloy, bronze) i) Charcoal and salt ii) Iron and sulphur i) Common salt in water ii) Iodine in alcohol i) Sulphur in water ii) Sand in water i) Alcohol in water ii) Petrol in kerosene oil i) Oil in water ii) Carbon disulphide in water

Homogeneous

Liquid-Gas mixture

i) Sulphur dioxide gas in water ii) Ammonia gas in water

Homogeneous

Gas-Gas mixture

i) Air (oxygen and nitrogen) ii) Hydrogen and ammonia.

`

6. Separation of mixtures Principles involved in the separation of components of a mixture 1. 2.

The method/methods necessary to separate the components of a mixture depend upon : The physical state of the constituents of the mixture. The difference in one or more physical properties of the constituents of the mixture. Following physical properties are considered in the separation of constituents of a mixture. a) Density of the constituents. b) Melting point and boiling point of the constituents of the mixture. c) Property of volatility of one or more constituents of the mixture. d) Solubility of the constituents of the mixture in various solvents. e) Ability of the constituents of the mixture to sublime. f) Magnetic properties of the constituents of mixture. g) Ability of the constituents of the mixture to diffuse.

Separation of solid - solid mixtures The various methods or techniques employed in separation of solid-solid mixtures are: 1. Magnetic separation 2. Gravity Method 3. Using solvent 4. Fractional crystallization 5. Sublimation.

I.

Magnetic separation : Physical property involved in separation: One of the components of the mixture is a magnetic substance (iron, cobalt; nickel and steel or their oxides are magnetic in nature). Example: Let us consider a mixture of iron filings and sulphur.

Magnetic Separation of Constituents of a Iron + Sulphur Mixture

Method : 1. Spread the mixture evenly in the form of a thin layer over a piece of paper. 2. Place another sheet of paper over the mixture. 3. Place a powerful horse shoe magnet over the paper and then lift. Some iron filings will cling to paper. 4. Remove the magnet from the paper. The iron filings will fall down. 5. Repeat the process a number of times, till all the iron filings are removed. II. Using gravity method : Physical property involved in separation: One of the components is heavier than water, whereas the other components are either lighter or soluble in water. www.betoppers.com

9th Class Chemistry

38 In this method one of the components of mixture is either lighter than the other or is soluble in water. This method is suitable for mixtures given below: Solid-Solid mixture

Decant off or filter the water along with lighter or soluble component.

Heavier Lighter component or component soluble component

Sand and saw dust

sand

Salt and sand

sand

salt (soluble)

limestone

charcoal (lighter)

Charcoal and limestone

3.

saw dust (lighter)

Method : 1. Stir the mixture in water or any other suitable solvent. 2. Allow the mixture to stand, so that the heavier component settles down.

III. Using solvents : Physical property involved in separation: One of the components is soluble, but the other is insoluble in a specific solvent. The following shows the examples of the mixtures that can be separated by the above method.

Sand and sulphur

Carbon disulphide

Soluble Component Sulphur

Charcoal and sulphur Sand and wax

Carbon disulphide Turpentine oil

Sulphur Wax

Charcoal Sand

Water

Common salt

Marble powder

Water

Nitre

Charcoal

Solid-Solid mixture

Common salt and marble powder Nitre and charcoal

Solvent

Gun powder (nitre; carbon (i) Water Nitre and sulphur) (ii Carbon disulphide ii) Sulphur

Following is the list of some important solvents and the substance it dissolves

Substance

Solvent

Chlorophyll

Methylated spirit

Grease Petrol Iodine Ethyl alcohol Nail polish Acetone Nitre Water Oil Petrol Paraffin wax Turpentine oil Phosphorus Carbon disulphide Rust Oxalic acid Rubber Benzene Sulphur Carbon disulphide Shellac Ethyl alcohol Paint

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Turpentine oil

Insoluble Component Sand

Sulphur and carbon Carbon

Method : 1. Choose the solvent, such that only one particular component of the mixture is soluble in it, and other component is insoluble. 2. Dissolve the mixture in a good amount of solvent such that the soluble component of the mixture completely dissolves. 3. The above solution is filtered. The insoluble component of the mixture is left on the filter paper. The soluble content collects as filtrate. How to recover the components in the above method ? 1. The insoluble component is left on filter paper. It is dried either in hot air or in the folds of filter paper. 2. The filtrate is evaporated either on slow heat or in the sunlight. The solvent evaporates, leaving behind a soluble component.

Elements, Compounds & Mixtures

Formative Worksheet 18. Assertion:Air is a mixture, not a compound Reason - I: The composition of air is not fixed. Reason -II: Air shows no similarity with any compound in its constituents. a) Assertion is correct and only reason I is the correct explanation of assertion. b) Assertion is correct and only reason II is correct explanation of assertion. c) Assertion is correct but reason I and II are both incorrect. d) Assertion is correct and reasons I and II are correct and are the correct explanation of assertion. 19. Assertion:Water is a compound and not a mixture Reason -I: The constituents of water cannot be separated by physical methods. Reason -II: In water, hydrogen and oxygen are present in a fixed ratio (1:8) by weight. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 20. Match the following: Column ‘I’ Column ‘II’ a) Solid-solid mixture i) Alcohol in water b) Solid-liquid mixture ii) Hydrogen and ammonia c) Liquid-liquid mixture iii) Bronze d) Liquid-gas mixture iv) Ammonia gas in water e) Gas-gas mixture v) Iodine in alcohol 21. Potassium nitrate and sodium chloride can be separated by using which of the following techniques? a) Sublimation b) Using gravity c) Fractional crystallisation d) Decantation 22. If both the components are soluble in water, but their solubilities are different, i.e., one is more soluble than the other, which type of separation techniques can be employed? 23. What is the principle behind separation of charcoal from limestone?

39 24. Assertion:Ammonium chloride and iodine can be separated by using certain solvents. Reason: One of the components is soluble, but the other is insoluble in a specific solvent. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 25. Sulphur and sand can be separated by 26. Which one is the soluble component in the mixture of Nitre and charcoal? a) Charcoal b) Nitre c) Both charcoal and Nitre d) Neither charcoal nor nitre 27. We can separate certain solid-solid mixtures by using solvents. The principle behind this method of separation is: a)

One of the components is soluble in specific solvents. b) One of the components is not soluble in specific solvents. c) Both the components are soluble in specific solvents. d) None of the components are soluble in any solvent. 28. Choose the correct classification of solvents and substances. Column ‘A’ Column ‘B’ p) Oxalic acid i) shellac q) Benzene ii) Rust r) Ethyl alcohol iii) Paint s) Turpentine oil iv) Rubber

Conceptive Worksheet 16. Choose the correct statement: a) When two or more elements unite chemically, a compound is formed. b) Mixtures are always homogeneous. c) Mixtures have variable composition. d) The components of a mixture retain their individual chemical and physical properties. 17. Choose the false statement: a) All gases are partially miscible with liquids. b) Air entrapped in the pores of soil particles is an example of liquid-gas mixture. c) A mixture of sugar and sand is an example of solid-solid mixture. d) All gases present in air chemically react with each other. www.betoppers.com

40 18. Which of the following is a heterogeneous liquid-liquid mixture ? a) Ammonia gas in water b) Petrol in kerosene oil c) Carbon disulphide in water d) Mixture of oxygen and nitrogen 19. Charcoal and salt is an example of a) Homogeneous solid-solid mixture b) Heterogeneous solid-solid mixture c) Homogeneous solid-liquid mixture d) Homogeneous liquid-liquid mixture 20. Tincture of iodine is a a) Compound b) Solution c) Mixture d) Element 21. Choose the false statement: a) All the constituents of a mixture take part in a chemical reaction. b) All the constituents of a compound do not take part in a chemical reaction. c) The constituents of a compound cannot be separated from each other by physical means. d) The constituents of a mixture do not retain their physical and chemical properties. 22. Oil in water is a: a) Homogeneous liquid-liquid mixture b) Heterogeneous liquid-liquid mixture c) Homogeneous solid-liquid mixture d) Homogeneous liquid-gas mixture 23. Which out of the following is a heterogeneous mixture? a) Air b) Brass c) Salt and sand d) Steel 24. Which of the following is a homogeneous liquidgas mixture? a) Sulphur dioxide gas in water b) Ammonia gas in water c) Hydrogen and ammonia d) Alcohol in water 25. Which of the following is true statements? a) Fusible alloy is an example of solid-solid heterogeneous mixture. b) Iron and sulphur is a homogeneous solid-solid mixture. c) Sulphur in water is a heterogeneous mixture. d) Common salt in water is a homogeneous solidliquid mixture. 26. Charcoal and limestone can be separated by gravity method, because: a) Limestone is a heavier component. b) Charcoal is a lighter component. c) Limestone is a lighter component. d) Charcoal is a heavier component. www.betoppers.com

9th Class Chemistry 27. Gunpowder is soluble in: a) Carbon disulphide b) Water c) Benzene d) Ethyl alcohol 28. Choose the correct statement: a) Grease is soluble in petrol b) Nitre is soluble in ethyl alcohol c) Paraffin wax is soluble in turpentine oil d) Phosphorous is soluble in water IV. Fractional crystallisation Physical property involved in separation: Both the components are soluble in water, but their solubilities are different, i.e., one is more soluble than the other. Furthermore, they do not sublime. The process of separation of two different soluble substances from their solution by crystallisation at controlled temperature, such that one of the solid crystallises, is called fractional crystallisation. This method is suitable for mixtures mentioned below: Solid-Solid More soluble Less soluble mixture component component Potassium nitrate Potassium Sodium and sodium chloride nitrate chloride Potassium chloride Potassium Potassium and potassium chlorate chlorate chloride Sodium nitrate Sodium Sodium and sodium chloride nitrate chloride

Method : 1. Choose the solvent (generally water) and warm it to around 60 °C. 2. Add the mixture in solvent, till it stops dissolving. 3. Allow the mixture to cool. Large amount of more soluble solid crystallises out along with some amount of less soluble solid. 4. Filter the crystals and redissolve them in minimum amount of warm solvent. 5. Recrystallise the crystals, when large amount of more soluble salt crystallises out. 6. Concentrate the filtrate containing less soluble solid. On cooling, the crystals of less soluble solid separate out. Saturated solution of sodium chloride

Crystals of potassium ni trate

Saturated sol ution of s odium chloride and potassium nitrate above 60 °C

On cooling to room temperature the crystals of more soluble potassium nitrate separate out.

Elements, Compounds & Mixtures V.

41

By sublimation Physical property involved in separation: One of the components of the mixture sublimes on heating. Solid-Solid mixture

Sublimable solid

Ammonium chloride and common salt

Ammonium chloride

Iodine and sand

Iodine

Iodine and common salt . Sodium sulphate and benzoic acid

Iodine Benzoic acid

Method : 1. The mixture is placed in a china dish and heated by a low flame. 2. An inverted dry funnel is placed over the china dish and its stem is closed with cotton wool. 3. The sublimable component of the mixture sublimes and its vapours condense on the sides of the funnel to form fine powder. 4. The fine powder (sublimable component) is scrapped from the sides of the funnel. 5. The residue left behind is a non-sublimable component.

Naphthalene and iron filing Naphthalene Cotton wool plug Glass funnel Resolidified Ammonium Chloride Mixture of Ammonium chloride + Sodium chloride

Vapours of NH4Cl

China dish

Wire gauze

Burner

Separation of the constituents of mixture by sublimation

Summary of Techniques for the separation of solid-solid mixtures Technique Physical property involved in Examples employed separation Magnetic One of the components of the mixture (i) Separation of iron ore from separation is a magnetic substance (iron, cobalt; rocky material (gangue) nickel and steel or their oxides which (ii) Separation of nickel from are magnetic in nature). mixture of nickel and lead. Using gravity One of the components is heavier i) A mixture of sawdust and sand than water, whereas the other ii) A mixture of common salt and co mponents are either lighter or sand. soluble in water. Using solvents One of the components is soluble, but i) Sulphur and sand [sulphur the other is insoluble in a specific dissolves in carbon disulphide] solvent. ii) Ammonium chloride and iodine [ammonium chloride dissolves in water.] Fractional crystallization

Sublimation

Both the components are soluble in water, but their solubilities are different, i.e., one is more soluble than the other. Furthermore, they do not sublime. Both the components are soluble in water, but of them one can sublime but not the other, and both the co mponents are insoluble in water.

i)

Potassium nitrate and sodium chloride.

ii)

Potassium chlorate and potassium chloride.

i)

Ammonium chloride and common salt.

ii)

Iodine and sand.

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42

Formative Worksheet 29. Statement-A : A solid - solid mixture can be separated through fractional crystallisation. Statement-B : Potassium chlorate and potassium chloride can be separated through fractional crystallisation. Statement-C: Potassium chloride is more soluble component and potassium chlorate is less soluble component. a) Statement ‘A’ is true b) Statement ‘B’ is true c) Statement ‘C’ is false d) Statement ‘A’, ‘B’, ‘C’ are all false 30. Assertion: Potassium nitrate and sodium chloride can be separated through fractional crystallisation. Reason: Fractional crystallisation is based on the principle that different solids have different solubilities at a given temperature. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 31. Match the following: Column ‘I’ Column ‘II’ (P) Sublimation (I) Potassium chlorate and potassium chloride (Q) Fractional (II) Iodine and sand crystallisation (R) Magnetic (III) A mixture of sand separationdust and sand (S) Using gravity (IV) Iron and sand a) P  (III), Q  (II), R  (IV), S  (I) b) P  (II), Q  (III), R  (IV), S  (I) c) P  (IV), Q  (I), R  (II), S  (III) d) P  (II), Q  (I), R  (IV), S  (III) 32. We can separate certain solid-solid mixtures by using solvents. The principle behind this method of separation is: a) One of the components is soluble in specific solvents. b) One of the components is not soluble in specific solvents. c) Both the components should be soluble be in specific solvents. d) None of the components should be soluble in specific/solvents. www.betoppers.com

9th Class Chemistry 33. Statement A: The method of sublimation is used to separate a mixture consisting of two or more solids such that one of the solids sublimes on heating. Statement B: From the mixture of ammonium chloride and sodium chloride, sodium chloride can sublime on heating. Statement C: From the mixture of iodine and sand, iodine can sublime. a) Statement ‘A’ is false but statements ‘B’ and ‘C’ are true. b) Statement ‘A’ is true but statements ‘B’ and ‘C’ are false. c) Statements ‘A’ and ‘C’ are true but statement ‘B’ is false. d) Statements ‘A’, ‘B’ and ‘C’ are all false.

Conceptive Worksheet 29. The process of separation of two different soluble substances from their solutions by crystallisation at controlled temperature, such that one of the solid crystallises, is called ___________. a) Crystallisation. b) Filtration. c) Distillation. d) Fractional crystallisation. 30. A solid-solid mixture of sodium nitrate and sodium chloride can be separated through the method of: a) Dissolution b) Fractional crystallisation c) Magnetic separation d) Fusion 31. We can separate sodium sulphate and benzoic acid through the method of sublimation because a) Sodium sulphate is sublimable. b) Benzoic acid is sublimable. c) Sodium sulphate and Benzoic acid are both sublimable. d) Neither sodium sulphate nor benzoic acid are sublimable. 32. Generally sublimable solids sublime on heating. Which of the following sublime without heating? a) Naphthalene b) Ammonia c) Iodine d) None of these

Separation of Solid-Liquid Mixtures The solid-liquid mixtures can be separated by the techniques like: 1. Sedimentation and Decantation 2. Filtration 3. Evaporation 4. Distillation.

Elements, Compounds & Mixtures 1.

43

Separation by sedimentation and Decantion Sedimentation: The process in which a suspension of insoluble fine particles suspended in a liquid are allowed to stand undisturbed, such that the solid particles settle down, leaving the clear liquid above is called sedimentation. Sediment: The insoluble solid material which settles down when a suspension is allowed tostand undisturbed is called sediment.

Supernatant liquid : The clear liquid above the sediment, when a suspension is allowed to stand undisturbed is called supernatant liquid. Decantation: The process of pouring out the clear supernatant liquid above the sediment, thus helping the separation of solid particles from liquid is called decantation. Drawbacks of Decantation i) The constituents of the mixture of a solid and a liquid do not get separated completely. ii) The constituents of a solid lighter than liquid cannot be separated as they float on the surface of liquid, rather than settling down. 2. Separation by filtration : Filtration : The process of separation of insoluble solid constituent of a mixture from its liquid constituent, by passing it through some porous material is called filtration. Filtrate : The clear liquid obtained from a mixture of a solid and a liquid by the process of filtration is called filtrate. Residue : The insoluble solid constituent left on the filter paper when a mixture of an insoluble solid and a liquid is filtered is called residue. The method of filtration is employed for the following solid-liquid mixtures: Solid-Liquid mixture

Residue

Filtrate

Silver chloride and water

Silver chloride

Water

Barium sulphate and water Barium sulphate Chalk and water

Chalk

Water Water

Glass rod Residue Double Fold here fold

Four fold

Cone shape Filter paper Residue

Circular filter paper

Filtrate

To make filter paper cone. Method of filtration

Method: 1. A filter paper generally available in the form of a circular disc is folded to form a cone as illustrated above. 2. A glass funnel is moistened with water. The filter paper cone is inserted in the cavity of the funnel and is pressed on the sides. This expels out the air and the filter paper cone sticks tightly to the walls of the funnel. 3. The funnel is clamped in an iron stand and under its stem is placed a beaker, such that the wall of the beaker is in contact with the stem of the funnel. 4. The suspension of the solid-liquid is poured in the funnel slowly with the help of a glass rod, as shown in the figure. 5. The filtrate collects in the beaker. The residue is left on filter paper. The residue is dried either in hot air or in the folds of filter paper. Advantagesof Filtration over Sedimentation and Decantation 1. It is a quicker process than sedimentation and decantation. 2. The insoluble solid is completely removed, which is not possible in the case of decantation. 3. Separation by evaporation The process of changing a liquid into a gaseous state, below its boiling point by the supply of external heat, is called evaporation. The process of evaporation is suitable for the separation of non-volatile soluble solid from its liquid solvent. This method of evaporation is employed for the following solid-liquid mixtures. Solid-Liquid mixture Non-volatile solid Common salt and water Sodium sulphate and water Carbon disulphide and sulphur

Liquid

Common salt

Water

Sodium sulphate

Water

Sulphur

Carbon disulphide

Method : 1. Heat the sand in an iron vessel by placing it over a tripod stand. This arrangement is called sand bath.

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9th Class Chemistry

44 2.

3.

4.

Take the clear solution of soluble non-volatile substance in a china dish. Place the china dish on the sand bath. Heat gently, such that water (liquid) evaporates, but does not boil. Continue heating till liquid completely evaporates. When almost dry solid is left, reduce the flame, but go on heating for another five minutes. This helps in forming (i) completely dry solid (ii) will prevent the spurting (jumping out) of solid from the china dish due to excessive heat. Water vapour China dish

Sand bath

Wire gauze

Bunsen burner Tripod stand

Separation of non-volatile solid by evaporations.

4.

Note : Do not heat the mixture of sulphur and carbon disulphide, as carbon disulphide is highly inflammable. Instead, evaporate the solution in sunshine. Separation by distillation : The process of conversion of a liquid into gaseous state on boiling and then recondensing the gas so formed into liquid by condensation in another vessel, is called distillation. It is used in the situations where the liquid component of solid-liquid mixture is required in pure state. The solid-liquid mixtures which can be

separated by distillation are as follows:

Salt and water (sea water)

Pure water

Non-volatile solid Salt

Iodine and methyl alcohol

Methyl alcohol

Iodine

Chloroform

Iodine

Solid-Liquid mixture

Iodine and chloroform

Liquid

Method : 1. The solid-liquid mixture is placed in a distillation flask. The distillation flask is connected to Liebig’s condenser, at the end of which is placed a receiver to collect distilled liquid (distillate) as shown in figure. 2. When the distillation flask is heated, the liquid starts boiling. The vapour of the liquid passes through the Liebig condenser, where they condense to form the liquid. The liquid so formed trickles into the receiver. 3. The solid component of mixture forms residue in the flask.

Distillation of pure water from salt solution

Summary of separation techniques of solid-liquid mixtures Technique employed for Physical property involved in separation of mixture separation

Examples

Sedimentation and Decantation

One of the components is heavier than Muddy water. Water containing the liquid and is insoluble. sand.

Filtration

One of the components is a solid and is Silver chloride precipitates in insoluble in the liquid. water. Barium sulphate precipitates in water.

Evaporation

One of the components is nonvolatile. It may or may not be soluble in water.

Distillation

One of the components is soluble solid Iodine in chloroform. in the liquid.

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Common salt solution, sodium sulphate solution.

Elements, Compounds & Mixtures

Formative Worksheet 34. Statement(S): The technique employed to separate sodium sulphate solution is evaporation. Explanation(E) : One of the components is a solid soluble in liquid. a) Statement (S) is wrong b) Statement (S) is wrong but explanation (E) is correct c) Statement(S) is correct but explanation (E) is wrong d) Statement (S) and (E) are both correct. 35. Which of the following is the drawback of decantation technique? a) It is useful in the separation of the clay and sand particles from the muddy water. b) The constituents of a solid lighter than liquid cannot be separated as they float on the surface of the liquid, rather than settling down. c) One of the components is heavier than the liquid and is insoluble. d) The constituents of the mixture of a solid and a liquid do not get separated completely 36. Assertion:Filtr ation over sedimentation and decantation is more advantageous. Reason I: It is a slower process than sedimentation and decantation. Reason II: The insoluble solid is completely removed, which is not possible in the case of decantation. a) Assertion is correct and only reason I is the correct explanation of assertion. b) Assertion is correct and only reason II is the correct explanation of assertion. c) Assertion is incorrect but reasons I and II are both correct d) Assertion, reason I and II are all incorrect. 37. The principle behind the filtration technique of separation of solid-liquid mixture is. a) Insoluble component is removed from the liquid mixtures. b) Removal of insoluble lighter solids. c) Liquid changes into vapour state on gentle heating. d) Cooling the gaseous mixture to obtain components in the liquid. 38. Which of the following is/are non-volatile in nature? a) Water c) Carbon disulphide

b) Common salt d) Sulphur

45

Conceptive Worksheet 33. Which of the following mixtures can be separated by the method of filtration? a) Barium sulphate precipitate in water b) Common salt solution c) Silver chloride precipitate in water d) Iodine in Chloroform 34. Water containing sand can be separated by the method of sedimentation by Decantation. The principle behind sedimentation and decantation techniques is:

a) b) c) d) 35.

36.

37.

38.

39.

One of the components is heavier than the liquid and is insoluble.

One of the components is non-volatile. One of the components is soluble in the liquid. One of the components is a solid and is insoluble in the liquid. Iodine in chloroform can be separated by the method of: a) Filtration b) Evaporation c) Distillation d) Sedimentation and Decantation The insoluble solid material which settles down when a suspension is allowed to stand undisturbed is called : (A) . The clear liquid above the sediment, when a suspension is allowed to stand undisturbed is called (B) . a) A  Sediment, B  Supernatant liquid. b) A  Filltrate, B  Residue c) A  Decantate, B  Sediment d) A  Residue, B  Filtrate A mixture of chalk and water can be separated through the method of a) Separation by evaporation. b) Separation by crystallisation. c) Separation by filtration. d) Separation by distillation. The process of separation of insoluble solid constituent of a mixture from its liquid constituent by passing it through some porous material is called ______. a) Evaporation b) Crystallisation c) Filtration d) Decantation Barium sulphate and water can be separated by filtration. Which of the following is left as a residue from the above mixture? a) Water b) Barium sulphate c) Both water and barium sulphate d) Neither water nor barium sulphate www.betoppers.com

9th Class Chemistry

46

Separation of liquid-liquid mixtures

1.

The liquid-liquid mixtures can be separated by the techniques given belows: 1. Separating funnel 2. Fractional distillation. Separation of liquid-liquid mixtures by separating funnel: Separating funnel is a long glass tube provided with a tap as shown. The liquid-liquid mixture of immiscible components is poured into the funnel and allowed to stand. The liquids separate out on account of difference in their densities. Immiscible liquid-liquid mixture

Heavier liquid

Lighter liquid

Benzene and water Kerosene oil and water Turpentine oil and water Carbon disulphide and water Chloroform and water

Water

Benzene

Water

Kerosene oil

Water

Turpentine oil

Water

Carbon disulphide

Chlor oform

Water

Mercury

Alcohol

Mercury and alcohol

Method : 1. The tap of the separating funnel is closed. The separating funnel is clamped in the vertical position in an iron stand. 2. The immiscible liquid-liquid mixture is poured into the separating funnel. The mixture is allowed to stand for half an hour or more. 3. The immiscible components of the mixture separate out into two distinct layers. The heavier and denser liquid forms the lower layer. The lighter and less denser liquid forms the upper layer. Separating funnel

4.

5.

2.

A conical flask is placed under the nozzle of separating funnel. The tap is gently opened so that the heavier liquid trickles in to the flask drop by drop. Once the denser liquid is drained out, the tap is closed. Another conical flask is placed under the nozzle of separating funnel. The tap is opened to drain the lighter liquid. Separation of liquid-liquid mixtures by fractional distillation : In case two liquids have very close boiling points, both the liquids tend to distil over in different proportions. It means lesser the boiling point of a liquid, the more is the proportion of its distilling over. The above problem can be avoided by using a fractionating column. It gives the effect of repeated distillation by offering resistance to the passage of vapours. The process of separation of two miscible liquids by the process of distillation, making use of their difference in boiling points, is called fractional distillation. This process is useful only if the difference in the boiling points of the two miscible liquids is between 10°C to 20°C or more. Table given shows various miscible liquids, which can be separated by fractional distillation. Miscible liquid-liquid mixture

Ethyl alcohol

 B.P.= 64.5°C 

Ethyl alcohol

Water (heavier liquid) Immiscible mixture Stop cork

Conical flask Water Separating funnel.

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Water

 B.P =100°C

Methyl alcohol + +

Acetone

+

Acetone

+

 B.P. = 56.5°C 

Ethyl alcohol

Ethyl alcohol Methyl alcohol  B.P. = 78°C  Chloroform  B.P. =61°C

 B.P. =78°C 

 B.P. =56.5°C  Benzene (lighter liquid)

+

 B.P =78°C

Component which distils over

Water

Chloroform

 B.P. =100°C 

Acetone

Ethyl alcohol

Acetone

 B.P. = 78°C 

Method : 1. The process of fractional distillation is similar to the process of distillation, except that a fractionating column is attached in fractional distillation. 2. The design of a fractionating column is such that the vapour of one liquid (with a higher boiling point) is preferentially condensed as compared to the vapour of the other liquid (with lower boiling point) 3. Thus, the vapours of the liquid with low boiling point pass on to Liebig’s condenser where they condense. The liquid so formed is collected in the receiver.

Elements, Compounds & Mixtures 4.

47

The thermometer shows a constant reading as long as the vapour of oneliquid are passing to Liebig’s condenser. As soon as the temperature starts rising, the receiver is replaced by another receiver to collect the second liquid. Stand Fractionating column More volatile Vapour

Thermometer

Stand

Water outlet Liebig’s condenser

Less volatile Vapour Liquid mixture Mesh Burner

Receiver Water inlet Distillate

Fractional distillation of mixtures of liquids

Summary of techniques used in the separation of liquid-liquid mixtures Technique employed

Physical property involved

Examples

The liquid components 1. Kerosene oil i) do not dissolve in and water. 1. Separating one another 2. Carbon funnel (immiscible) disulphide and ii) have different water densities The liquid components 1. Ethyl alcohol i) dissolve in each (b.p. 78°C) and other (miscible) water 2. Fractional ii) have different (b.p. 100°C) distillation boiling points 2. Methyl alcohol (64.5°C) and acetone (b.p. 56.5°C)

Formative Worksheet 39. Assertion:A liquid mixture of kerosene and water can be separated by using a separating funnel. Reason : Kerosene, being heavier, will form the upper layer and water, being lighter, will form the lower layer. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 40. Assertion:Chloroform and water can be separated by using a separating funnel. Reason: The given mixture is a miscible liquid - liquid mixture. a) Both assertion and reason are correct and reason is the correct explanation of assertion.

b)

Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 41. Statement(S): A liquid mixture of mustard oil and water can be separated by using separating funnel. Explanation (E): As mustard oil is lighter than water, it will form the upper layer while water will form the lower layer in the funnel. a) Statement (S) is correct, Statement (E) is incorrect b) Statement (S) and (E) are both correct c) Statement (S) is incorrect, Explanation (E) is correct d) Statement (S) and Explanation(E) are both incorrect 42. Which of the following statements is true for the given mixture? Carbon disulphide and water a) Water is a heavier liquid. b) Carbon disulphide is a lighter liquid. c) It is an immiscible liquid-liquid mixture. d) The mixture can be separated by using a separating funnel. 43. The process of distillation is carried in a distillation flask that is provided with a fractionating column. The purpose of fractionating column is: a) It gives the effect of repeated distillation b) It supports combustion c) It slows down the repeated distillation d) It does not ofter resistance

Conceptive Worksheet 40. Statement A: Ethyl alcohol and water can be separated by fractional distillation. Statement B: Ethyl alcohol and water have different boiling points. Statement C: The boiling point of ethyl alcohol is lower compared to water. a) Statement ‘A’ is false b) Statement ‘B’ is true c) Statement ‘C’ is false d) Statements ‘A’, ‘B’, ‘C’ are all true 41. Which method of separ ation do you suggest to separate methyl alcohol and acetone? a) Fractional Crystallisation b) Fractional distillation c) Sedimentation followed by Decantation d) Separating funnel

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9th Class Chemistry

48 42. Statement A: Methyl alcohol and acetone can be separated by fractional distillation. Statement B: Methyl alcohol completely dissolves in acetone. a) Statement ‘A’ is false but Statement ‘B’ is true. b) Statement ‘A’ is true but statement ‘B’ is false. c) Statements ‘A’ and ‘B’ are both true. d) Statements ‘A’ and ‘B’ are both false. 43. What is the principle used in separation of liquidliquid mixtures, using separating funnel ? a) Difference in melting points b) difference in boiling points c) Difference in solubilities d) Difference in their densities. 44. Choose the correct statement(s): a) Benzene and water form a miscible liquidliquid mixture. b) Compared to benzene, water is a heavier liquid. c) Benzene and water can be separated by using separating funnel. d) We cannot separate benzene and water. 45. The process of separation of two miscible liquids by the process of distillation, making use of their difference in boiling points is called: a) Fractional distillation b) Fractional crystallisation c) Chromatography d) Boiling 46. Out of the liquid-liquid mixture of methyl alcohol and ethyl alcohol the component that distills over first is: a) Methyl alcohol. b) Ethyl alcohol. c) Neither methyl alcohol nor ethyl alcohol. d) Both ethyl alcohol and methyl alcohol. 47. The process of fractional distillation is useful only if the difference in the boiling points of the two miscible liquids is between _______ to ________ or more. a) 100°C to 200°C b) 50°C to 60°C c) 70°C to 80°C d) 10°C to 20°C 48. The boiling point of ethyl alcohol is: a) 78°C b) 100°C c) 61°C d) 64.5°C 49. The boiling point of chloroform is___. a) 78°C b) 61°C c) 78°C d) 100°C

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Separation of liquid-gas mixtures : The solution of a gas in water (liquid) is called liquid-gas mixture. The separation of gas from water is based on the principle that solubility of a gas decreases with the rise in temperature. Following gases can be separated from a liquid-gas mixture. L iquid-Gas mixture Air-wate r mixture

Gas collected on heating Air

Carbon dio xide-wate r mixture

Carbon dioxide

Sulph ur d ioxide -water mixture

Sulphur dioxide

Stand

Delivery tube filled with water

Tap water (water – air mixture)

Graduated tube Dissolved air or boiled off air Beehive shelf

Mesh Burner

Cold water Trough

Separation of liquid-gas mixture

Method: 1. The liquid-gas mixture is filled in a flask and the apparatus is set up as shown in figure. 2. On heating gently (do not boil), the solubility of gas decreases. The bubbles of gas collect over water. Note : Mixture of ammonia in water or HCl gas in water cannot be separated hy this process because of their extreme solubility in water.

7. Separation of gas – gas mixtures

I.

The various techniques used in the separation of gas-gas mixtures is as follows: 1. Diffusion 2. Dissolution in a suitable solvent 3. Preferential liquefaction 4 Fractional evaporation of mixture of liquefied gases. Separation of a gas-gas mixture by diffusion: The rate of diffusion of any gas through a porous partition is inversely proportional to the square root of its vapour density (or molecular weight).

Elements, Compounds & Mixtures Porous partitions Mixture of CO 2 and H2

1. Hydrogen

denotes heavier molecules (say CO 2)

2. 3.

denotes lighter molecules (say hydrogen) Separation of gases by diffusion

Thus, if a mixture of two gases of different densities is passed through porous partitions, then the lighter gas (having less vapour density) will diffuse out more rapidly than the heavier gas. The various gaseous mixtures that can be separated by diffusion are as follows:

Mixture of nitrogen and carbon dioxide Nitrogen Water

Lighter component of gas which diffuses out first

Gas-Gas mixture Carbon dioxide and hydrogen Sulphur dioxide and nitrogen Carbon monoxide and carbon dioxide

Carbon monoxide

Ammonia and nitrogen

Ammonia

Hydrogen Nitrogen

KOH solution Separation of nitrogen from carbon dioxide gas

Method: If a mixture of carbon dioxide and hydrogen is passed through a long tube having a number of porous partitions, hydrogen molecules will diffuse more rapidly as compared to carbon dioxide molecules. Thus, if there is a sufficient number of partitions, in the end hydrogen comes out, as illustrated in figures. II. Separation of gas-gas mixture by dissolution in suitable solvents: The constituents of two gases can be separated if : 1. One of the constituents is soluble in some particular liquid (generally water). 2. One of the constituents reacts chemically with a liquid from which the constituent can be recovered by chemical action. Gas-Gas mixture

49 Method: Let us imagine there is a mixture of nitrogen and carbon dioxide gas. Pass the mixture slowly through potassium hydroxide solution contained in a conical flask. Carbon dioxide reacts with KOH solution chemically to form potassium hydrogen carbonate. However, nitrogen, being insoluble, bubbles out. Collect nitrogen over water, as shown in figure The carbon dioxide can be recovered from potassium hydrogen carbonate solution, by treating it with dilute hydrochloric acid.

Solvent used

Soluble Insoluble gas gas

Formative Worksheet 44. Statement A: The solution of a gas in water (liquid) is called liquid - gas mixture. Statement B: The separation of gas from water is based on the principle that solubility of a gas decreases with the rise in temperature. a) Statement ‘A’ is true but ‘B’ is false. b) Statement ‘A’ is false. c) Statement ‘B’ is false. d) Statements ‘A’ and ‘B’ are both true. 45. On heating we can separate air from air-water mixture. The principle behind the above technique is a) Solubility of a gas decreases with decrease in temperature. b) Solubility of gas decreases with rise in temperature. c) Solubility of a gas depends on the nature of the solvent. d) Solubility of a gas depends on the boiling point of the solvent. 46. Assertion:A mixture of ammonia in water cannot be separated by solubility technique. Reason: Ammonia is insoluble in water.

N 2 and CO 2

KOH solution

CO2

N2

NH3 and N2

Water

NH 3

N2

a)

Cl2 and HCl

Water

HCl

Cl2

Both assertion and reason are correct, and reason is the correct explanation of assertion.

b)

SO2 and O2

KOH solution

SO 2

O2

Both assertion and reason are correct, but reason is not the correct explanation of assertion. Assertion is correct and reason is incorrect. Assertion is incorrect and reason is correct. www.betoppers.com

c) d)

9th Class Chemistry

50 47. Assertion: The constituents of two gases can be separated by dissolution in suitable solvents. Reason I: One of the constituents is soluble in some particular liquid (generally water) Reason II: One of the constituents react chemically with a liquid from which the constituent can be recovered by chemical action.

a)

Assertion is correct and only reason I is the correct explanation of assertion.

b)

Assertion is correct and only reason II is correct explanation of assertion.

c)

Assertion is correct, but reasons I and II are both incorrect.

d)

Assertion is correct and reasons I and II are correct and are the correct explanation of assertion.

48. Statement A: The rate of diffusion of any gas through a porous partition is inversely proportional to the square root of its vapour density (or molecular weight). Statement B: The lighter gas having less vapour density will diffuse out more rapidlythan heavier gas. a)

Statement ‘ A’ is true, but ‘ B’ is false.

b)

Statement ‘ A’ is false.

c)

Statement ‘ B’ is false.

d)

Statements ‘ A’ and ‘ B’ are both true

Conceptive Worksheet 50. Which gas will diffuse out first in separation of mixture of SO2 and NO2? a)

Sulphur dioxide

b)

Nitrogen

c)

Both sulphur dioxide and Nitrogen at the same time

d)

Neither sulphur dioxide nor Nitrogen

51. If a mixture of sulphur dioxide gas and oxygen are given to you, which of the following solvents would you choose to separate the contents? a) Carbon disulphide

b) Water

c) KOH solution

d) Petrol

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III. Separation of gas-gas mixture by preferential liquefaction This method is generally employed for industrial separation of a homogeneous mixture of two gases, such that one of the components of the mixture under high pressure liquefies when the gases are suddenly allowed to expand. The component which escapes on liquefaction is separated from the other component. For example, when a mixture of hydrogen and ammonia under a very high pressure is suddenly allowed to expand in another vessel, the ammonia liquefies and separates from hydrogen. Mixture of gases Ammonia + nitrogen

Component which liquefies Ammonia

Sulphur dioxide + oxygen

Sulphur dioxide

Chlorine + nitrogen Carbon dioxide + oxygen

Chlorine Carbon dioxide

Carbon monoxide + carbon dioxide Carbon dioxide

IV. Separation by fractional evaporation of liquefied mixture of two gases: Sometimes, when a mixture of two gases under extremely high pressure is allowed to expand, both the gases liquefy. For example, when cold air under very high pressure is suddenly allowed to expand, both the constituents of air, i.e., nitrogen and oxygen liquefy. The boiling point of liquid oxygen is –183 °C and that of liquid nitrogen is –196 °C. When the above liquid is maintained at –196 °C, nitrogen starts boiling to produce nitrogen gas. It is collected separately. Oxygen is left in liquid state as it does not boil off.

Hydrogen and oxygen

Component which boils off Hydrogen

Sulphur dioxide and chlorine

Sulphur dioxide

Components of liquefied gas

Elements, Compounds & Mixtures

51

Summary of the techniques of separation of gas-gas mixtures Technique employed for the separation of gas-gas Physical property involved in separation Examples mixture Diffusion The rate of diffusion of less dense gas (lighter Hydrogen and carbon dioxide; gas) is higher as compared to a heavier gas. Nitrogen and chlorine. Dissolution in a suitable solvent Preferential liquefaction

One of the components of gas is soluble in a particular solvent. One of the gaseous components can be easily liquefied as compared to other components. Fractional evaporation of The component of liquefied gas having lower mixture of liquefied gases. boiling point evaporates first.

Formative Worksheet 49. The mixture of sulphur dioxide and oxygen can be separated by the method of preferential liquefaction. The principle behind the preferential liquefaction is: a) One of the components has high boiling point. b) One of the components has low melting point. c) One of the components of the mixture under high pressure liquefies when the gases are suddenly allowed to expand. d) One of the components is soluble in water. 50. Chlorine and Oxygen can be separated by the method of preferential liquefaction: Choose the physical property involved in separation:

a) b)

The rate of diffusion of less dense gas is higher as compared to a heavier gas.

One of the components of gas is soluble in a particular solvent. c) One of the gaseous components can be easily liquefied as compared to other components. d) The component of liquefied gas having lower boiling point evaporates first. 51. Match the following: Column I Column II (Technique employed) (Example) p) Diffusion i) Liquefied air q) Preferential ii) Ammonia and liquefaction hydrogen r) Fractional iii) Nitrogen and evaporation of chlorine mixture of liquefied gases s) Dissolution iv) Chlorine and in a suitable solvent Oxygen a) p  (iii), q  (iv), r  (i), s  (ii) b) p  (i), q  (iii), r  (iv), s  (ii) c) p  (iv), q  (i), r  (ii), s  (iii) d) p  (ii), q  (iv), r  (i), s  (iii)

Ammonia and hydrogen; HCl gas and chlorine Chlorine and oxygen; Carbon dioxide and hydrogen Liquefied air; Liquefied nitrogen and hydrogen gases.

52. When cold air under very high pressure is suddenly allowed to expand, what would happen to the constituents of air? a) Nitrogen and Oxygen would liquefy simultaneously. b) Oxygen would have diffused faster. c) Nitrogen would liquefy faster. d) Oxygen would liquefy faster. 53. Match the following: Column I Column II (i) Fractional distillation p) Benzoic acid + Sodium chloride (ii) Fractional q) Chlorine + crystallisation Nitrogen (iii) Sublimation r) Carbon dioxide and hydrogen (iv) Liquefaction s) Ethyl alcohol + Chloroform (v) Diffusion t) Sodium nitrate and sodium chloride a) i  s, ii  t, iii  p, iv  q, v  r b) i  r, ii  q, iii  q, iv  s, v  t c) i  t, ii  s, iii  p, iv  q, v  r d) i  r, ii  q, iii  s, iv  p, v  t 54. Statement A: If salt is soluble in water, it can be separated by simple distillation. Statement B: Water distils after leaving the salt behind in the distillation flask. a) Statement ‘A’ is false but Statement ‘B’ is true. b) Statement ‘A’ is true but Statement ‘B’ is false. c) Statement ‘A’ and ‘B’ are both are true. d) Statement ‘A’ and ‘B’ are both false.

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9th Class Chemistry

52

55. Choose the correct option to complete the table given below based on separation technique. Mixture Ammonia + Hydrogen Charcoal + Sulphur + Nitre

a) b) c) d)

Type of mixture A

Separation t echnique B

Heterogeneous

C

A  Using solvents., B  fractional homogeneous A  Fractional evaporation of liquefied A  Homogeneous, B  preferential A  Using solvents., B  homogeneous, mixture

evoporation of liquefied mixture, C  mixture.,B  usingsolvents,C  homogeneous liquefaction, C  using solvents. C  fractional evoporation of liquefied

Conceptive Worksheet 52. Statement A: The components of mixture of two gases hydrogen and oxygen can be separated by fractional evaporation.

Summative Worksheet 1.

Element

Statement C: Both Hydrogen and Oxygen boil off. Statement ‘A’ is false

b)

Statement ‘B’ is true

c)

Statement ‘C’ is false

d)

A and B are true, C is false

Metals

53. Chloroform+Water mixture can be separated by:

a)

Fractional distillation

b)

Using separating funnel

c)

Sedimentation

d)

Fractional crystallisation

54. The rate of moment of each adsorbed material depends upon: a)

The relative solubility of the constituent of mixture in a given solvent.

b)

The relative affinity of the constituents of mixture for the adsorbed medium.

c)

The relative solubility of the constituent of mixture in a given solute.

d)

None of these.

2.

3.

55. Iodine and methyl alcohol can be separated by distillation because: a)

Methyl alcohol is non-volatile in nature.

b)

Iodine is a non-volatile solid

c)

Both methyl alcohol and iodine are volatile.

d)

Iodine acts as a solvent

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Matter X

Statement B: Hydrogen boils off. a)

Principle involved Difference in boiling points One of the components is soluble in a specific solvent.

4.

Y

Compound

C

Metalloids

A

Heterogeneous

E

a) X  Pure substance, Y  Mixtures, A  Homogeneous mixtures, C  Non-metals, E  Noble gases. b) X  Non-metals, Y  Mixtures, A  Homogeneous mixtures, C  pure substance, E  Noble gases. c) X  Homogeneous mixtures, Y  Mixtures, A  Noble gases, C  pure substance, E  Non-metals. d) X  Pure substance, Y  Mixtures, A  Noble gases, C  Homogeneous mixtures, E  Non-metals. 9g of pure water will always contain: a) 1g of hydrogen and 8g of oxygen. b) 2g of hydrogen and 3g of oxygen c) 8g of hydrogen and 1g of hydrogen d) 12g of oxygen and 4g of hydrogen Classify the following elements into metals, nonmetals, noble gases: Na, N, Ne, Mg, F, Ar, Ca, Cl, Kr Metals Non-metals Noble gases a) Na, N, Ne Mg, F, Ar Ca, Cl, Kr b) Na, Mg, Ca N, F, Cl Ne, Ar, Kr c) Na, N, Mg Ne, F, Ar Ca, Cl, Kr d) None Choose the correct order of the symbols for the following given elements: (i) Mercury (ii) Tin (iii) Lead (iv) Copper a) Cu, Sn, Hg, Pb b) Pb, Hg, Cu, Sn c) Sn, Hg, Cu, Pb d) Hg, Sn, Pb, Cu

Elements, Compounds & Mixtures 5.

6.

Statement A : Normal elements give harmful radiations. Statement B: Radioactive elements do not produce harmful radiations. a) ‘A’ is true, ‘B’ is false. b) ‘A’ is false, ‘B’ is true. c) Both ‘A’ and ‘B’ are true. d) Both ‘A’ and ‘B’ are false. The approximate number of normal elements is _____ and that of radioactive elements is ______.

a) 115, 33 b) 82, 33 c) 92, 12 d) 75, 35 7. Elements are further classified into: a) Metals b) Non-metals c) Metalloids d) Noble gases 8. Match the following: Column ‘A’ Column ‘B’ (i) Solid (p) Chlorine (ii) Liquid (q) Phosphorous (iii) Gas (r) Bromine (s) Iodine (t) Oxygen a) (i)  qs, (ii)  r, (iii)  pt b) (i)  st, (ii)  rq (iii)  p c) (i)  pqr, (ii)  s, (iii)  t d) (i)  p, (ii)  qr, (iii)  st 9. A pure chemical compound can be: a) Only heterogeneous b) Only homogeneous c) Both heterogeneous and homogeneous d) Neither heterogeneous nor homogeneous 10. Which of the following statements is/are false? a) The nature of elements constituting a chemical compound may not remain the same. b) Formation of a compound involves energy changes. c) A chemical compound cannot be broken into two or more different elements. d) A chemical compound has fixed composition. 11. Choose the correct statements: a) A compound is a pure substance. b) A compound is made of two or more elements, combined chemically in a fixed proportion by weight. c) A compound can be broken into smaller parts of elements by only chemical methods. d) Formation of a compound involves energy changes.

53 12. The process of breaking down of a chemical compound into its elements by chemical means is called, a) Cracking b) Analysing c) Decomposition d) Catalysis 13. Which of the following statement(s) is/are true? a) C60 is a kind of Carbon with no tensile strength. b) The density of Carbon, Sulphur and Phosphorus is more than 1. c) Bromine is a soft metal d) Arsenic and Antimony are non ductile. 14. Choose the false statements. a) Composition of a mixture is always fixed. b) A mixture has definite melting and boiling points. c) In the formation of a mixture, no chemical reaction occurs. d) Durng the formation of a mixture, the constituents undergo a major change in their composition. 15. Choose the correct statements: a) In air, the main constituents, i.e., oxygen, nitrogen and carbondi oxide do not bind each other with chemical bonds. b) A mixture of sand and salt can be in the ratio of 1:2 or 5:6, etc, by weight. c) In the mixture of iron and sulphur, attimes iron is more and/or sulphur may be more. d) On mixing iron and sulphur, heat energy is neither absorbed nor evolved. 16. Which one of the following substances is soluble in Acetone? a) Iodine b) Chlorophyll c) Nail polish d) Oil 17. Rust is soluble in: a) Benzene b) Oxalic acid c) Carbon disulphide d) Petrol 18. Chlorophyll is soluble in a) Petrol b) Methylated spirit c) Water d) Ethyl alcohol 19. Choose the solvent that separates common salt and marble powder. a) Turpentine oil b) Water c) Carbon disulphide d) Benzene 20. Which of the following properties are considered for separation of constituents of a mixture? a) Density b) Solubility c) Volatility d) Sublimation 21. Which method of separation would you suggest to separate ammonium chloride and common salt? a) Distillation b) Sublimation c) Filtration d) Crystallisation www.betoppers.com

9th Class Chemistry

54 22. Choose the correct statement(s) : a) Ink is a mixture made of water and one or more dyes. b) Milk is a pure substance. c) Soft drinks are homogeneous mixtures. d) Alloys form a useful class of mixtures. 23. The process of evaporation is suitable for the separation of (i) a solid from its (ii). a) (i)  Volatile soluble solid, (ii)  Solute b) (i)  Insoluble solid, (ii)  Solution c) (i)  Non-volatile solid, (ii)  Liquid solvent d) (i)  Sublimate, (ii)  Filtrate 24. Distillation method of separation can be used to separate: a) Homogeneous mixture. b) Heterogeneous mixture. c) Both homogeneous and heterogeneous mixtures. d) None of the above. 25. Statement-A: We use dissolution method of separation when one of the constituents is soluble in the suitable solvent. Statement-B: We can separate salt or sugar from sand by using dissolution method.

3.

Liquefied air Fractional distillation

4.

5.

HOTS Worksheet

2.

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Low boiling

High boiling

A

c) Both ‘A’ and ‘B’ are true. d) Both ‘A’ and ‘B’ are false. Match the following: Column ‘A’ Column ‘B’ (a) CuSO4 .5H2 O (i) Washing soda (b) Na 2 CO3.10H2 O (ii) Blue vitriol (c) NaHCO3 (iii) Common salt (d) NaCl (iv) Baking soda a) a  (ii), b  (i), c  (iv), d  (iii) b) a  (i), b  (iii), c  (iv), d  (ii) c) a  (i), b  (iii), c  (iv), d  (ii) d) a  (iv), b  (i), c  (iii), d  (ii) Match the following: Column ‘A’ Column ‘B’ (a) Seawater (i) Element (b) Ice (ii) Mixture (c) Mercury (iii) Homogeneous mixture (d) Soil (iv) Compound (e) Filtered tea (v) Solution a) a  (iv), b  (v), c  (iii), d  (ii), e  (i) b) a  (v), b  (iv), c  (i), d  (ii), e  (iii) c) a  (ii), b  (iv), c  (i), d  (v), e  (iii) d) a  (i), b  (iv), c  (iii), d  (ii), e  (v)

Pressure increased, Temperature decreased

Air

a) ‘A’ is true, ‘B’ is false. b) ‘A’ is false, ‘B’ is true.

1.

Assertion: We can separate nickel from mixtureof nickel and lead by magnetic separation. Reason: One of the components of the mixture is a magnetic substance. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct.

B

Choose the correct option: a) A  Nitrogen, B  Oxygen b) A  Nitrogen, Oxygen c) B  Nitrogen, Oxygen d) A  Oxygen, B  Nitrogen Identify the gas that diffuses first in the following mixtures: p) Carbon dioxide and hydrogen q) Sulphur dioxide and nitrogen r) Carbon monoxide and carbon dioxide s) Ammonia and nitrogen

p

q

r

s

a)

Hydrogen

Nitrogen

Ammonia

b)

Carbon dioxide Carbon dioxide None

Sulphur dioxide Nitrogen

Carbon monoxide Carbon dioxide Carbon dioxide

c) d)

6.

Nitrogen

Iodine can be dissolved in: Gas – Gas mixture Nitrogen gas and Carbon dioxide g as I and nitrogen gas Chlorine gas and II Sulphur dio xide gas and oxygen gas

a) b) c) d) 7.

Nitrogen

I HCl NH 3 CO 3 NH 3 Name

Solvent used III

Soluble gas CO 2

Insoluble gas IV

Water Water KOH solution

NH3 HCl VI

N2 V O2

II III KOH NH 3 HCl KOH Cl2 N2 HCl KOH the following:

IV N2 N2 KOH N2

V SO2 Cl2 N2 Cl2

VI CO 2 SO2 SO2 SO2

Elements, Compounds & Mixtures (i) (ii) (iii) (iv) (v) (vi) (vii) 8.

55

The most abundant element in the earth’s crust. A non-malleable, non ductile metal. A lustrous non-metal other than graphite A metalloid other than arsenic A liquid non-metal A liquid metal A non-metal, good conductor of electricity

Match the following i.

Column – I Pure substances which cannot be broken down into two or more simpler substances.

ii. Elements which generally have lustre, are good conductors of heat and electricity and are malleable and ductile.

Column – II A. Metals B. Noble gases C. Metalloids

iii. Elements which generally have properties mid-way between metals and non-metals.

D. Elements

iv. Elements which are chemically inactive, and occur in traces in the atmosphere.

E. Non-metals

v. Elements which generally lack lustre, are non-malleable and non-ductile and bad conductors of heat and electricity.

9.

State which of the characteristics below pertains to a mixture and which of them to a compound. i. Is a pure substance ii. Consists of elements, compounds or both. iii. Is composed of two or more elements. iv. Constituents in it are mixed in any proportion. v. Is composed of two or more substances. vi. Retains the properties of its constituents. vii. Constituents combined chemically in a fixed proportion. vii. Properties are different from the properties of the constituents elements. 10. Milk is regarded as a mixuture, while sodium chloride as a compound. Explain. 11. State whether the following mixtures are homogeneous or hetergeneous. i. Duralumin (alloy) ii. Sugar in water iii. Sulphur dioxide in water iv. Ammonia in air v. Potassium and sodium chloride mixture 12. How would you separate: i) pure water from sea water? ii) kerosene oil from a mixture of kerosene oil and petrol iii) lead sulphate from a mixture of lead sulphate and lead chloride

13. From the techniques (or methods) of distillation, filtration, fractional distillation, evaporation, crystallisation, select and write down the technique you would use to separate. a. A mixture of acetone and water. b. Carbon disulphide from sulphur c. Barium sulphate from water. 14. How would you: i) Obtain sulphur from a mixture of sulphur and iron filings. ii) Separate a precipitate of lead sulphate obtained by adding sulphuric acid to a solution of lead nitrate. iii) Obtain oil from a mixture of oil and water. iv) Obtain iodine from a mixture of powdered iodine and ammonium chloride. v) Obtain powdered charcoal from a mixture of copper oxide and powdered charcoal. 15. Match the following Mixture Method of Separation i) Sand + water ( ) a) Separating funnel ii) Iron + copper ( ) b) Sedimentation iii) Sand + Iodine ( ) c) Evaporation iv) Oil + water ( ) d) Magnetic separation v) Common salt ( ) e) Sublimation in

sea water 

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56

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9th Class Chemistry

B y t h e e n d o f t h i s c h a p t e r , yo u w i l l un de r s t an d • • • • • •

Ancient ideas of atom Cathode rays and anode rays Discovery of neutrons Thomson’s atomic model Rutherford’s model of atom Quantum theory

• • • • •

Bohr’s atomic model Atomic number & Mass number Electronic configuration Geometric representation of an atom Radioactivity

1. Introduction In our earlier classes of chemical classification of matter, we have learnt that matter is made up of atoms and molecules. Atoms are the basis of chemistry. They are the basis for everything in the Universe. Atoms and the study of atoms are a world unto themselves. In this chapter, we are going to explore atom and its constituents. Now, let us learn a little more about this idea that has brought about a revolution in science.

2. Ancient ideas of an atom The views of Kanad Way back as the sixth century BC, the Indian philosopher Maharshi Kanad came forward with the following idea. Matter is not continuous, and made up of tiny particles, named paramanus. (In Sanskrit, param means final or ultimate and anu means particle.) Kanad further said that two or more paramanus combine to form bigger particles.

The Views of Democritus and Leukiposs In the fifth century BC, the Greek philosophers Democritus and Leukiposs came up with a similar idea. They thought that on dividing a piece of a substance, one would ultimately get a particle that could not be divided any further. They gave the name atomos (in Greek, atomos means indivisible) to these ultimate particles.

Dalton’s Theory The theories of Kanad as well as of Democritus and Leukiposs remained forgotten for more than two thousand years. But when experimental chemistry developed, it became necessary to explain the observed facts.

Chapter -3

ATOMIC STRUCTURE

Learning Outcomes

In this connection, in 1803, an English Chemist, John Dalton, put forward his Atomic Theory. The main postulates of this theory are as follows: 1. Composition of matter: Matter is made of very small particles called atoms. 2. Indivisibility of atoms: Atoms are indivisible. They cannot be further broken down. 3. Invincibility of atoms: Atoms can be neither created nor destroyed in a chemical reaction. 4. Atoms of similar elements and dissimilar elements: Atoms of a given element are identical in all respects - same mass, same size and same properties. For example, atoms of hydrogen element are identical in all respects. Atoms of different elements are different. For example, atoms of hydrogen and oxygen elements are different in all respects. 5. Combination of atoms: Atoms of different elements combine in a whole number ratio to form compounds. For example, the ratio of combination of hydrogen and oxygen to form water molecule (H2O) is 2:1 which is a whole number ratio. 6. Role of atom in a chemical reaction: Atom is the smallest particle of matter that takes part in a chemical reaction. Contradiction of Dalton’s atomic theory by modern atomic theory Further experiments conducted by scientists have provided the evidence that atoms are further divisible and made of fundamental particles called electrons, protons and neutrons. Hence Dalton’s theory does not give an explanation regarding further divisibility of atoms and also the internal structure of an atom. Let us see the contradictions of Dalton’s theory by the modern atomic theory.

9th Class Chemistry

58 S.No.

Dalton's atomic theory

Modern atomic theory

1.

Matter consists of small indivisible Atom is no longer indivisible, but particles called atoms. consists of neutrons, protons and electrons.

2.

Atoms of same element are alike in all respects.

All atoms have isotopes. It means some of the atoms of same element have different atomic weights.

3.

Atoms of different elements are different in all respects.

Atoms of different elements are sometimes similar in some respects. For example, atoms of argon and calcium have same atomic weight.

4.

Atoms combine in small whole Atoms in organic compounds do not numbers to form compound atoms combine in small whole number ratio. (molecules). The molecules of proteins are highly complex.

Similarity : Atom is the smallest unit of matter that takes part in a chemical reaction.

3. Discovery of Cathode rays Although many of the pioneers of 19 th century physics, including Faraday, were convinced on the basis of chemistry and the phenomena observed in electrolysis that electric current consisted of the flow of particles of charge, the nature of these charges was not understood. Even the basic question of whether the charge of the particles was positive or negative remained undetermined. The answers to these questions, and to the basic structure of matter, were resolved by experiments that began with the study of electric discharges in evacuated tubes. Along the way a series of discoveries were made which led to the technological revolution of the 20th Century. In 1855, The German inventor Heinrich Geissler developed the mercury pump and produced the first good vacuum tubes. These tubes, as modified by Sir William Crookes, became the first to produce cathode rays, leading eventually to the discovery of the electron.

Experiments in discharge tube: The electrodes are connected to high voltage for the current to flow. The changes that occur in the discharge tube at different pressures were observed. The high voltage provides energy for the atoms of a gas to further split or break up . When both electrodes are connected to high voltage, current starts flowing. At high pressure no electricity flows through the air in the discharge tube, so low pressure is used. Low pressure helps in conduction of electricity. At high voltage of 10,000 volts and at normal atmospheric pressure there is no effect. But keeping the same voltage if pressure is reduced to 0.01 mm of Hg, a greenish glow was observed at anode. The rays are emitted from the direction of the cathode, and are called cathode rays.

Description of discharge tube The discharge tube consists of a glass tube from which most of the air has been evacuated having two metal plates sealed at both the ends. These metal plates are called electrodes. These electrodes are connected to positive and negative terminals of a battery. The electrode connected to the positive terminal is known as anode and the electrode connected to the negative terminal is known as cathode. www.betoppers.com

Properties of cathode rays: The experiments on cathode rays by J.J Thomson led to the discovery of electrons. The following are the properties of cathode rays :

Atomic Structure (i)

59

Cathode rays travel in straight lines. (iv) Cathode rays are negatively charged and Explanation: When an object is placed opposite to affected by magnetic field. the direction of the cathode rays, a sharp shadow is formed. This concludes that cathode rays travel in straight lines.

Explanation: When the magnetic field is applied perpendicularly to the path of cathode rays, they get deflected towards the north pole of the magnet (ii) Cathode rays contain some material particles. which is expected of negatively charged particles. This further confirms that cathode rays are negatively charged. (v) Cathode rays generate heating effect Explanation: When cathode rays are focused on a thin metal foil, it gets heated up. This proves that cathode rays have some kinetic energy. (vi) Cathode rays affect ZnS screen. Explanation: When cathode rays are allowed to strike ZnS screen, they produce a faint greenish Explanation: When cathode rays are allowed to fluorescence. fall on a paddle wheel, it rotates. This is possible (vii) Cathode rays ionises the gases. only when the rays striking it have some material Explanation: When cathode rays are allowed to particles. From this, it can be concluded that cathode pass through gases, different glows are seen in the rays consist of some material particles. tube. These different glows are due to the ionisation (iii) Cathode rays are negatively charged and affected by the electric field. of gases. (viii)Cathode rays have penetrating power. Explanation: When cathode rays are allowed to pass through thin metal foils, a glow is seen behind the metal foil indicating that they have good penetrating power. (ix) Cathode rays produce X-rays Explanation: When cathode rays are allowed to fall on metals such as tungsten, copper, X-rays are observed. (x) The e/m ratio of cathode ray particles is constant Explanation: When the electric field is applied When cathode rays were obtained by using perpendicularly to the path of cathode rays, they different gases and vapours of different materials, deflect towards positive plate. As opposite charges the ratio of charge and mass (e/m ratio) of the attract, the cathode ray particles are negatively cathode ray particles of these gases were obtained. charged. It was observed that the e/m ratio of the cathode ray particles remained the same irrespective of the nature of the gas or vapour entering the discharge tube. As the e/m ratios is same, we can conclude www.betoppers.com

9th Class Chemistry

60 that the cathode ray particles obtained from different gases and vapours have the same charge and mass. Further, the cathode ray particles are similar for any gas and vapour of any matter. Thus, a cathode ray-particle is a common constituent of any matter with same mass and charge everywhere. So, it is called the universal constituent of matter. This common constituent of matter was considered as the fundamental particle called electron.

Determination of charge and mass of the electron. The charge of the electron was determined by R.A Millikan with the help of oil drop experiment. The value was found to be 1.602 × 10–19 coulomb (C). This negative charges on the electron was regarded as unit negative charge (–1). The mass of the electron was also determined by a suitable method. It was found to be 9.1 × 10–28g. This was regarded as negligible and was nearly

1 times the mass of hydrogen atom. The charge 1837 (e) of the electron was divided by its mass (m) to calculate the charge/mass ratio also called e/m ratio.

e 1.6 1019 C   1.76  108 C / g . 28 m 9.1 10 g

All about electrons JJ Thomson 9.1 × 10-28 g 9.1 × 10-31 kg 0.0005 486 amu Charge - 1.602 × 10-19 C - 4.8 × 10-10 e.s.u Relative Charge -1 e/m ratio 1.76 × 108 C/g Note: amu is called atomic mass unit and is the smallest unit of mass. 1 amu = 1.66 × 10–24g or 1.66 × 10–27kg. Discoverer Mass

4. Discovery of anode rays We know that in science many discoveries and inventions had their origin from small ideas. One such discovery was the existence of proton, by Goldstein. Goldstein repeated the cathode-ray experiment, using a perforated cathode. He observed that there was a glow on the wall opposite the anode. So, some rays must be travelling in the direction opposite to that of the cathode rays, i.e., from the anode towards the cathode. These rays were called anode rays or canal rays (as they moved through the perforations, or canals, in the cathode). It was found that these rays contained positively charged particles, and so J.J.Thomson called them positive rays. Positive rays from anode

Cathode rays from cathode

Anode

Positive rays

Red glow

Cathode

High-voltage source

Anode rays

How anode rays are produced When an electric discharge is passed through the gas, some of the molecules of the gas ionised and produce cathode rays. Cathode rays consist of electrons. These electrons move with high speed towards the anode. As they move, they collide with the remaining molecules of the gas in the tube causing them to electrons. The positive ions formed are attracted by the performed cathode. The stream of these positive ions pass a glow on the glass wall of the discharge tube. (On the other side of the discharge tube, the cathode rays produce a green light). The stream of positive ions so formed constitute the positive or anode rays. www.betoppers.com

Atomic Structure

61

Properties of anode rays: i) Anode rays travel in straight lines. ii) Anode rays contain positively charged particles and hence cause mechanical motion. iii) Anode rays are deflected both in electric (towards negative plate) and magnetic fields (towards South pole). iv) They cause heating effect and have some kinetic energy. v) The e/m ratio of the particles in the anode rays is not constant and depends on the nature of gas taken in the discharge tube.

Discovery of protons: So the experiments showed that the e/m ratio for positive rays depends on the gas used in the discharge tube. Since hydrogen is the lightest element, its mass ‘m’ is the lowest and e/m ratio is the highest.

Goldstein gave the name proton for the particles which comprised the positive rays, produced by hydrogen gas. Thus the proton is a hydrogen ion which is produced from the hydrogen atom. The hydrogen ion is formed by loss of an electron from a hydrogen atom. Note: i) The charge of proton +1.602  10 –19 C or 4.8  10–10 e.s.u (electro static unit) . ii) The mass of proton is1.672 × 10 -24 g or 1.00727 a.m.u.

All about protons Goldstein 1.672 × 1024g 1.672 × 1027kg 1.00727 amu + 1.602 × 10-19 C Charge + 4.8 × 10-10 e.s.u Relative Charge +1 9.58 × 104 C/g e/m ratio Discoverer Mass

Comparison of cathode rays and anode rays Cathode rays Cathode rays travel in straight lines from cathode to anode and cast the shadow of the object placed in their path. Cathode rays contain material particles (electrons) which are negatively charged. These rays are deflected in both magnetic and electric fields. These rays have kinetic energy and raise the temperature of a metallic object on which they fall. Cathode ray particles are common constituents of all matter and their e/m ratio is constant for all gases.

Anode rays Anode rays travel in straight lines from anode to cathode and cast the shadow of the object placed in their path. Anode rays contain material particles which are positively charged. These rays are deflected in both magnetic and electric fields. These rays have kinetic energy and raise the temperature of a metallic object on which they fall. The e/m ratio of anode ray particles is different for different gases and is maximum for hydrogen.

5. Discovery of Neutrons By 1920, with the discovery of electron and proton, it was thought that the inner structure of the atom was complete. The mass of the electron is negligible; hence the mass of atom should be equal to the mass of protons concentrated inside the nucleus. Different atoms have different number of protons, hence different atomic masses. However, it was observed that there was a discrepancy between the actual atomic mass and the calculated atomic mass. For example, an atom of carbon has 6 protons; therefore its mass should be six times the mass of hydrogen atom which has one proton. But, experimentally, it was found that the mass of carbon atoms is twelve times the mass of hydrogen atom. A similar problem was encountered with regard to the mass of other atoms. Then, what was the reason for this discrepancy? Rutherford was the first scientist to predict the reason for this discrepancy. He predicted that along with protons, there were some other neutral particles present inside the nucleus. These particles were discovered as neutrons by James Chadwick in 1932. Now, let us study the observations and conclusions that led to the discovery of neutrons. www.betoppers.com

9th Class Chemistry

62 Chadwick’s observations and conclusions During an experiment it was found that, when alpha particles bombarded Beryllium nuclei, some radiations were observed. These radiations were found to be undeviated in an electric field. So, the earlier scientist thought these radiations could be electromagnetic radiation and hence discarded it. But Chadwick repeated the experiment and made the following observations.

2. 3. 4. 5.

6. 7. 8. 1. A paddle wheel was placed behind the Beryllium nucleus and the nucleus was bombaded with  -particles. It was observed that the paddle wheel rotates. From this, it was concluded that the beryllium nucleus emits some invisible 1. radiations having material particles. 2. When these invisible radiations were allowed to pass through an electric field, there was no deviation seen. This confirmed the fact that these rays contained neutral particles. These neutral particles were called neutrons by James Chadwick. All about neutrons

Discoverer Mass

James Chadwick -24

1.675 × 10

2.

g

1.675 × 10-27 kg 1.00866 amu Charge

No charge

Formative Worksheet 1.

Assertion : The specific charge of anode ray 3. particles depends on nature of the gas taken in the discharge tube. Reason : The particles in anode rays carry positive charge. 1) Both assertion and reason are correct and reason is the correct explanation of assertion. 2) Both assertion and reason are correct and reason is not the correct explanation of 4. assertion. 3) Assertion is correct and reason is incorrect. 5. 4) Assertion is incorrect and reason is correct. 6.

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What is the ratio of mass of electron to that of proton? How many number of electrons make 1 cgs unit of mass? What is the mass of one mole (6.023× 10 23 ) electrons ? Faraday is the biggest unit of charge and is equal to 96500 coulomb. Find the number of electrons whose charge equal to the biggest unit of charge. What is the ratio of specific charge (e/m) of an electron to that of a hydrogen ion ? The charge on an electron is 4.8 × 10–10 esu. What is the value of the charge on Li+ ion? What is the ratio of specific charge of proton to that of an alpha particle? (Hint : mass of  – particle = 4 × 1.66 × 10–24 g)

Conceptive Worksheet Which of the following is true according to Dalton’s atomic theory? 1) Matter consists of small indivisible particles called atoms. 2) Atoms of same element are alike in all respects. 3) Atoms combine in small whole numbers to form compound atoms (molecules). 4) Atom is the smallest unit of matter which takes part in a chemical reaction. Which of the following is not true according to modern atomic theory? 1) Matter consists of small indivisible particles called atoms. 2) Atoms of same element are alike in all respects. 3) Atoms combine in small whole numbers to form compound atoms (molecules). 4) Atom is the smallest unit of matter which takes part in a chemical reaction. Which of the following is true according to modern atomic theory? 1) Atom is no more indivisible. 2) All atoms of same element are not always similar. 3) Atom is not the smallest particle that takes part in chemical reaction. 4) Atoms of different elements are different in all respects What can be concluded from the constancy of e/m ratio ? What is the pressure and voltage at which cathode rays are produced ? Find the number of protons whose mass is equal to 1 S.I unit of mass.

Atomic Structure

63

7.

Write charge and mass of electron, proton and neutron. 8. Does the mass and charge on positive rays remain same with the change in nature of gas in a discharge tube? 9. Name the gas that produces positive rays consisting only protons. 10. The specific charge for positive rays is much less than the specific charge for cathode rays. Explain. 11. What is the e/m ratio of a neutron? 12. The e/m ratio is not constant for 1) Cathode rays 2) Positive rays   rays 3) 4)   rays

been given different names: Water Melon Model (or) Plum Pudding Model (or) Raisin Pudding Model. He believed that an atom was made up of positively charged substances in the form of a sphere in which electrons were embedded. This can be compared to a water melon, which has evenly distributed red spongy mass with black seeds embedded in it. He could not explain how the positively charged particles were shielded from the negatively charged particles without getting neutralized. Hence this model of atom was discarded. + + +

6. The structure of an atom After the discovery of fundamental particles, scientists started thinking of the arrangement of these fundamental particles inside the atoms. This resulted in the development of various atomic models by different scientists. People are acquainted with the revolving air symbol used to depict the atom.

Electrons

+ +

+ +

Positively Charged sphere

+ +

8. Rutherford’s Model of Atom

Nucleus Ernest Rutherford (1871-1937)

Electrons

The symbol shows a glowing nucleus enmeshed within the intertwined loops of orbiting electrons. What is the truth behind this symbol? Did the atom ever sit for its portrait? The truth, however, is that no one has ever seen an atom. Nevertheless, bit by bit, the minute terrain of the atom has been mapped. Taken as a whole, this airy symbol has built up a picture of the atom with proven usefulness.

7. Thomson’s Atomic Model

Rutherford overturned Thomson’s model in 1911 with his famous gold-foil experiment in which he demonstrated that the atom has a tiny, massive nucleus Five years earlier Rutherford had noticed that alpha particles beamed through a hole onto a photographic plate would make a sharp-edged picture, while alpha particles beamed through a sheet of mica only 20 micrometre (or about 0.002 cm) thick would make an impression with blurry edges. For some particles the blurring corresponded to a two-degree deflection. Remembering those results, Rutherford had his postdoctoral fellow, Hans Geiger, and an undergraduate student, Ernest Marsden, refine the experiment. Let us now know about the experiment.

Experimental Setup: J.J. Thomson (1856-1940)

One of the first attempts in predicting the arrangement of subatomic particles with in the atom was made by JJ Thomson. His model of atom has

The experimental arrangement was very simple. First, a lead block with a cavity containing the radioactive element, radium was taken. In front of the lead block, a very thin gold foil (0.0004 mm) was kept. At the other end, a screen of zinc sulphide was adjusted. www.betoppers.com

9th Class Chemistry

64 Alpha rays from the radioactive radium passed through the golden foil and hit the ZnS screen. Scintillations were produced when alpha particles struck the zinc sulphide screen.

Rutherford’s observations and conclusions 1. Most of the  -particles went straight through the foil. This is explained by the fact that they were not attracted to or repelled by any particle. In other words, the atom is mostly empty. -particles

Rutherford’s explanation of -particle scattering on the basis of his nuclear model

2. Some of these particles deviated slightly from their path. They were repelled to a small extent by a positive charge. Very few of the particles, the ones at the centre, almost retraced their path. This meant that they were strongly repelled by a small positively charged body at the centre of the atom. This positively charged body is called the nucleus. Due to this reason this model of atom is also called Planetary model of atom. The mass of the atom is concentrated in the nucleus. 3. Rutherford also theorised that electrons revolve round the nucleus at large distances from it just like planets round the sun. For this reason, this model of atom is also called Planetary model or solar model. Rutherford estimated the diameter of the nucleus to be of the order of 10–13 cm and that of the atom to be of the order of 10–8 cm. Thus, the diameter of the nucleus is about 105(= 1,00,000) times smaller than that of the atom. www.betoppers.com

Rutherford’s explanation for the stability of nucleus and atom: After the success of bringing out a successful atomic model, Rutherford was cornered by the other seniors by a question: “Why does the nucleus not disintegrate inspite of repulsion among the protons?” To explain the stability of the nucleus, Rutherford predicted the presence of neutral particles known as neutron. The presence of these neutrons between the protons neutralises the repulsion among the protons. Rutherford predicted the presence of the neutron even before it was discovered. Coming to the stability of the atom, he explained that the revolving electron is under the influence of two types of forces. Electron Centrifugal force 

+

Electrostatic force

i) The electrostatic force of attraction between the nucleus and the electron and ii) The centrifugal force directed away from the revolving electron. These two forces are equal and opposite and hence keep the electron in equilibrium in the path. This is the reason why electrons do not fall into the nucleus in spite of inward nuclear pull.

Drawbacks of Rutherford’s atomic model 1. Spiral path of electron: According to the law of electro-dynamics, a charged particle revolving round another oppositely charged particle should lose energy continuously. So the electron moving round the positively charged nucleus should continuously emit radiation and lose energy. As a result of this , a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus. e–

+

Atomic Structure

2.

It was calculated that the electron should fall into the nucleus in less than 10–8 sec. But it is known that electrons keep moving outside the nucleus. Discrete lines in atomic spectra: Experimentally, the atomic spectra consist of discrete lines. If the electron loses energy continuously, the atomic spectra should be a continuous band. Rutherford could not explain this.

Formative Worksheet Statement A: Size of the nucleus is very small as compared to size of the atom. Statement B : Almost all the mass of the atom is concentrated in the nucleus. 1) Statement A is correct , statement B is incorrect 2) Statement A is incorrect and statement B is correct 3) Both the statements are correct 4) Both the statements are incorrect 10. The nuclear radius is of the order 10–13 cm while atomic radius is of the order 10–8 cm. Assuming the nucleus and the atom to be spherical, what fraction of atomic volume is occupied by the nucleus? 11. A moving particle should radiate energy continuously; likewise a revolving electron should also radiate energy continuously and fall into the nucleus and the atom should collapse. Explain. 9.

Conceptive Worksheet

65 18. What are the other names for Rutherford’s atomic model? 19 Rutherford’s atomic model is also called nuclear model and solar model. Give reason. 20 What was the purpose of ZnS in Rutherford’s gold foil experiment? 21. What is the composition of  – rays? 22. Name the scientist who predicted the presence of neutrons inside the nucleus.

9. Quantum Theory In 1900, a German scientist, Max Planck proposed a theory to explain the energy absorbed or emitted by light (an electromagnetic radiation). This theory is the main basis for Bohr’s atomic model. The main points of the quantum theory are: i) Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy. ii) The smallest packet of energy is called quantum. In the case of light the quantum is known as photon. iii) The energy of a quantum is directly proportional to the frequency of the radiation. E   (or) E = h  where ‘  ’ is the frequency of radiation and h is Planck’s constant having the value 6.626 × 10–27 erg-sec or 6.626 –34 × 10 J-sec. iv) A body can radiate or absorb energy in whole number multiples of a quantum i.e., h  , 2h  , 3h  ….nh  where ‘n’ is the positive integer.. Note: For any electromagnetic radiations,

13. How did Rutherford explain the stability of the atom hc E  h  ?  14. What are the drawbacks of Rutherford’s atomic Since all electromagnetic radiations move with same model? speed. i.e., speed of light in air 15. J.J Thomson’s atomic model is also called (3 × 108 ×m/s) 1) Water melon model 10. Bohr’s atomic model 2) Apple pie pudding model In 1913, Neils’ Bohr presented a model of the 3) Nuclear model atom, called Bohr model. This model was the first 4) Solar model in the series of modern concept which explains 16. When alpha particles are sent through a thin metal many properties of an atom. It is based on certain foil, most of them go straight through the foil. This assumptions (usually called postulates) which we is because: shall discuss now. 1) Alpha particles are much heavier than electrons. Important Postulates 2) Alpha particles are positively charged. i) Circular Orbits: An atom consists of a dense 3) Most part of the atom is empty space. nucleus situated at the centre with the electron 4) Alpha particles move with high velocity. revolving round in circular paths called 17. What is the approximate thickness of the golden circular orbits. foil used in  - rays scattering experiment? 1) 1m 2) 0.05 cm 3) 0.0004 mm 4) 0.4 mm www.betoppers.com

9th Class Chemistry

66 ii) Stationary Orbits: As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence these orbits are called stationary orbits or stationary states. iii)Energies of the Orbits:

v)Amount of energy emitted or absorbed: Since the excited state is less stable, an atom will lose its energy and come back to the ground state. Energy absorbed or released in an electron jump,

 E  is given by E  E 2  E1  h

Each stationary state is associated with a definite amount of energy and it is also known as energylevels. The greater the distance of the energy level from the nucleus, the more the energy associated with it. The different energy levels are numbered as 1, 2, 3, 4, (from nucleus onwards) or K, L, M, N etc. Orbits

Where E 2 and E1 are the energies of the electron in the two different energy levels, and  is the frequency of radiation absorbed or emitted. vi) Quantization of angular momentum: Of the finite number of circular orbits possible around the nucleus, and an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor h / 2 .

Nucleus N M L K

Electrons

iv)Energy emissions or absorption: Ordinarily an electron continues to move in a particular stationary state without losing energy. Such a stable state of the atom is called as ground state or normal state. If energy is supplied to an electron, it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2, 3, 4, etc.) by absorbing one quantum of energy. This new state of electron is called as excited state. On the other hand, energy is emitted when an electron jumps from higher orbit to lower orbit. This is called de-excitation. n=2 +energy

energy is absorbed +

Absorption of energy during excitation n=2 energy

energy is released +

Release of energy during de-excitatior

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nh 2 where, m = mass of the electron v = velocity of the electron n = orbit number in which electron is present r = radius of the orbit mvr 

1 2 3 4

Formative Worksheet 12. Find the ratio of energies of two radiations, one with a wavelength of 400 nm and the other with that of 800 nm. 13. Find the number of photons of light with wavelength 4000pm that provide 1J of energy. 14. 3 × 1018 photons of a certain light radiation are found to produce 1.5 J of energy. Calculate the wavelength of light radiations. (h = 6.63 × 10–34 Js). 15. An electron cannot occupy an intermediate orbit between n = 2 and n = 3. Give reason. 16. Find the angular momentum of first orbit of hydrogen atom. 17. Find the momentum and kinetic energy of an electron revolving in the first orbit of hydrogen atom, whose radius is 0.529Å. 18. The radii of first and second orbit of Helium atom are r1 and r2 respectively. The ratio of the velocities of the electrons revolving in their orbits is _____. 19. 1 S.I. unit of ‘h’ = ____________C.G.S. units of ‘h’ .

Atomic Structure

Conceptive Worksheet

3)

L

M

N

Size of the orbits

K

M

N

11. Atomic number number K

4)

K L

Size of the orbits

2)

L

M

N

K L

M

N

Size of the orbits

1)

Size of the orbits

23. Name the scientist who introduced the concept of orbits. 24. What are the names given by Neils’ Bohr to 31. different orbits of an atom ? 25. Why are Bohr’s orbits called Stationary orbits? 26. Theoretically, the number of orbits that are possible 32. for the electron to revolve in fixed paths is: 33. 1) 10 2) 1010 3) 10100 4) infinite 27. Which of the following is true regarding the size of the orbits? 34.

67 orbit radiates energy in the form of electromagnetic radiations containing photons. If E2 and E1 are the energies of higher and lower orbits, respectively, find the frequency of radiation. If ‘n’ is the no. of photons needed to generate an energy equivalent to ‘E’, find the wavelength of each photon. Find the total energy equivalent to 1000 photons whose wavelength is 106 A0. Calculate the ratio of energies of two radiations whose wavelengths are 6000A0 and 12000A0 respectively. Find the wavelength of a photon whose energy is equal to 931.5 MeV (MeV is million electro volt and 1eV = 1.602 × 10–19 J)

28. Can an electron shift its position from one orbit to another. If so, how? 29. What is the energy of ‘n’ quanta whose frequency is ‘ ’? 30. An electron jumping from a higher orbit to a lower

Element

Symbol

Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon Potassium Calcium

H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

Atomic number (Z) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Mass number (A) 1 4 7 9 11 12 14 16 19 20 23 24 27 28 31 32 35 40 39 40

and

mass

The atomic number, or the proton number, of an element is the number of protons present in the nucleus of an atom of the element. The sum of the numbers of protons and neutrons in an atom is known as the mass number of the atom. You also know that an atom is electrically neutral and so it has the same number of electrons as protons. So, in a neutral atom, the number of electrons is equal to the number of protons (Z) and the number of neutrons = A – Z. The numbers of fundamental particles in atoms with atomic numbers 1 to 20 are given in the following table

No. of electrons

No. of protons

No. of neutrons

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 2 4 5 6 6 7 8 10 10 12 12 14 14 16 16 18 22 20 20 www.betoppers.com

Nuclide symbol

9th Class Chemistry

68

The nuclide symbol of an atom is the symbol of the element with its atomic number as the subscript and mass number as the superscript, which are set to the left of the symbol of the element. The nuclide symbol is expressed as For example, the nuclide symbol

A Z

35 17

X.

Cl represents a chlorine atom, whose atomic number is 17 and mass

number is 35. You can immediately guess that there are 17 electrons, 17 protons and 35 – 17 (= 18) neutrons in the atom.

Formative Worksheet 20. Five atoms are labeled as follows:

Atoms A Z

A 40 20

D 19 9

C 7 3

L S 16 20 8 10

i) Find the ratio of their no. of neutrons ii) Identify the atoms in which

A 2 Z

iii) Identify the atoms in which

A 1 2Z

21. Except for the symbols of the elements, all the remaining information for the following elements is correct. Fill in the remaining blanks.

Symbol

Z

F A I T H

8

A

No. of Protons

No. of Electrons

40

12 6

C would be approximately 1) same 2) doubled 3) halved 4) reduced by 25% 26. The HD+ ion contains 1) 1 proton, 1 neutron, 1 electron 2) 2 protons, 1 neutron, 2 electrons 3) 2 protons, 2 neutrons, 0 electron 4) 2 protons, 1 neutron, 1 electron

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Name of the element

22 13 12 20

22. Find the number of neutrons present in di-positively charged zinc ion with mass number 70. 23. What is the ratio of number of neutrons in silicon to phosphorus? 24. Find the number of electrons, protons and neutrons present in: a) Nitrate ion b) Sulphate ion 25. Atoms consist of electrons, protons and neutrons. If the mass attributed to neutrons were halved and that attributed to the electrons were doubled, atomic mass of

No. of Neutrons

:

20

Conceptive Worksheet 35. An atom has net charge of –1. It has 18 electrons and 20 neutrons. Its mass number is 1) 372) 35 3) 38 4) 20 36. The total number of protons present in all the elements up to Zn in the periodic table is 1) 300 2) 350 3) 465 4) 450 37. The atomic mass of lead is 208 and its atomic number is 82. The atomic mass of bismuth is 209 and its atomic number is 83. The ratio of neutrons/protons in the atom 1) higher of Pb 2) higher of Bi 3) same of both 4) None of these 38. Atomic number is always equal to the number of electrons present in an atom. True / False. Explain. 39. Identify the element whose atomic number and mass number are same. 40. The atoms of all the elements contain electrons, protons and neutrons. True / False. Explain. 41. Name a few elements whose mass number is double their atomic numbers.

Atomic Structure

69

42. The mass number of an element is 23 and atomic number is 11. Find the number of protons, electrons and neutrons respectively present in the atom . 43. Why does atomic number not change during a chemical reaction? 44. The mass number of an element is 52 and its atomic number is 24. Find the number of electrons, protons and neutrons in an atom of this element . 45. Which of the following statements is not correct about calcium atom? 1) The number of electrons is same as the number of neutrons. 2) The number of nucleons is double the number of electrons. 3) The number of protons is half the number of neutrons 4) The number of nucleons is double the atomic number. 235 46. If three neutrons are added to the nuclei of 92 U the new particle will have an atomic number of

1) 89

2) 95

3) 90

4) 92

12. Electronic Configuration The arrangement of electrons in the different shells of an atom is called electronic configuration. The electron shells of an atom are not filled arbitrarily. The number of electrons in a shell follows a set of rules called the Bohr-Bury rules. Bohr-Bury Rules Of the various Bohr-Bury rules, we need to know the following in order to write the electronic configuration of elements up to calcium (Z = 20). 1. The maximum number of electrons that can be accommodated in a shell is given by 2n2 , where n is the orbit number or shell number. For example, n = 1 denotes the first (K) shell, and so on. The K, L, M, N,... shells can accommodate a maximum of 2, 8,18, 32,... electrons respectively.

3. The penultimate shell (last but one) cannot have more than 18 electrons. Applying Bohr-Bury rules By now, it should be clear to you that the only electron in a hydrogen atom (lH) occupies the K shell, and so do the two electrons in a helium atom (2He). As the K shell cannot have more than two electrons (Rule 1), the third electron of the lithium atom (3Li) must go to the next shell, i.e., the L shell. So, the arrangement of electrons in a lithium atom can be shown as K L. However, there is a 2 1

convention that the names of the shells are not mentioned in electronic configuration. The order in which the numbers of electrons are mentioned indicate the order of the shell, i.e., K, L, M, N,... respectively.

13. Geometric representation This representation helps us to know the number of electrons present in each orbit round the nucleus. In this representation, the nucleus is shown at the centre with the no. of protons and neutrons it contains. The orbits are shown in form of circles round the nucleus. The circle nearest to nucleus represents first orbit. The next immediate circle represents the second one and so on. Based on electronic configuration, the electrons are placed in the respective orbits. The electrons are shown with small circles ( ). Now, let’s see the geometric representation of oxygen (8O16) Mass number of oxygen (A) = 16 ; Atomic number of oxygen (Z) = 8  Number of protons = Z = 8  Number of electrons = No. of protons = 8

 Number of neutrons = A – Z = 16 – 8 = 8

And the electrons in the first shell = 2 [K-shell] Remaining electrons in second shell = (8 – 2) = 6 [L - shell] So, the geometric structure of oxygen atom is:

2. The outermost shell of an atom cannot contain more than 8 electrons in any case. A new shell is formed as soon as the outermost shell attains 8 electrons. www.betoppers.com

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Atomic Structure

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Atomic Structure

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Formative Worksheet

49. Write the electronic configuration of Mg2+ and Cl–

50. An element has 2 electrons in the M shell. What is 27. Write the electronic configuration of phosphorus. the atomic number of the element? 28. Write the electronic configuration of calcium. 51. Find the number of 'M' electrons in 13Al27. 29. According to Bohr–Bury rules The electronic configuration of a tripositively charged ion (X+3) is 2, 8. 14. Radioactivity a) Identify the element and find the number of After the discovery of X-rays, many scientists were neutrons in it. fascinated and were doing further studies on it. One b) Write the formula of the compound that it such scientist was Henri Becquerel . He knew that forms with chlorine and oxygen respectively. Rontgen had discovered X-rays with the property 30. The electronic configuration of a di-positive metal exhibited by X-rays called fluorescence. (When high ion M 2+ is 2, 8, 14 and its mass number is 56. Find frequency electromagnetic radiation like X-rays fall the number of neutrons in its nucleus. on certain substances they glow. This is called 31. A tripositively charged ion of an element 'X' has fluorescence) the same number of electrons as in trinegatively –3 Henri Becquerel wondered if the reverse might also charged N . Identify 'X'. be true. Could fluorescent bodies produce X32. Assume that the formula to calculate the maximum number of electrons has been changed from 2n2 to rays? To check this, he started experiments with 3n3. If an orbit can accommodate a maximum of certain Uranium salts which exhibit fluorescence 81 electrons, then identify the orbit number. when exposed to sunlight. He took the Uranium 33. Match the following: salt and wrapped it in a photographic plate in black Element Electronic configuration paper. He then placed the Uranium salt on the black p) Potassium 1) 2, 5 paper and exposed it to sunlight. q) Nitrogen 2) 2, 8, 8, 1 He thought that if X-rays were emitted during the r) Chlorine 3) 2, 8, 7 fluorescence of Uranium salts, they should 34. Which of the following species has the same number penetrate into the black paper (only X-rays have of electrons in the outermost and penultimate shells? 2– + this capacity to pass through black paper) and affect a) O b) K c) Al3+ d) F– the photographic plate. Photographic plates were 35. Find the maximum number of electrons that can be affected, indeed. Becquerel thought that he had accommodated in M-shell. discovered the means of producing X-rays from 36. Name the shell that can accommodate 32 electrons. uranium salts through fluorescence. 37. Find the total number of electrons that can be To confirm this, he repeated the experiment. But it accommodated in 7th orbit. was a cloudy day and the sun was not seen. He 38. The total number of electrons in NO3 ion is kept the whole setup in his drawer. As the cloudy 1) 152) 32 3) 314) 46 days continued, he lost his patience and decided to test the photographic plate. onceptive orksheet To his surprise, the photographic plate was affected 47. Write the electronic configuration of the following (exposed) in spite of the absence of sunlight and fluorescence from Uranium salts. Some different elements: 35 type of radiation, which is not produced by i) 17 ii) 12 Cl 6 C fluorescent substance, was passing through the 39 31 iii) 19 iv) 15 K P black paper and effecting the photographic plate. 48. The electronic configuration 2, 8, 8, 2 represents Becquerel made an exciting discovery purely by the element accident as often happens in science. Later these 1) Argon 2) Potassium radiations were called “radioactive rays” by Madame Curie. 3) Calcium 4) Chlorine

C

W

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Atomic Structure

Nature of radioactive radiations Rutherford and Villard proved that radioactive radiations are of three kinds. Rutherford performed the following experiment to observe the nature of radioactive rays. A small piece of radium was placed in a thick lead box with a small opening. The rays emitted by radium were passed through a slit to get a thin beam. This beam of radioactive radiations was then passed between the positive and negative plates of a strong electric field. The beam was found to split up into 3 parts showing the existence of three kinds of radiations. One type of radiation bent towards negative plate of electric field showing that it has positive charge. These positively charged rays are called alpha    rays.

75

 -particles:  -particles consist of fast moving electrons. It is interesting to note that these electrons are not the ones in the orbit around the nucleus, but are the electrons which have originated inside the nucleus. So,  -particles are the electrons from the nucleus. We have been reading that the nucleus consists of protons and neutrons. Then how can, during  emission, electrons be emitted from the nucleus? A neutron in the nucleus splits into a proton and an electron as follows:n = p + + e-Thus a neutron converts itself into a proton and emits and electron during  - emission.  - radiations:  - rays are pure electromagnetic radiations like light, infrared rays etc.  - rays consist of high energy photons (energy packets) and have highest frequency among the electromagnetic radiations

Cause for radioactivity As the atomic number and mass number increases, the size of the nucleus also increases. The increase in size leads to weakening of nuclear force and the repulsive force starts playing a prominent role. In the element with atomic number Z = 83 (Bismuth) the repulsive force just exceeds the attraction nuclear force, making the nucleus unstable. The second type of radiation bent towards positive plate showing that it has negative charge. These negatively charged rays are called beta    rays. The third type of radiation was not deflected by electric field showing that it has no charge, i.e., it is neutral. These neutral rays are called gamma   

Emissions start occurring from such unstable nucleus and they try to attain stability by emitting particles from inside it. This process of emission is called radioactivity. Definition: The spontaneous disintegration of the

rays. Thus, radioactive radiations are of three types.

nuclei with the emission of certain particles and

These are called alpha    rays, beta    rays

radiations is called radioactivity.

and gamma    rays.

Composition of radioactive emission  -particles: This is same as helium nucleus and is represented as 2He4. The atomic number (Z) is 2 and its mass number (A) is 4.  -particles consist of two protons and two neutrons.

The word ‘spontaneous’ is used because we are not sure of the moment at which the emission occurs from the nucleus. For a given radioactive nucleus it can happen after this moment or a moment later or after a million years. Radioactive emission has three types of radiations – ,  and  .

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Properties of , β and  rays Property

  rays

  rays

  rays

Nature

Alpha rays are made up of a positively- charged alpha particles. An alpha particle has 2 units positive charge and 4 units mass just like nucleus of helium atom. Thus an alpha particle is helium nucleus. It is represented as  or 42 He

Beta rays are made up of a negatively- charged beta particles.  particle has one unit negative charge and negligible mass just like an electron. Thus beta particle is, in fact, an electron. It is represented as  or 10 e .

Gamma rays are electromagnetic waves of short wavelength. Since gamma rays are not made up of particles, they have no charge and no mass. They are represented as .

Velocity

  particles are given out by unstable nuclei at a very high velocity of about 3 × 109 cm/sec. Thus, speed of an   particle is 1/10 of that of light.

  rays are emitted at a greater speed. The velocity of   rays is almost 3 × 1010 cm/s.

  rays have velocity equal to that of light.

Penetrating power

Alpha particles have the least penetrating power. They can be stopped by even a thin layer of solid matter. Alpha particles can penetrate through a 0.002cm thick aluminium sheet.

Penetrating power of   rays is 100 times greater than that of   rays. They can penetrate through a 0.2cm thick aluminium sheet.

  rays have high penetrating power. They can penetrate through a 100cm thick aluminium sheet.

Ionising power

Alpha rays ionize gases. Their ionizing power is 100 times greater than that of   rays and 10,000 times greater than that of   rays.

  rays ionize gases. Their ionizing power is 100 times less than that of   rays but 100 times greater than that of   rays.

  ray ionize gases but their ionizing power is much less. It is 100 times less than that of   rays.

Effect of photographic plate

  particles affect photographic plate. They darken it feebly.

  particles affect photographic plate. They darken it appreciably.

  rays affect photographic plate. They darken it deeply.

Effect of electric field

In an electric field,   rays are deflected towards negative plate.

In an electric field   rays are deflected towards positive plate.

 rays are not deflected in an electric field.

15. Isotopes We know that atoms of given elements are similar. But later studies have proved that, for a few elements this similarity is confined to only number of protons and electrons and they differ in number of neutrons. Such atoms are called isotopes. Isotopes are atoms of the same element having the same atomic number but different mass numbers. Isotopes of an element have same number of protons but differ in the number of neutrons in their nuclei. www.betoppers.com

Atomic Structure

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For example the two isotopes of chlorine atoms contain 17 protons. So the atomic number of all the chlorine atoms is 17. But they contain different number of neutrons i.e., one atom has 18 neutrons and another atom has 20 neutrons. Hence they have different mass number of 35 (17 + 18) and 37(17 + 2. 20) the isotopes of chlorine can be written as 17Cl35 and 17Cl37. The complete composition of the two isotopes of chlorine is given below

Examples of Isotopes: Isotopes 35 17 Cl 37 17

1.

12 6

Protons 17

Neutrons 18

Electrons 17

17

20

17

Cl

Isotopes of Hydrogen: The hydrogen element has three isotopes having the same atomic number 3. of 1 but different mass numbers of 1, 2 and 3 respectively. The three isotopes of hydrogen can be represented as: 1 1

should be noted that the ordinary hydrogen isotope (protium) does not contain any neutron; the heavy hydrogen isotope (deuterium) contains 1 neutron; whereas the very heavy hydrogen isotope (tritium) contains 2 neutrons. Isotopes of Carbon The carbon element has three isotopes having the same atomic number of 6, but different mass numbers of 12, 13 and 14. The three isotopes of carbon can be written as:

H, 12 H and 13H

The three isotopes of hydrogen, 11 H, 12 H and 13H , have been given special names like protium, deuterium and tritium respectively i) Protium is the ordinary isotope of hydrogen with mass number 1. Protium is represented as 11 H . Protium does not have a special symbol. ii) Deuterium is a heavy isotope of hydrogen with mass number 2. Deuterium is represented as 2 1

H . The special symbol of deuterium is D. iii)Tritium is a very heavy isotope of hydrogen with mass number 3. Tritium is represented as

14 C, 13 6 C and 6 C These three isotopes of carbon contain 6 protons and 6 electrons each, but they contain an unequal number of neutrons. The C – 12 isotope contains 6 neutrons, C – 13 isotope contains 7 neutrons, whereas the C – 14 isotope contains 8 neutrons. Isotopes of Oxygen The oxygen element has three isotopes: 16 8

18 O, 17 8 O and 8 O All the isotopes of oxygen have the same atomic number of 8, but they have different mass numbers (or atomic masses) of 16, 17 and 18 respectively.

Chemical properties of isotopes The chemical properties of an atom of the element depend on the number of protons and electrons, but not on the number of neutrons. Since all the isotopes of an element contain the same number of protons and electrons, they have identical electronic configuration having the same number of valence electrons. Hence, isotopes have similar chemical properties. For example, the two isotopes of chlorine,

Deuterium

2 1

H

1

1

1

37 Cl and 17 Cl , both have the same number of 17 electrons in them due to which both of them have the same electronic configuration of 2, 8, 7. Since both the isotopes of chlorine, Cl – 35 and Cl – 37 have identical electronic configurations (having the same number of 7 valence electrons), they show identical chemical properties. The physical properties of an element depend on the mass of the atom and its mass number. The isotopes of an element have different mass number due to difference in number of neutrons. Hence, isotopes differ in their physical properties. For example, the two isotopes of chlorine,

Tritium

3 1

H

1

2

1

35 17

3 1

H . The special symbol of tritium is T. Thus, we can now say that hydrogen element has three isotopes: protium, deuterium and tritium, having the same atomic number of 1, but different mass number of 1, 2 and 3 respectively. The complete composition of the three isotopes of hydrogen is given below: Name Protium

Isotope Protons 1 1 1H

Neutrons 0

Electrons 1

It is clear from the above table that all the isotopes of hydrogen contain 1 proton and 1 electron each, but they contain 0, 1 and 2 neutrons respectively. It

35 17

37 Cl and 17 Cl , have slightly different physical properties because they have slightly different atomic masses of 35 and 37 respectively.

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16. Isobars Observe the following atoms of different elements: 17 17 17 7 N , 8O , 9 F

In the above atoms, the atomic number is different, but their mass number is same. This means, number of protons in the atoms are different, but the total number of nucleons i.e., sum of protons and neutrons number is same. So these nuclides are called isobars. Isobars are the nucleus with same mass number but different atomic number. Isobars have different number of protons and neutrons, but the sum of protons and neutrons i s same. Isobars differ in all except in mass number. Some more examples of isobars: i)

40 18

Ar,

40 20

32 ii) 15 P,

Ca

32 16

S

17. Isotones Let us understand by observing the number of neutrons in the following species: 17 7

N,

18 8

O,

19 9

F,

Species 17 7N 18 8O 19 9F 20 10Ne

20 10

Z

Ne A 17 18 19 20

n=A-Z 17 – 7 = 10 18 – 8 = 10 19 – 9 = 10 20 – 10 = 10

7 8 9 10 We observe that in the above set of species, the number of neutrons are same. Such species are called isotones. Isotones are nuclides with different mass numbers and atomic numbers but have same number of neutrons. Example: i)

30 14

Si,

31 15

P,

32 16

S (number of neutrons - 16)

18. Isoelectronic species How many electrons are present in the species of the following set? N–3, O2–, F–, Ne, Na+, Mg+2 Let us find the same. Species Number of electrons N–3 7 + 3 = 10 2– O 8 + 2 = 10 F– 9 + 1 = 10 Ne 10 + 0 = 10 Na + 11 – 1 = 10 Mg +2 12 – 2 = 10

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We observe that, each of the species in the above set contain same number of electrons i.e., 10. Such set of species are called isoelectronic species. The species of different elements or compounds containing the same number of electrons are called isoelectronic species. Some more examples of isoelectronic species: i) H–, He, Li+, Be+2 (each of the species contains 2 electrons) ii) P–3, S2–, Cl–, Ar, K+, Ca+2 (each of the species contains 18 electrons)

Formative Worksheet 39. Select the isotopes, isobars and isotones from the following: (Note: the symbols are hypothetical) A1, 3D7, 1C3, 6L12, 5B12, 4E8 6S14, 2T4 1 40. The ion that is iso-electronic with CO is: 1) O2 2) N2 3) CN– 4) O2 . 41. Which of the following triads represents isotones? a)

12 6

14 C, 13 6 C, 6 C

b)

40 18

42 43 Ar, 20 Ca, 21 Sc

40 40 41 16 18 c) 18 Ar, 20 Ca, 21 Sc d) 14 7 N, 8 O, 9 F 42. X3– is isoelectronic with argon. It has electrons and neutrons in 1 : 1 ratio. The mass number of ‘X’ is 1) 302) 323) 33 4) 35 43. The set of nuclei that are isotonic is 1) 6C12, 7N14, 8O16 2) 6C14, 7N15, 8O16 3) 6C14, 7N14, 8O18 4) 6C14, 7N15, 8O18 44. Among the following groupings which represents the collection of isoelectronic species?

1) NO , C22 , O 2 , CO

2) N2 , C22 , CO, NO

3) CO, NO , CN  , C22

4) NO, CN  , N 2 , O 2

Conceptive Worksheet 52. Identify the element which has three isotopes with the neutrons, 6, 7, 8 respectively. 1) Nitrogen 2) Carbon 3) both (1) and (2) 4) Neon 53. The species having same number of protons are called: 1) Isobars 2) Isotones 3) both (1) and (2) 4) Isotopes 54. Mass numbers of atoms of an element are 1, 2, 3, respectively. Identify the element. 1) Beryllium 2) Lithium 3) Hydrogen 4) Helium 55. The isotopes have : 1) same mass number 2) same valency 3) same number of protons 4) same number of electrons

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56. Which of the following is true for isotopes? 1) They have same mass number 2) They have same number of protons 3) They have same number of neutrons 4) Both (1) and (2) 57. Most elements have fractional atomic masses because 1) they have isotopes 2) their isotopes have different masses 3) their isotopes have non-integral masses 4) the constituent neutrons, protons and electrons combine to give fractional masses. 58. Species Protons Neutrons Electrons A B

17 17

18 20

17 17

A and B are: 1) Isotopes 2) Isobars 3) Isotones 4) None 59. Isotopes have : 1) Similar chemical properties and physical properties. 2) Dissimilar chemical properties and physical properties. 3) Similar chemical properties and dissimilar physical properties. 4) Similar physical properties and dissimilar chemical properties. 60. Which of the following statements is not correct? 1) Isotones are atoms of different elements having same number of neutrons. 2) Isotopes are atoms of different elements having same number of nucleons. 3) Isobars are atoms of different elements having same number of nucleons. 4) Isotones and isobars are atoms of different elements. 61. Which of the following statements is not correct? 1) 7N14 and 7N15 are isotones 2) 7N14 and 6C14 are isobars 3) 6C13 and 6C14 are isotopes 4) 6C13 and 7N14 are isotones 62. An isotone of

76 32

Ge is

77 77 77 1) 32 Ge 2) 33 As 3) 34 Se 4) 81 36 Kr + 63. Na ion is isoelectronic with 1) Li+ 2) Mg2+ 3) Ca2+ 4) Ba2+ 64. Which pair is isoelectronic ? 1) Ar and Cl 2) Na+ and Ne + 3) Na and Mg 4) Mg and Ne

Summative Worksheet 1.

2.

3. 4. 5. 6. 7. 8.

9.

Identify the following from their description. A) These are the rays produced in the discharge tube. These rays deflect towards the +ve plate in an electric field, and contain the universal constituent of matter. B) This is a subatomic particle present in all the atoms except Hydrogen. C) These are rays used by James Chadwick and Rutherford for their important experiments to understand the structure of an atom. Find the number of electrons whose mass is equal to the total mass of neutrons present in one calcium atom. Why is Rutherford’s atomic model called nuclear model, Solar model respectively? What is the purpose of lead block and ZnS in Rutherford’s gold foil experiment? ‘x’ is the unit used to measure the atomic particles and ‘y’ is the S.I. unit of mass. Find y/x. What is the amount of charge in coulomb’s present on stable oxide ion? 1 Coulomb = ________ electrons. An atom has mass number A and atomic number Z: a) How many protons are present in the nucleus? b) How many electrons revolve around the nucleus? c) How many neutrons are present in the nucleus? The electronic configuration of a tripositively charged ion ( X+3 ) is 2,8. Then a) Identify the element and find the number of neutrons in it. b) Write the formula of the compound that it forms with chlorine and oxygen respectively. c) Name the element whose dinegatively charged

ion is isoelectronic with X 3 . d) What is the valency of ‘X’ ? Name the group in periodic table to which it belongs. e) Find the number of ‘M’ electrons present in its neutral atom. 10. What is the formula of the compound formed by P and Q if the number of electrons present in their outermost shells ( M-shell ) is 3 and 7 respectively ? 11. Why does atomic number not change during a chemical reaction?

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80 12. The ratio of mass of one atom of nitrogen to its atomic weight is x. The ratio of mass of one atom of calcium to its atomic weight is y. Find x / y. 13. What is the charge present on an atom containing 8 protons, 10 neutrons and 10 electrons ? 14. The atomic number of elements P, Q and R are 7, 11, 15 respectively. State with a reason which two elements are likely to have same chemical properties. 15. 1) The valence electrons in three elements R, A, J was found to be 1, 2, 3 respectively. Then, predict the nature of R, A, J . 2) The valence electrons in four elements A, R, T, I was found to be 4, 5, 6, 7 respectively. Then, predict the nature of A, R, T, I. 3) Identify an element which has one electron in K-shell but is a non-metal. 4) Identify an element that has two electrons in K-shell (other shells are empty, but is a noble gas.

HOTS Worksheet 1.

2.

The symbol of an element is

16

X 32 .

i) How many protons, electrons and neutrons are present respectively ? ii) How many nucleons are present ? iii) What is the total mass of electrons, protons and neutrons present in 1 atom of it ? iv) How many atoms of X are present in 32 grams of X ? v) Write the electronic configuration of X. vi) How many valence electrons are present in X? vii) What is the valency of X? viii) Write the formula of the compound formed by the combination of ‘X’ and ‘O’. ix) How many electrons are present in L shell of X ? x) Identify the element. Atomic numbers of elements with hypothetical symbols A, D, C, L, S are 3, 9, 11, 17, 19 respectively. i) Identify set of elements which have similar chemical properties. ii) Which set is likely to have metallic character? iii) Which set is likely to have non-metallic character ? iv) Write the formula of the compound formed by the combination of E and A. v) Write the formula of the compound formed by the combination of atoms of (a) D (b) L

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Three elements S, A, I have atomic numbers 4, 12, 20 respectively. i) State the number of valence electrons present in each of the elements. ii) Are these elements going to have similar or dissimilar properties? iii)Are the elements, metals or non-metals? iv)Name the groups to which they belong. 4. Find i) the total number of electrons, protons, neutrons and ii) total mass of electrons, protons and neutrons present in 0.12 mg of carbon. 5. Two elements A and B contain equal number of protons , electrons and neutrons. B contains double the number of electrons, protons and neutrons than in A. A contains 8 electrons in L-shell, B contains 2 electrons in N-shell. Identify A and B. 6. In an element ‘X’ , 5th electron of 3rd shell is its last electron. In another element ‘Y’ , 7th electron of 3rd shell is its last electron. Then 1) Identify ‘X’ and ‘Y’. 2) Write the formula of the compound formed by the combination of ‘X’ and ‘Y’. 3) What is the atomicity of ‘X’ and’ ‘Y’? 4) Find the ratio of number of neutron of ‘Y’ to ‘X’. 7. Which of the following statements is false ? Rewrite it correctly. a) Mass of a neutron is more than mass of a proton. b) There is no difference between atomic mass and mass number c) Isobars differ in the number of neutrons as well as protons. 8. Identify the following: A) A uni-negative ion whose configuration is 2, 8, 8 D) A di-positive ion whose configuration is 2, 8, 8 C) A di-negative ion whose configuration is 2, 8,8 L) A neutral atom whose configuration is 2, 8, 8 9. There are 10 electrons in each of the species A+1, D+2, C+3. If the number of neutrons in them are 12, 12 and 14 respectively, then choose the right option regarding their atomic number and mass number. 10. Which of the following statement(s) is(are) correct? 1) Isotopes are the atoms of same element. 2) Isobars are the atoms of different elements. 3) Isotones are the species having same number of neutrons. 4) Isoelectronic species contain same number of electrons. 1) A 2) A, B 3) A, B, C 4) All of these

Atomic Structure 11. In which of the following, the ratio of number of 4. electrons present in K-shell to L-shell is 1:2? 1) Hydrogen 2) Beryllium 3) Carbon 4) Oxygen 12. The electronic configuration of a tripositively charged ion ( X+3 ) is 2,8. Then a) i) Identify the element. ii) Find the number of neutrons in it. b) Write the formula of the compound that it forms with chlorine and oxygen respectively. c) Name the element whose dinegatively charged

81

Rutherford’s experiment which established the nuclear model of the atom used a beam of 1)  - particles which impinged on a metal foil and got absorbed 2)  - rays which impinged on a metal foil and ejected electrons 3) Helium atoms, which impinged on a metal foil and got scattered 4) Helium nuclei, which impinged on a metal foil and got scattered ion is isoelectronic with X +3 . 5. A neutral atom of an element has a nucleus with a d) What is valency of ‘X’? nuclear charge 13 times and mass 27 times that of e) Find the number of electrons present in ‘M’ shell hydrogen nucleus. How many electrons would be of its neutral atom. in its stable positively charged ion ? (a) (b) (c) (d) (e) 1) 27 2) 14 3) 13 4) 10 i ii 6. An element M has an atomic mass 19 and atomic 1) Boron 10 BCl3 & B2O3 Oxygen 2 2 number 9; its ions are represented by 2) Aluminium 14 AlCl3 & Al2O3 Oxygen 3 3 1) M+ 2) M2+ 3) M– 4) M2– 3) Aluminium 13 AlCl3 & Al2O3 Fluorine 2 3 7. When alpha rays are sent through a thin metal foil, 4) Boron 9 BCl3 & B2O3 Nitrogen 3 2 most of them go straight through the foil because 13. In an element ‘X’ , 5th electron of 3rd shell is its last 1) alpha rays are much heavier than electron rays th electron. In another element ‘Y’, 7 electron of 2) alpha rays are positively charged 3rd shell is its last electron. Then 3) alpha rays move with high velocity 1) Identify ‘X’ and ‘Y’. 4) most part of the atom is empty 2) Write the formula of the compound formed by 8. Rutherford’s alpha ray scattering experiment the combination of ‘X’ and ‘Y’. eventually led to the conclusion that 3) What is the atomicity of ‘X’ and’ ‘Y’? 1) mass and energy are related 4) Find the ratio of number of neutrons of ‘Y’ 2) electrons occupy space around the nucleus to ‘X’. 3) neutrons are buried deep in the nucleus A 4) the point of impact with matter can be  1  A  2Z 2Z precisely determined 9. Which of the following is not isoelectronic? a b c d + X Y X Y 1) Na 2) Mg2+ 3) O2– 4) Cl– 1) Phosphorus Chlorine XY3 4 2 9:8 10. ‘Ca’ has atomic number 20 and atomic weight 40. 2) Phosphorus Chlorine XY2 3 2 7:8 Which of the following statements is not correct 3) Boron Fluorine XY3 4 2 9:8 about ‘Ca’ atom? 4) None 1) The number of electrons is same as the number of neutrons orksheet 2) The number of nucleons is double the number 1. Cathode rays are made up of of neutrons 1) positively charged particles 3) The number of protons is half the number of 2) negatively charged particles neutrons 3) neutral particles 4) The number of nucleons is double the atomic 4) none of these number 2. The value of Planck’s constant is –27 –27 1) 6.6256 × 10 erg sec 2) 66.256 × 10 erg sec 11. The radius of an atomic nucleus is close to 1) 10–2cm 2) 10–5cm 3) 10–8cm 4) 10–12cm 3) 6.02 × 10–15 erg sec 4) 3.01 × 10–23 erg sec 3. The electronic configuration 2, 8, 8, 2 represents 12. Which is not true with respect to cathode rays? 1) A stream of electrons 2) Charged particles the element 3) Move with same speed as that of light 1) argon 2) potassium 3) calcium 4) chlorine 4) Can be deflected by magnetic fields 5) Can be deflected by electric fields www.betoppers.com

IIT JEE W

82 13. Number of protons, neutrons and electrons in the element

231

X is 1) 89, 89, 242 2) 89, 142, 89 3) 89, 71, 89 4) 89, 231, 89 14. A neutral atom (atomic number > 1) has 1) electron and proton 2) neutron and electron 3) neutron, electron and proton 4) neutron and proton 15. The hydride ion is isoelectronic with 1) H+ 2) He+ 3) He 4) Be 16. Ratio of mass of proton and electron is 1) infinite 2) 1.8 × 103 3) 1.8 4) none of these 17. Which of the following statements is correct ? 1) the charge on an electron and on a proton are equal and opposite 2) neutron has no charge 3) electrons and protons have the same weight 4) the mass of proton and a neutron are nearly identical 18. Anode rays were discovered by 1) Goldstein 2) Rutherford 3) G.J.Stoney 4) J.J. Thomson 19. The charge on the atom having 17 protons, 18 neutrons and 18 electrons is 1) +1 2) –1 3) –2 4) zero 20. Which pair is isoelectronic? (AFMC 1983) 1) Ar and Cl 2) Na+ and Ne 3) Na+ and Mg 4) Mg and Ne 21. Maximum number of electrons in any orbit is 1) n2 2) 2n2 3) 1/2 n2 4) none of these 22. Experimental evidence for the existence of atomic nucleus comes from 1) Millikan’s oil drop method 2) Atomic absorption spectroscopy 3) The magnetic bending of cathode rays 4) Alpha rays scattering by a thin metal foil 23. 22Na contains 1) 22 protons 2) 11 neutrons 3) 22 neutrons 4) none of these 24. Which of the following atoms has no neutron in its nucleus? 1) Helium 2) Lithium3) Protium 4) Tritium 25. An element has atomic number 11 and mass number 24. What does the nucleus contain? 1) 11 protons and 13 neutrons 2) 11 protons, 13 neutrons and 13 electrons 3) 13 protons and 11 neutrons 4) 13 protons and 11 electrons 26. The number of electrons and neutrons of an element is 18 and 20 respectively. Its mass number is 1) 2 2) 17 3) 37 4) 38 www.betoppers.com 89

9th Class Chemistry 27. Which is different in isotopes of an element? 1) Atomic number 2) Mass number 3) Number of protons 4) Number of electrons 28. Na+ ion is isoelectronic with 1) Li+ 2) Mg+2 3) Ca2+ 4) Ba2+ 29. Neutron is a fundamental particle carrying 1) a charge of +1 and a mass of 1 unit 2) no charge and a mass of 1 unit 3) no charge and no mass 4) a charge of –1 and mass of 1 unit 30. The number of electrons in the atom which has 20 protons in the nucleus is 1) 20 2) 10 3) 304) 40 31. Number of electrons in the outermost orbit of the element of atomic number 15 is 1) 7 2) 5 3) 3 4) 2 32. The discovery of neutron became very late because 1) neutrons are present in nucleus 2) neutrons are chargeless 3) neutrons are fundamental particles 4) all 33. The mass of the neutron is of the order of 1) 10–23 kg 2) 10–24 kg 3) 10–26 kg 4) 10–27 kg 34. An element has atomic weight 2Z + 6 where Z is the number of extra nuclear electrons in the atom. Then the number of neutrons in the nucleus is ___. 35. Cathode rays are 1) electronegative waves 2) radiations 3) stream of  -particles 4) stream of electrons 36. An atom of Mn, atomic number 25 and atomic weight 55 contains in its nucleus 1) 55 protons 2) 55 neutrons 3) 25 neutrons 4) 25 protons 37. Which is the correct statement about proton? 1) proton is nucleus of deuterium 2) proton is  particle 3) proton is ionized hydrogen molecule 4) proton is ionized hydrogen atom 38. Number of neutrons is heavy hydrogen atom is __. 39. Neutron was discovered by 1) Rutherford 2) Langmuir 3) Chadwick 4) Austin 40. As we move away from nucleus, the energy of orbit 1) decreases 2) increases 3) remains unchanged 4) none of these 41. The mass of atom is constituted mainly by 1) neutrons and neutrions 2) neutrons and electrons 3) neutrons and protons 4) protons and electrons 42. The electronic configuration of elements A, B, C and D are (2, 8, 1) (2, 8, 2) and (2, 8, 7) respectively. Which of them can make an ion with two negative charges. 1) A 2) B 3) C 4) D

By the end of this chapter, you will understand

1.

•

Earlier attempts in classifying the elements

•

Major contributions leading to the development of modern periodic table

•

Mendeleev’s Periodic classification

•

Moseley’s modification and Modern periodic table

•

IUPAC Nomenclature of elements with atomic numbers > 100

• • • • • • • •

Introduction You must have visited a library. There are thousands of books in a large library. In spite of this if you ask for a particular book, the library staff can locate it easily. How is it possible? In a library, the books are classified into various categories and sub-categories. They are arranged on shelves accordingly. Therefore location of books becomes easy. Let us come back to chemistry. Most of the matter that we see, touch and feel is made up of compounds. There are millions of such compounds existing presently. You will be surprised to know that compounds are formed as a result of various permutations and combinations of only about 110 odd elements. To study properties of these elements and their compounds individually is a tremendous 2. task. How then was this task simplified? i) This task was simplified by simple classification of elements into a few groups. Instead of studying each and every element or compounds, we just learn the properties of groups. The attempts were made by different scientists to classify elements ii) based on their properties. Necessity for classification of elements Following are the reasons for the classification of elements. 1. The classification may help to study them better. 2. The classification may lead to correlate the properties of the elements with some fundamental property that is characteristic of all the elements. 3. The classification may further reveal relationship between the different elements.

Merits of the long form of periodic table Anomalies of the Long form of Periodic table Periodicity and Periodic properties Atomic radius Ionisation energy Metallic character Non-metallic character Some interesting and important points

Chapter -4

PERIODIC CLASSIFICATION

Learning Outcomes

Origin of the periodic table Chemists would be overwhelmed with isolated pieces of information if they did not have some way of relating the facts they know about the more than 100 known elements. The Periodic Table provides a means of organizing information so that relationships among elements can be clearly seen and understood. Even before so many elements were known, scientists were searching for relationships among elements. Three important attempts to determine such relationships were made during the nineteenth century by the English physician William Prout, the German chemist Johann Döbereiner, and the English chemist John Newland. All three of these scientists based their work on the model of the atom proposed in 1803 by the English scientist John Dalton.

Earlier attempts in classifying the elements Greeks classification: The ancient Greeks erroneously suggested that all matter consisted of four elements only - Earth, air fire and water. But their idea could not be supported by the experiments.

Dalton’s contribution: The following are some important points of Dalton’s contribution. 1. All matter consists of simple bodies (elements) and compound bodies (compounds). The smallest part of a simple body is the atom. The smallest part of a compound body is the compound atom (later called the molecule). 2. All atoms of the same element have the same properties, such as shape, size, and mass. Atoms of different elements have different properties.

9th Class Chemistry

84 3. When matter undergoes chemical change, atoms of different elements either combine or separate from one another. 4. Atoms cannot be destroyed, even during chemical change. In 1808, John Dalton published the first list of atomic weights in his ‘Table of the relative atomic weights of the ultimate particles of gaseous and other bodies’. He changed chemists’ ideas from a qualitative to a quantitative basis, and started the chemical revolution during the 19th century.

v) Lavoisier’s classification: By the late 1860’s, more than 60 chemical elements had been identified. Based on similar physical and chemical properties, Lavoisier and early chemists classified the elements into metals and non-metals. 1. The elements which were malleable and ductile, good conductors of heat and electricity and possessed characteristic metallic lustre were named as metals. 2. The elements which were brittle, bad conductors of heat and electricity and did not possess metallic lustre were named as non-metals. Certain elements such as antimony, arsenic, boron, silicon and tellurium resembled metals in some respects and non-metals in certain respects and were therefore metalloids. Reasons for rejection: i) Some of the elements behave both as metals and non-metals. ii) The elements were divided only into two broad categories which does not help much in the study of elements.

iii) Prout’s Hypothesis In 1814, Prout suggested that all of the other elements are developed from hydrogen. In other words, Prout proposed that hydrogen is the fundamental element. Prout reached this conclusion because he observed that the masses of atoms are whole number multiples of the atomic mass of hydrogen. The notion of atomic mass as the relative mass of an atom on a scale where hydrogen has a mass of one unit grew out of this idea. Later, oxygen became the standard. Later still, the concept of atomic mass unit (amu), with carbon-12 as a standard, was developed. At the time, Prout’s idea seemed revolutionary. Now, it appears that Prout may have been close to the truth. Evidence from studies of radioactive changes led modern scientists to believe that all elements may be derived from hydrogen.

Formative Worksheet 1.

iv) Classification on the Basis of Valency Realising the importance of valency in chemistry, an attempt was made to classify elements on this basis. The monovalent elements were classed together and so were the divalent ones, the trivalent ones and so on. However, such classification suffers from the following drawbacks. 1. Several elements have variable valency, e.g., iron has a valency of 2 and 3, copper 1 and 2, tin 2 and 4, lead 2 and 4, etc. This makes the position of such elements uncertain. 2. Such classification does not explain the diverse nature of elements having the same valency. For example, both sodium and chlorine are monovalent, but they are quite different from each other in chemical behaviour. Sodium is a strongly electropositive metal whereas chlorine is a strongly electronegative nonmetal.

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2.

Match the following : Column – I Column – II (Classification) (Scientist) A) Earth, Air, i) Dalton’s Fire and Water classification B) Metal and ii) Lavosier’s non-metals classification C) Atomic iii) Greeks’ weight table classification State True or False: i) Classification of elements is not useful to reveal the relationship between the different elements. ii) Classification of elements may lead to correlate the properties of the elements with some fundamental property that is characteristic of all the elements.

Conceptive Worksheet 1.

2.

Which of the following is a correct statement? 1) Generally, metals have 1 to 3 electrons in their valence shell. 2) Generally, non-metals have 4 to 8 electrons in their valence shell. 3) Hydrogen is a metal. 4) Chlorine is a strong electropositive metal. The elements that are brittle, bad conductors of heat and electricity and do not possess metallic lustre were named as __________.

Periodic Classification 3.

4.

5.

85

Which of the following are the reasons for the 3 classification of elements? 1) The classification may help to study them better. 2) The classification may lead to correlate the properties of the elements with some 1. fundamental property that is characteristic of all the elements. 3) The classification may further reveal relationship between the different elements 4) All the above Which of the following are true based on the classification based on valency? 1) The monovalent elements were classed together. 2) The diavalent elements were classed together. 3) The trivalent elements were classed together. 4) All the above. The elements that are malleable and ductile, good conductors of heat and electricity and possess characteristic metallic lustre are named as

Major contributions leading to the development of modern periodic table Dobereiner’s Triads In 1817, Dobereiner noticed that certain groups of three elements have related properties. Dobereiner called these groups triads. If the elem ents of a triad are arranged in order of increasing atom ic m ass, the atom ic m ass of the m iddle elem ent is the average of the atom ic m asses of the other two elem ents.

The properties of the middle element of a triad are also approximately midway between the properties of the other two elements. Thus bromine is less reactive than chlorine but more reactive than iodine.

Examples of Dobereiner’s Triads Element Atomic weight Average of the atomic weights of the two extremes

Set-1 Li 7

Na K 23 39 7  39  23 2

Significance of Dobereiner Triads: This classification of elements in triads had greater significance in predicting the atomic mass and properties of the middle element. However, only a few elements could be arranged in such triads. Defects of Triad Classification: i) Quite a large number of similar elements could not be grouped into triads. Example: Iron, manganese, nickel, cobalt, zinc and copper are similar elements but cannot be placed in the triads. ii) It was possible that quite dissimilar elements could be grouped into triads.As Dobereiner failed to arrange all the known elements in the form of triads, his classification was not very successful. Example: For example, carbon (12), nitrogen (14) and oxygen (16) can form a triad but their properties are entirely different from each other.

2. Newland’s classification John Alexander Reina Newland was a chemist as well as a lover of music.He arranged many of the known elements in the increasing order of their atomic masses. It was noticed that the eighth element was similar in properties to the first element,

Set-2 Cl Br I 35.5 80 127 35.5  127  81.25 2

Set-3 Ca Sr Ba 40 87.5 137 40  137  88.5 2

just like the eighth note in music - Western as well as Indian. Western Indian Elements Do Sa Lithium Sodium Re Re Beryllium Magnesium Me Ga Boron Aluminium Fa Ma Carbon Silicon So Pa Nitrogen Phosphorus La Da Oxygen Sulphur Ti Ne Fluorine Chlorine Do Sa The eighth element after lithium is sodium. It is similar to lithium in many of its chemical properties. Similarly, the eighth element after sodium is potassium, whose properties are similar to sodium. The eighth element from fluorine is chlorine both of which are similar in their properties. The eighth element from nitrogen is phosphorus and both these elements are similar in properties. Based on this observation, Newland stated his law of octaves. When elements are arranged in increasing order of their atomic mass the eighth element resembles the first in physical and chemical properties just like eighth note on a musical scale resembles the first note www.betoppers.com

9th Class Chemistry

86 However, a very important conclusion was made 3. that there is some systematic relationship between the order of atomic masses and the repetition of properties of elements. This gave rise to a new term called Periodicity. It is the recurrence of characteristic properties of elements arranged in a table, at regular intervals of a period. Achievements of the Law of Octaves i) The law of octaves was the first logical attempt to classify elements on the basis of atomic weights. ii) The periodicity of elements was recognised for the first time. Defects of Law of Octaves i) This law could be best applied, only up to the element calcium. ii) The newly discovered elements could not fit into the octave structure. iii) It failed to exhibit this feature with heavier elements.

Lother Meyer classification In 1869, Lother Meyer, a German chemist, studied physical properties like atomic volume, melting point, boiling point etc. of various elements and plotted a graph between the atomic volumes (atomic volume of an element is the atomic mass divided by the density of the element) and atomic masses (in amu) of the elements. From the atomic volume-atomic mass curve shown in figure. Lother Meyer observed that the elements with similar properties occupy approximately similar positions on the curve.

This fact can be illustrated by studying the following examples: i)

The alkali metals (Li, Na, K, Rb) having the largest atomic volumes occupy the peaks on the curve.

ii)

The alkaline earth metals (Be, Mg, Ca, Sr) lie at about mid points on the descending positions on the curve.

iii)

The halogens (F, Cl, Br) occupy the ascending positions on the curve.

iv)

The transition metals occupy the troughs of the curve. On the basis of his observation, Lother Meyer proposed that the physical properties of the elements are a periodic function of their atomic masses (atomic weights). Lother Meyer arranged all the known elements in the tabular form in the increasing order of their atomic weights.

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Periodic Classification

Formative Worksheet 3.

87 3) Relation between the properties of same elements. 4) Relation between the atomic mass of all elements.

A, D, C are the correct symbols of right elements of the periodic table arranged in the increasing order of their atomic weights. The atomic weight of ‘A’ Be B C N O F is 40 and that of D is 137. If A ,D, C are the 12. Li Na elements of a Dobereiner triad, then find the atomic weight of ‘D’. The above table is related to: 4. Which of the following is not a Dobereiner triad? 1) Mendeleev’s law 1) Cl, Br, I 2) Ca, Sr, Ba 2) Newland’s law 3) Li, Na, K 4) Fe, Co, Ni 3) J.W. Dobereiner’s law 5. ‘X’ and ‘Y’ are two elements with similar properties 4) Moseley’s law and obey the Newland’s law of Octaves. How many elements are there in between ‘X’ and ‘Y’? 4 Mendeleev’s Periodic 6. The discovery of _________ gave a death-blow classification to Newland’s law of octaves. With the failures of many attempts, there was a choatic mess in the arrangement of elements. An onceptive orksheet end to this chaotic mess of elements was put by 6. Dobereiner’s law was rejected due to Mendeleev.Dmitri Ivanovich Mendeleev, a Russian Statement A : Quite a large number of elements chemist, was the first to put forward the successful cannot be grouped into triads. arrangement of elements. In 1869, he published a Statement B : It was possible to group quite periodic table of elements. dissimilar elements into triads. Meaning of Periodic Table 1) Statement ‘A’ is correct but ‘B’ is incorrect 2) Statement ‘B’ is correct but ‘A’ is incorrect A periodic table is a chart in which the 3) Both ‘A’ and ‘B’ statements are correct elements are arranged in such a way that : 4) Statement ‘A’ and ‘B’ are incorrect i) The elements having similar 7. The law of triads was proposed by: properties are placed in the same 1) Dobereiner 2) Newland vertical column, called group. 3) Lother Meyer 4) Chancourtois ii) In the term “periodic table”, the 8. The law of triads is applicable to word “periodic” means that the 1) Lithium, beryllium, boron elements having similar properties 2) Fluorine, chlorine, bromine are repeated after certain definite intervals or periods. The word 3) Chlorine, bromine, iodine “table” means that the elements 4) Sodium, potassium, rubidium have been arranged in tabular form. 9. Select the following pair of elements in which their arithmetic mean of atomic weights is equal to the Mendeleev’s periodic law: Mendeleev studied the chemical properties of all atomic weight of strontium. 63 elements known at that time. On the basis of 1) Lithium, Barium 2) Sodium, Calcium their properties, he proposed that when elements 3) Calcium, Barium 4) Sodium, Barium are arranged in the increasing order of their atomic 10. Which of the following is a wrong triad? masses, the elements with similar properties appear 1) Chlorine, bromine, iodine at regular intervals or periods, i.e., physical and 2) Lithium, sodium, potassium chemical properties of the elements are a periodic 3) Carbon, nitrogen, oxygen function of their atomic masses (atomic weights). 4) Calcium, strontium, barium Later on, when Mendeleev came to know about 11. Which of the following is an achievement of the the work of Lother Meyer, he integrated the two triads classification? statements (Lother Meyer’s and his own statement) 1) Relation between all properties of an element. in the form of a law called Mendeleev Lother 2) Relation between only atomic weights of an

C

W

element.

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9th Class Chemistry

88 Meyer periodic law or simply Mendeleev’s periodic law. This law states that: The physical and chemical properties of the elements are a periodic function of their atomic weights i.e., when the elements are arranged in the increasing order of their atomic weights, the elements with similar properties are repeated after certain regular intervals.

Group  Periods  1 2 3 4 5 6 7 8 9 10 11 12

What is Mendeleev’s periodic table? Mendeleev (1871) arranged all the then known 63 elements in the increasing order of their atomic masses. The arrangement of elements was made in horizontal rows (called periods) and vertical columns (called groups). This arrangement showed that the elements having similar chemical properties came directly under one another in the same group. This arrangement of elements was called Mendeleev’s periodic table, Which can thus be defined as an arrangement of elements in the increasing order of their atomic masses in different groups and periods.

I

II

III

IV

V

VI

VII

H 1 Li 7 Na 23 K 39 Cu 63 Rb 85 Ag 108 Cs 133 ? ?

Be 9.4 Mg 24 Ca 40 Zn 65 Sr 87 Cd 112 Ba 137 ? ?

O 16 S 32 Cr 52 Se 78 Mo 96 Te 125 ?

F 19 Cl 35.5 Mn 55 Br 80 ? 100 I 127 ?

? Ta 182 Bi 208 ?

? W 184 ?

? ?

Hg 200 ?

C 12 Si 28 Ti 48 ? 72 Zr 90 Sn 118 Ce 140 ? La 180 Pb 207 Th 231

N 14 P 31 V 51 As 75 Nb 94 Sb 122 ?

Au 199 ?

B 11 Al 27.3 ? 44 ? 68 Yt 88 In 113 Di 138 ? Er 178 Tl 204 ?

VIII

Fe Co Ni 56 59 59

Ru Rh Pd 104 104 106

?

Os Ir Pt 195 197 198

?

U 240

Simplified version of Mendeleev’s periodic table (1871) The number indicate the atomic masses of elements

Main features of Mendeleev’s periodic table 1. 2. 3. 4.

In Mendeleev’s table, the elements were arranged in vertical columns called groups. There were in all eight groups: Groups I to VIII. The group numbers were indicated by Roman numerals. i.e., I, II, III, IV, V, VI, VII & VIII. Except VIII group, every group is further divided into subgroups i.e., A and B. Groups VIII occupy three triads of three elements each, i.e., in all nine elements The properties of the elements in same group or subgroup are similar.

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Periodic Classification

89

5.

There is no resemblance in the elements of subgroups A and B of same group except valency. 6. The horizontal rows of the periodic table are known as periods. 7. There were seven periods, represented by Arabic numerals 1 to 12. To accommodate more elements, the periods 4, 5, 6 and 7 were divided into two halves. The first half of the elements are placed in the upper left corner and second half in the lower right corner. For example, the elements occupying the box corresponding to group I and period 4 are potassium (K) and copper (Cu). K is written in the top left corner, while Cu is written in the lower right corner. 8. A period comprises the entire range of elements after which the properties repeat themselves. 9. In a period, the properties of the elements gradually change from metallic to nonmetallic while moving from left to right. 10. There were gaps left in the periodic table. Mendeleev left these gaps knowingly, as these elements were not discovered at that time.

Property Atomic weight Oxide Specific gravity Sulphate Property Atomic weight Oxide Specific gravity Melting point Solubility in acid and alkali

Scandium 44.9 Sc2O3 3.864 Sc2 (SO4)3

Eka aluminium 68

Gallium 69.9

Ea2O3 5.9

Ga2O3 5.94

Low

303.15K

Dissolves slowly in both acid and alkali

Dissolves slowly in both acid and alkali

Property Eka-silicon Atomic 72 weight Specific 5.5 gravity Melting High point Valency 4 Reaction Slightly with acid attacked by and alkali acids,

How Useful was Mendeleev’s Periodic Table? Mendeleev’s periodic classification brought about a revolution in chemistry. His periodic table classifies elements in a more systematic manner than the earlier methods of classification. Contributions of Mendeleev’s Periodic Table i) Systematic study of elements Mendeleev’s Periodic Table simplified the study of elements. It became useful in studying and understanding the properties of a large number of elements, in a simpler manner. This is because the elements showing similar properties belonged to the same group. ii) Prediction of new elements While arranging the elements in increasing order of atomic mass, Mendeleev left three blanks for elements that were not discovered at that time. He was able to predict the properties of these unknown elements more or less accurately. He named them eka-boron, eka-aluminium and ekasilicon. He named them so, as they were just below boron, aluminium and silicon in the respective subgroups. Later, eka-boron was named as scandium, eka-aluminium as gallium and eka-silicon as germanium.7 A comparative study of the properties of the elements predicted and later discovered

Eka boron 44 Eb2O3 3.5 Eb2 (SO4)3

resists attack by alkali

Germanium 72.32 5.47 958°C 4 Dissolves neither in HCl nor in NaOH

iii) Correction of atomic masses Mendeleev’s periodic table helped in correcting the atomic masses of some of the elements, based on their positions in the periodic table. For example, atomic mass of beryllium was corrected from 13.5 to 9. Atomic masses of indium, gold, platinum were also corrected.

Defects of Mendeleev’s Periodic Table i)

ii)

Position of hydrogen: The position of hydrogen is not correctly defined. It is still not certain whether to place hydrogen in group I A or group VII A. Anomalous pairs of elements: In certain pairs of elements like Ar (40) and K (39); Co (58.9) and Ni (58.6); Te (127.6) and I (126.9) the arrangement was not justified. For example, argon was placed before potassium, whereas its atomic mass is more than potassium. In this case, the periodic law is violated.

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9th Class Chemistry

90 iii) Position of isotopes Isotopes are atoms of the same element having different atomic mass but same atomic number. For example, there are three isotopes of hydrogen with atomic mass 1, 2, and 3. According to Mendeleev’s periodic table, these should be placed at three separate places. However, isotopes have not been given separate places in the periodic table. iv) Grouping of elements Certain chemically dissimilar elements have been grouped together. Elements of group IA, such as lithium, sodium and potassium, were grouped with dissimilar elements, such as copper, silver and gold. v)

Cause of periodicity Mendeleev’s table was unable to explain the cause of periodicity among elements. That is by the elements with similar properties fall one below the other, if they are arranged in the increasing order of their atomic weights.

vi) Position of lanthanides and actinides Fourteen elements that follow lanthanum called lanthanides and fourteen elements following actinium called actinides were not given proper places in Mendeleev’s periodic table.

Formative Worksheet 7.

a) What is the Mendeleev’s Periodic law? b) How many elements were known at the time of Mendeleev’s classification? c) When Mendeleev placed the element titanium below silicon in his period i.e., table on the basis of their similar properties, a gap was left below aluminium. Name the element which has subsequently filled this gap. d) Name two elements whose properties were predicted on the basis of their positions in Mendeleev’s periodic table. e) Name two elements whose atomic weights were corrected on the basis of their positions in Mendeleev’s periodic table. 8. Match the following : Column - I Column - II i) eka-boron p) Germanium ii) eka-aluminium q) Scandium iii) eka-silicon r) Gallium s) Indium www.betoppers.com

9.

Which of the following elements position cannot be justified on the basis of atomic weights? 1) Rare earth elements

2) Alkali metals

3) Transition elements

4) Actinides

Conceptive Worksheet 13. Assertion (A): According to Mendeleev, periodic properties of elements is a function of their atomic masses. Reason (R): Atomic number is equal to the number of protons. 1) Both assertion and reason are correct and reason is the correct explanation of assertion. 2) Both assertion and reason are correct but reason is not the correct explanation of assertion. 3) Assertion is correct and reason is incorrect. 4) Assertion is incorrect and reason is correct. 14. What are the indefinite positions of hydrogen given in Mendeleev’s periodic table ? 1)I B, III B

2) I A, II B

3) I A, VII A

4) VIIA, III B

15. The properties of the following elements which were predicted by Mendeleev before their isolation are : 1) Co and Ni

2) I and Te

3) Sc, Ga and Ge

4) Cl, Ar and K

16. The anomalous pairs of elements in Mendeleev’s table are : 1) Ar, K

2) Co, Ni

3)Te, I

4) All the above

17. Mendeleev’s periodic law was useful in correcting the atomic weights of : 1) Al and Sc

2) Be and In

3) U and Si

4) N and O

Periodic Classification

5

91

Moseley’s modification and Modern justification for Moseley’s modification : Modern periodic table Basis for Moseley’s classification : Discovery of radioactivity, isotopes, isobars and atomic nuclei led Moseley (in 1913) to change the periodic law as given by Mendeleev. He observed regularities in the characteristic X-ray spectra of the elements and found that plot

ν vs. Z (atomic number) is a

straight line while

ν vs. A (atomic weight) is not,

and

  a  Z  b  , where a and b are constants

that are same for all elements. Thus he concluded that atomic number is a more fundamental property than atomic weight.

Later, Henry Gywn-Jeffreys Moseley showed that the atomic number of an element is numerically equal to the number of electrons round the nucleus. The number of electrons in turn is equal to the number of protons in the nucleus. He suggested that atomic number is a more fundamental property of an element than its atomic mass. When the elements are arranged in the increasing order of their atomic number, most of the defects of Mendeleev’s classification get rectified. Relation between Mass Number and Atomic Number : Atomic Number (Z) is the number

of protons in the nucleus of an atom. It is also equal to the number of electrons since the atom is electrically neutral. Mass Number (A) is the total number of neutrons and protons present in the nucleus of an atom. A = Z + N    Mass Atomic Number number number of neutrons The periodic law given earlier is now modified and followed today. “The properties of the elements are periodic functions of their atomic numbers” The modern periodic table is also known as the long form of the periodic table or the extended form of the periodic table.

Automatic Removal of Some Discrepancies in Mendeleev’s Periodic Table

Moseley’s experiment Modern periodic law : The physical and chemical properties of the elements are periodic functions of their atomic numbers.

When the basis of arrangement is changed from atomic mass to atomic number, the following discrepancies of Mendeleev’s periodic table automatically vanish. 1. Position of isotopes: All the isotopes of an element have the same atomic number, no matter what the mass number is. Hence the different isotopes of an element do not require separate positions in the periodic table. 2. Anomalous pairs of elements: When arranged in increasing order of atomic number, (i) Ar(18) should precede K(19), (ii) Co(27) should precede Ni(28) and, (iii) Te(52) should precede I(53). Thus the positions of these elements stand justified and they no longer remain anomalous pairs of elements. www.betoppers.com

9th Class Chemistry

92

Main features of Modern periodic table In order to make the periodic table more useful, another form has been evolved by separating the subgroups. The modern periodic table thus developed is also known as the long form or extended form of the periodic table (see the table in next page) as against the short form, i.e., Mendeleev’s periodic table. 1.

Basis for classification: The elements are arranged in the increasing order of their atomic numbers.

2.

Periods : The seven horizontal rows of the periodic table are known as periods. Each period begins with the outermost electron entering into a new principal quantum number and completes after the outermost shell’s psubshell is complete. The period number denotes the outermost orbit’s number of that element. The first element of each period (except 1st period) is an alkali metal and the last element is an inert gas. The description of periods are given below : Period

Length

No. of elements 1st period Very short period 2 2nd and 3rd periods Short periods 8 4th and 5th periods Long periods 18 6th period Very long period 32 7th period Incomplete period --

To avoid inconvenience, 14 elements, which do not include lanthanum and actinium belonging to 6th and 7th period are placed in two separate rows at the bottom of the periodic table (now called lanthanides and actinides respectively). Criterion for placement of an element in a period: The number of elements in each period of the periodic table is equal to the number of electrons filled in the corresponding electronic shell and a new period begins with an element that has one electron in a new main energy level (i.e., new shell)

Starting element of the period Period

Last element of the period

Name

Electronic configuration

Z

First

Hydrogen

1

1

2

Second

Lithium

2, 1

3

Third

Sodium

2, 8, 1

Fourth

Potassium

Fifth

No. of elements =No of electrons filled

Name

Electronic configuration

Helium

2

2

10

Neon

2, 8

8

11

18

Argon

2, 8, 8

8

2, 8, 8, 1

19

36

Krypton

2, 8, 18, 8

18

Rubidium

2, 8, 18, 8, 1

37

54

Xenon

2, 8, 18, 18, 8

18

Sixth

Cesium

2, 8, 18, 18, 8, 1

55

86

Radon

2, 8, 18, 32, 18, 8, 1

32

Seventh

Francium

2, 8, 18, 32, 18, 8, 1

87

-

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Z

-

Incomplete

Periodic Classification

3.

93

Groups : The elements present in vertical columns are called groups. The groups are again classified into two subgroups - A and B. There are sixteen groups under the headings: IA to VII A, 0 and IB – VII B and VIII. 4. Normal or Typical elements: The elements placed in ‘A’ subgroups, IA and IIA on the left and IIIA to VIIIA (0) on the right are called typical elements. They are also called representative, normal or main group elements. 5. Transitional elements: The ‘B’ subgroup elements present between IA and IIIA are called transition elements as they show transition properties from metals to non-metals. 6. Active metals: The strong metallic elements are placed in IA and IIA groups on the left hand side of the periodic table. 7. Active nonmetals: The strong nonmetallic elements are placed in VII A and VIA groups on the right hand side of the periodic table. 8. Non-reactive elements: The rare gases (noble gases) that are inert, are placed in zero group at the end, extreme right hand side of the periodic table (He, Ne, Ar, Kr, Xe and Rn). 9. Lanthanides and Actinides: Lanthanides and Actinides are placed separately at the bottom of the periodic table due to their unique properties. Their details shall be learnt in pages ahead. Note: The detailed study of normal elements, transitional elements, lanthanides and actinides will be dealt in 10th class. www.betoppers.com

9th Class Chemistry

94

Common names of different groups: Group

General or Special Name

IA

Alkali metals

IIA IIIA

Alkaline earth metals Boron family

IVA

Carbon family

VA

Pnicogens

VIA

Chalcogens

VII

Halogens

Zero group VIII or VIIIB

IB

Rare gases Inert gases Noble gases Iron triad/ ferrous metals/Platinu m triads/Platinu m metals Coinage metals

Reason

Due to the formation of strong oxides and hydroxides with a strong alkaline character (Basic in nature), these are called alkali metals. These oxides are alkaline in nature and exist in the earth. Hence, these elements are called Alkaline Earth metals. As all the elements in this group represent similar properties and boron being the first of these elements, this group is called the Boron group. All the elements in this group have similar properties. Carbon being the first element of this group, this group is called the Carbon group. Pnicogen is a Greek word meaning ‘suffocation’. As the Hydrides of this group NH3, PH3, ASH3 have a pungent odour, and when inhaled causes suffocation, this group is called Pnicogen. Chalcogens in Greek means ore-forming. Oxygen and sulphur are two important elements of this group, and these elements are associated with ores of many metals in the form of their oxides and sulphides. In Greek ‘halogen’ mean salt producer. A salt consists of anion and cation. For example common salt (NaCl) consists Na+ (cation) and Cl(anion). The elements of this group form the anion of salt easily, hence they are called Halogens. Because of their presence in small quantities they are called rare gases. Due to their stable electronic configuration they are called Noble gases. As they have little tendency to react, they are also called Inert gases. VIII group consists of 3 triad series. The first triad series Fe, Co, Ni are called ferrous metals and second and third triad series Ru, Rh, Pd and Os, Ir, Pt are called Platinum metals.

As these elements (copper group metals) were used for the manufacture of currencies in the olden days, they are called Coinage metals.

Formative Worksheet 10. The alphabets in the word “ADCL IIT FOUNDATION” represent hypothetical symbols of the real elements. The atomic numbers and mass numbers of a few elements are shown below. Element Atomic number(Z) Mass number(A)

A

D

C

L

I

T

F

O

10

9

2

6

1

7

8

3

20

19

4

12

1

14

16

7

Identify the following: a) The number of neutrons in each element. b) The elements having same number of protons, electrons and neutrons. c) The element(s) having no neutrons. d) The different elements having same number of neutrons. e) The elements whose mass number is equal to twice their atomic number. 11. An element ‘A’ has atomic number 14. To which period does this element belong? How many number of elements are present in the period to which element ‘A’ belongs? 12. Elements, X, Y and Z have atomic numbers, 1, 11 and 17 respectively. State the nature of valencies shown elements of X and Z. Indicate the groups in the periodic table to which X, Y and Z belong. 13. The outermost shell is completely filled in ‘P’ and incompletely filled in ‘Q’ and the last two principal shells are incompletely filled in ‘R’. Identify the category of P, Q and R respectively. www.betoppers.com

Periodic Classification 14. The atomic number of an element is 36. (a) Name the group and period to which it belongs (b) What is the valency of the element? (c) Name the neighbouring elements. 15. The electronic configuration of A, D, C, L, B, E, S, T are given below (symbols are hypothetical): (A) 2, 6 (D) 2, 8, 1 (C) 2, 8, 8, 2 (L) 2, 8 (B) 2, 8, 4 (E) 2, 5 (S) 2, 8, 13, 2 (T) 2, 8, 7 Identify the following from the above: (a) an alkali metal (b) an alkaline earth metal, (c) a halogen, (d) a noble gas (e) belong to VA group, (f) belong to second period (g) has zero chemical reactivity. (h) is a transition element. 16. Three elements A, B and C have atomic number Z, Z + 2 and Z + 3 respectively. Among these, ‘C’ is an alkali metal. To which groups of the periodic table do the elements ‘A’ and ‘B’ belong respectively? 1) 16, 18 2) 14, 16 3) 15, 17 4) 12, 14 17. The atomic numbers of other elements which lie in the same group as tenth element in the periodic table are: 1) 2, 18, 30, 36 2) 8, 18, 36, 84 3) 18, 32, 54, 86 4) 2, 18, 36, 54 18. Match the following: Column - I Column - II 1) Pnicogens p) Fluorine family 2) Chalcogens q) Nitrogen family 3) Halogens r) Helium family 4) Aerogens s) Oxygen family 19. Match the following: i) Shortest period p) Cs to Rn ii) Short period q) Rb to Xe iii) Long period r) Li to Ne iv) Longest period s) H to He 20. Match the following: Column - I Column - II i) First transition series p) La to Hg ii) Second transition series q) Sc to Zn iii) Third transition series r) Incomplete s) Y to Cd

95

Conceptive Worksheet 18. Which is the fundamental property of elements according to Moseley? 19. If x is the atomic number of an element, y is its mass number, then the number of neutrons is equal to : 20. According to Moseley 21. 22.

23.

24.

25.

ν = a  z - b  , then ν is

__________. The middle portion of the periodic table includes : The inner transition elements constitute : 1) 58Ce to 71Lu 2) 90Th to 103Lr 3) Both 4) None Number of elements in each period = x × number of orbitals available in the energy level that is being filled. ‘x’ is equal to _________. Which of the following statements is incorrect? 1) The elements of subgroups ‘A’ are all normal elements. 2) The elements of subgroups ‘B’ are all transition elements. 3) The elements on the left of the periodic table are metals, on the right are non-metals and in the middle are metalloids. 4) d-block elements are also called transition elements. The atomic number of an element M is 20. (a) Name the group to which it belongs. (b) Name the period to which it belongs (c) What is the valency of the element? (d) What is the oxidation state of its ion? (e) Find the number of electrons in K, L, M and N shells respectively. (f) State if it is metallic / nonmetallic or an inert element.(g) Write the symbols of the neighbouring elements . (h) Name the fifth element in its period. (i) Identify the last element of the same period. (j) Name the first element of its group. (k) Name the last element of its group. (l) Write the formula of its chloride, oxide and nitrate. (m) Identify the neutral atom that is isoelectronic with its dipositive ion. (n) Number of neutrons (o) Identify the element/s having same number of neutrons as of M. (p) Identify the element whose mass number is same as that of M.

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9th Class Chemistry

96

6

IUPAC Nomenclature of elements with atomic numbers > 100 The naming of the new elements had been traditionally the privilege of the discoverer (or discoverers) and the suggested name was ratified by the IUPAC. In recent years this has led to some controversy. The new elements with very high atomic numbers are so unstable that only minute quantities, sometimes only a few atoms, are obtained. Their synthesis and characterisation, therefore, require highly sophisticated costly equipment and laboratory. Such work is carried out with competitive spirit only in some laboratories in the world. Scientists, before collecting reliable data on the new element, at times get tempted to claim its discovery. For example, both American and Soviet scientists claimed credit for discovering 104th element. The Americans named it Rutherfordium whereas Soviets named it Kurchatovium. To avoid such problems, the IUPAC has made recommendation that until a new element’s discovery is proved and

its name is officially recognized, a systematic nomenclature be derived directly from the atomic number of the element, using the numerical roots for 0 and numbers 1-9. These are shown in the table below. The roots are put together in order of digits which make up the atomic number and “ium” is added at the end. The IUPAC names for elements with atomic number (Z) above 100 are shown in the table. Digit Name Abbreviation 0 nil n 1 un u 2 bi b 3 tri t 4 quad q 5 pent p 6 hex h 7 sept s 8 oct o 9 enn e

Atomic IUPAC IUPAC Name Symbol number Official Name Symbol 101 Unnilunium Unu Mendelevium Md 102 Unnilbium Unb Nobelium No 103 Unniltrium Unt Lawrencium Lr 104 Unnilquadium Unq Rutherfordium Rf 105 Unnilpentium Unp Dubnium Db 106 Unnilhexium Unh Seaborgium Sg 107 Unnilseptium Uns Bohrium Bh 108 Unniloctium Uno Hassnium Hs 109 Unnilennium Une Meitnerium Mt 110 Ununnillium Uun Darmstadtium Ds 111 Unununium Uuu Rontgenium Rg 112 Ununbium Uub   113 Ununtrium Uut  114 Ununquadium Uuq   115 Ununpentium Uup  116 Ununhexium Uuh   117 Ununseptium Uus  118 Ununoctium Uuo    Official IUPAC name yet to be announced.   Elements yet to be discovered.

Thus, the new element first gets a temporary name, with symbol consisting of three letters. Later, permanent name and symbol are given by a vote of IUPAC representatives from each country. The permanent name might reflect the country (or state of the country) in which the element was discovered, or pay tribute to an eminent scientist. As of now, elements with atomic numbers up to 112, 114 and 116 have been discovered. Elements with atomic numbers 113, 115, 117 and 118 are not yet known. www.betoppers.com

Periodic Classification

7.

97

Merits of the long form of 8. periodic table

We have seen that the periodic table has developed in various stages. The classification evolving at every stage proved to be more useful than the previous one. We will now discuss the uses of the periodic table. Advantages of the Long Form of Periodic Table (i) This classification is based on the atomic number which is a more fundamental property of the elements. (ii) The position of elements in the periodic table is governed by the electronic configurations which determine their properties. (iii) The completion of each period is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached. (iv) It eliminates the even and odd series of IV, V and VI periods of Mendeleev’s periodic table. (v) The position of VIII group is also justified in this table. All the transition elements have been brought in the middle as the properties of transition elements are intermediate between s- and p-block elements. (vi) Due to separation of two sub-groups, dissimilar elements do not fall together. One vertical column accommodates elements with same electronic configuration thereby showing same properties. (vii) The table completely separates metals and nonmetals. Non-metals are present in upper right corner of the periodic table. (viii) There is a gradual change in properties of the elements with increase in their atomic numbers i.e., periodicity of properties can be easily visualised. The same properties occur after the intervals of 2, 8, 8, 18, 18 and 32 elements which indicates the capacity of various periods of the table. (ix) Since this classification is based on the atomic number and not on the atomic mass, the position of placing isotopes at one place is fully justified. (x) It is easy to remember and reproduce. (xi) The position of some elements which were misfit on the basis of atomic mass is now justified on the basis of atomic number. For example, argon precedes potassium because argon has atomic number 18 and potassium has 19. (xii) The lathanides and actinides which have properties different from other groups are placed separately at the bottom of the periodic table. (xiii) This arrangement of elements is easy to remember and reproduce.

Anomalies of the Long form of Periodic table The long form of periodic table has removed all anomalies of the Mendeleev’s periodic table except: i) Position of H is not fixed till now. ii) Lanthanides and Actinides still find a place at the bottom of the table. iii) It does not reflect the exact distribution of electrons among all the orbitals of the atoms of all the elements.

Formative Worksheet 21. Match the following: Column - I Column - II i) Fermium p) 100 ii) Lawrencium q) 101 iii) Mendelevium r) 102 iv) Nobelium s) 103 22. Match the following : Column - I Column - II A) Unnilunium p) Nobelium B) Unnilbium q) Mendelevium C) Unniltrium r) Lawrencium 23. What would be the IUPAC name and symbol for the element with atomic number 120? 24. State True/False. i) The Modern periodic table correlates the position of elements with their electronic configuration more clearly. ii) The position of VIII group is not justified in this table. iii)It removes the anomalies of inversions which existed in the Mendeleev’s table. iv)If does not separate metals and non-metals.

Conceptive Worksheet 26. The elements beyond uranium (Z = 92) are called ________. 27. The atomic numbers of a few elements are given below. Which of them can be considered as transfermium elements? 1) 101 2) 105 3) 93 4) 96 28. Write the respective names given by IUPAC to the elements named Unununnium, Ununnillium, Unnilennium. 29. The atomic number of the element named after the scientist who introduced the concepts of orbits or main energy levels: 30. The IUPAC symbol of the elements with atomic numbers 104, 105 and 106 respectively is : 1) Db, Rf, Hs 2) Rf, Db, Sg 3) Rf, Mt, Ds 4) Mt, Ds, Rg www.betoppers.com

9th Class Chemistry

98 31. The atomic number of the elements Hassnium, Meitnerium and Darmstadtium respectively is : 32. _________ and _________ have been assigned separate location in this form of periodic table. 33. The position of __(i)__ is fully justified, as __(ii)__ is basis for classification. 34. Which of the following is/are correct? 1) Position of hydrogen in unresolved. 2) Long form of periodic table fails to accommodate lanthanides and actinides in the body of the table. 3) Both 1 and 2. 4) None

9. 1.

2.

Periodicity properties

and

Periodic

When elements are arranged in increasing order of atomic numbers, elements with similar properties re-occur (due to similar outer electronic configuration) at regular intervals in the periodic table. This is known as periodicity. Elements coming at intervals of 2, 8, 8, 18, 18, 32 will have similar properties and thus grouped in one particular group. For example, elements with atomic number 1, 3, 11, 19, 37, 55 and 87. Elements with atomic number 4, 12, 20, 38, 56 and 88 will have similar properties. Note : Two adjacent elements in a group generally differ by atomic number 2, 8, 8, 18, 18, 32.

nuclear charge. This results in decrease in the value of atomic radius because protons attract the electronic orbits with greater force. (b) Number of orbits : The effect of increase in the number of orbits in an atom increases the atomic size.

Variation of atomic radius in a period and group: In a period , the number of orbits remains same, on going from left to right in a period while there is a unit increase in the atomic number. Thus the electron experiences more force of attraction towards nucleus. Hence atomic radius decreases from left to right in a period. In a period from left to right, atomic radius decreases as the nuclear charge increases. On moving from left to right across a particular period, the atomic radius decreases upto Halogens and increases to Inert gases. In a given period, alkali metal is the largest and halogen is the smallest in size. For atoms of Inert gases, only van der Waal radius is applicable. In a group, from top to bottom, the atomic radius increases gradually due to the increase in the number of orbits and it overweighs the effect of increased nuclear charge. De creasi ng ato mi c

10. Atomic radius

1A H siudar ci mot a gn i s aer c nI

In atoms, the electron cloud around the nucleus extends to infinity. The distance between the centre of the nucleus and the electron cloud of outermost energy level is called atomic radius. Atomic radius cannot be determined directly, but can be measured from the internuclear distance of combined atoms, using X-ray diffraction techniques. Units: Atomic radii is expressed in angstrom, nanometers, picometre units.

1Å = 10-1 nm ; 1Å = 102 pico.metres

Factors affecting atomic radius: (a) Effective Nuclear Charge : The effect of increase in the number of protons increases the effective www.betoppers.com

2A

radius

3A

4A

5A

6A

7A

B

C

N

O

F

50 Ne

98

91

92

73

72

70

32 Li

Be

152

112

S

Cl

8A He

Na

Mg

Al

Si

P

186

160

143

132

128

127

K

Ca

Ga

Ge

As

Se

Br

Kr

227

197

135

137

139

140

114

112

99

Ar 98

Rb

Sr

In

Sn

Sb

Te

I

Xe

248

215

166

162

159

160

133

131

Cs

Ba

Tl

Pb

Bi

Po

At

Rn

265

222

171

175

170

164

142

140

Ionic radius : When a neutral atom loses one (or) more electrons, a positive ion called cation is formed.

Na  Na + + e  The ionic radius of cation is less than that of neutral atom. It is because the cation has higher effective nuclear charge. For example, the size of Na > Na +

Periodic Classification

99

Among the cations as the positive charge increases, the ionic radius decreases. For example, the size of

Fe2+ > Fe3+

F 25.

26.

27.

28.

29.

Anion

Radius of anion

Cation

Radius of cation

30.

The size of the following species increases in the order: 1) Mg2+ < Na+ Na+ > K+ > Rb+ due to decrease in effective nuclear charge. 3) Li+ = Na+ = K+ = Rb+ For example, the size of Cl– > Cl 4) Rb+ > Na+ > K+ > Li+ Among the anions, as the negative charge increases, 37. O2– and Si4+ are isoelectronic ions. If the ionic radius the ionic radius increases. of O2– is 1.4 A0, the ionic radius of Si4+ will be For example, the size of O2– > O– 1) 1.4 A0 2) 0.41 A0 3) 2.8 A0 4) 1.5 A0 The decreasing order of the radii is : 37. For isoelectronic species, the greater the nuclear Anion > Atom > Cation ; charge, the lesser its size. Thus, the ionic radius of I– > I > I+ ; H – > H > H + Si+4 is less than O–2. The species (atoms or ions) having the same number atomic radii of two elements X and Y are 0.72 of electrons are known as iso - electronic species. 38. The 0 A and 1.6A0. Then the elements X and Y are : In isoelectronic species, the size decreases with 1) F and Ne 2) Ne and F increase of negative charge and decrease of positive 3) Li and Be 4) Fe and Fe2+ ion charge. 39. Among O, C, F, Cl and Br, the increasing order of Decreasing order of size atomic radii is C4– > N3–> O2– > F– > Ne > Na+ > Mg2+ > Al3+ > 1) F < O < C < Cl < Br 4+ Si 2) F < C < O < Cl < Br 3) F < Cl < Br < O < C ormative orksheet 4) C < O < F < Cl < Br In which of the following species, is the size of the 40. The radii of F–, F, O and O–2 are in the order first species not more than that of the second? 1) O2 –>F–>O>F 2) O2 –>F–>F>O 1) Li, F 2) Fe2+, Fe3+ 2 – – 3) F >O >O >F 4) O2 –>O–>F>F– 3) Na+, F– 4) S, O In the isoelectronic series: K+, Cl–, S2–, Ca2+, the species with largest size is 11 Ionisation energy The minimum amount of energy required to remove __________. the most loosely bound electron (i.e, outer most shell In which of the following compounds does the ratio electron) from an isolated neutral gaseous atom is of anion size to cation size have the maximum value? called ionisation potential. 1) CsI 2) LiF 3) LiI 4) CsF Units: IE is measured in eV/atom or KJ/mole or Which of the following sets of elements is arranged K.cal/mole. in the order of increasing atomic radii? 1 eV / atom = 23.06 K.Cal/mole = 96.45 KJ/mole 1) Na, Mg, Al, Si 2) C, N, O, F Note: Ionisation energy is determined by spectral 3) O, S, Se, Te 4) I, Br, Cl, F studies or discharge tube experiments. The correct order of atomic size of C, N, P, S follows Types of Ionisation energies: the order: i) First ionisation energy: The energy required to 1) N < C < S < P 2) N < C < P < S remove an electron from a neutral gaseous atom to 3) C < N < S < P 4) C < N < P < S

W

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9th Class Chemistry

100

M  g  + IE1  M +  g   e 

ii)

Second ionisation energy: The energy required to remove an electron from a unipostive ion to convert it into a dipositive ion is known as Second ionisation energy (IE2)

M   g  + IE 2  M 2+  g   e 

iii) Third ionisation energy: The energy required to remove an electron from a dipositive ion to convert it into a tripositive ion is known as third ionisation energy (IE3). M 2+  g  + IE 3  M3+  g   e 

Ionisation potential depends on : a) Atomic radius: With increase in the atomic size ‘IP’ decreases due to decrease in attractive force of nucleus on outermost orbit electrons. b) Nuclear charge: With increase in the effective nuclear charge, IP increases. c) Charge on a cation and anion: As the positive charge on cation increases, IP increases. As the -ve charge on anion increases, IP decreases.

Variation in IE: a) Along a period: Generally ionisation energy increases as we move from left to right along a period, because with increase in atomic number in a period effective nuclear charge increases and so electrons are more tightly held. Li

Be

B

IE1 (in ev)

5.4

9.3

8.2

C

N

O

F

Ne

11.2 14.54 13.64 17.42 21.56

Ne (2000)

2500 2000

N (1402)

1500

Be (899)

1000

Li (520)

F (1681)

C (1096)

O (1314)

2

4

6

8

10

However, there are number of exceptions that shall be discussed in the higher classes. b) In a group: On moving down in a group, the distance between valence shell and the nucleus increases and thus nuclear attraction towards test electron decreases which results in decrease in ionisation energy.

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Na

K

Rb

Cs

IE1 (eV)

5.4

5.1

4.3

4.2

3.9

550

Li (520) Na (496)

500

K (419)

450

Rb (403)

400 350 0

10

Ca (374)

20 30 40 50 Atomic number

60

i) Atoms of inert gases have highest IP values . ii) In any period an Alkali metal atom has lowest IP and Inert gas element has highest IP. iii) In periods from left to right, or IP increases due to decrease in atomic size and increase in effective nuclear charge. iv) In groups from top to bottom, IP decreases due to the increase in the atomic size and increase in the screening effect of inner electrons. v) IE order among 2nd period elements. IE1 - Li < Be > B < C < N > O < F < Ne IE2 - Li > Be < B > C < N < O > F < Ne vi) IE order among 3rd period elements IE1 - Na < Mg > Al < Si < P > S < Cl < Ar IE2 - Na > Mg < Al > Si < P < S > Cl < Ar vii) Element with Lowest IP - Cs viii)Knowledge of successive IE can be used to find the number of valence electrons ix) For alkali metals the IE2 shows sudden jump. x) For alkaline earth metals, the IE3 shows sudden jump. xi) The number of IE possible for an atom of an element is equal to its atomic number.

Formative Worksheet

B (801)

500 0

Li

Some more important points related to IP.

Factors affecting I.E

II period

I gp.

IE/kJ mol–1

convert it into a unipositive ion is known as First ionisation energy (IE1).

31. The successive ionisation potentials of an element are 7.1. 14.3, 34.5 and 162.2 ev. The number of electrons in its outermost shell is ____. 31. From the above values it is clear that there is large jump in energy from 3rd IP to 4th IP. This is possible only when the element attains stable configuration after the removal of 3 electrons. Hence, the element has 3 electrons in its outermost shell. 32. Arrange the following in the increasing order of their ionisation energies: Ne, Al+3, Mg+2, Na+.

Periodic Classification 32.

33. 33. 34.

34.

35.

101

Species No. of e’s E. C Ne 10 2, 8 Al+3 10 2, 8 +2 Mg 10 2, 8 Na+ 10 2, 8 From the above chart, it is clear that all the species have same electronic configuration. Hence, the IP depends on the extent of nuclear pull. The greater the nuclear pull, the greater the IP. Further, the extent of nuclear pull depends on the number of protons. Therefore, the species with greater number of protons have greater IP and vice-versa. Therefore, the increasing order of IP of the given species is as follows: IP of Ne < Na+2 < Mg+2 < Al+3 Out of cobalt and scandium which has higher IP and why? The IP of cobalt is more because of its small size. The atomic numbers of the elements A, B, C and D is 9, 10, 11 and 12 respectively. The correct order of ionization energies is: 1) A > B > C > D 2) B > A > D > C 3) B > A > C > D 4) D > C > B > A A and B belong to second period whereas C and D belong to third period. In general, IE increases along the period and decreases down the group. Therefore, the correct order is B > A > D > C. IE1 and IE2 of Mg are 178 kcal mol–1 and 348 kcal mol–1 respectively. The energy required for the reaction

Mg  g    Mg 2+  g  + 2e- is:

1) + 170 kcal 3) – 170 kcal

2) + 526 kcal 4) –525 kcal

35. The reaction, Mg  g   Mg 2  g   2e  occurs in two steps: IE1 Mg  g   Mg   g   e  ;

 H 178 kcal mol1 IE 2 Mg   g   Mg 2   g   e  ;

 H  348 kcalmol 1 Mg  g    Mg 2   g   2e  ;

 H  526 kcal mol 1

Conceptive Worksheet 41. Define ionisation potential. 42.

Ag   E   A  g   e  . What is E called?

43. What are the units of Ionisation potential ? 44. For any atom, the order of ionization potential values is 1) I1< I2< I3 2) I1> I2> I3 3) I1< I2> I3 4 I 1> I2< I3 45. The first, second, third, fourth, fifth ionization potential values of an element are 6.11, 11.87, 51.21, 67.0, 84.39 eV respectively. The element is 1) Calcium 2) Potassium 3) Aluminium 4) Carbon 46. The element with highest ionization potential is 1) Nitrogen 2) oxygen 3) Helium 4) Neon 47. As atomic number of elements increases, the IP value of the elements of the same period 1) decreases 2) increases 3) remains constant 4) first increases and then decreases 48. The ionization potential of elements in any group decreases from top to bottom. This is due to 1) Increase in size of atom 2) Increase in atomic number 3) Increase in screening effect 4) both increase in size of atom and increase in screening effect 49. The ionization potential values of an element are in the following order I1Si>C 2. The incorrect statement among the following is: 1) the first ionisation potential of Al is less than the first ionisation potential of Mg 2) the second ionisation potential of Mg is greater than the second ionisation potential of Na 3) the first ionisation potential of Na is less than the first ionisation potential of Mg 4) the third ionisation potential of Mg is greater than the third ionisation potential of Al 3. Which of the following is largest in size? 1) Br– 2) I 3) I– 4) I+ 4. Which of the following has lowest ionisation potential? – 1) Na+ 2) O2– 3) N34) F 5. The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is 1) C>N>O>F 2)O>N>F>C 3) O> F >N>C 4 )F>O>N>C 6. Which one of the following is smallest in size ? 1)N3 – 2)O2 – 3)F– 4)Na + 7. The first ionisation potentials of nitrogen and oxygen respectively in electron Volts 1) 14.6,13.6 2) 13.6,14.6 3) 13.6,13.6 4) 14.6,14.6 8. The first ionisation potential of Na, Mg, Al and Si are in the order 1) Na < Mg > Al < Si 2) Na > Mg > Al > Si 3) Na < Mg < Al > Si 4) Na > Mg > Al < Si 9. Atomic radii of fluorine and neon in angstrom units are respectively given by 1) 0.72,1.60 2) 1.60,1.60 3)0.72,0.72 4) none of these 10. The correct order of radii is 1)N K 3) Ba > K > Ca 4) K > Ba > Ca 12. Which one of the following is largest in size? 1)Na 2)Na + 3)Na– 4) none

Periodic Classification 13.

14.

15.

16.

17. 18.

19.

20. 21.

22.

23.

24.

25.

26. 27.

107

Which one of the following is the correct order of increase of size ? – – 1)MgMg>Be 3)Mg>Na>Be 4) Mg > be > Na The 100th element is named in honour of 1) Einstein 2)Bohr 3) Fermi 4) Madame Curie Which one of the following relations is correct with respect to first (I) and second (H) ionisation potentials of Na and Mg? 1) INa > I Mg 2) INa = I Mg 3) IImg > I Na 4) IINa > II Mg The element that has least value of first ionisation potential 1) neon 2) fluorine3) oxygen 4) nitrogen Which is largest in size? 1)O2 – 2)F 3)Mg2+ 4) Na+ The isoelectronic species with CO is – – – 1) CN 2) OH 3)N4 4) O

28. Which one of the isoelectronic ions has the largestradius? 1) Na+ 2) Mg2+ 3) O–2 4) None 29. The rare gas that is most abundant in the atmosphere is 1) He 2)Ne 3)Ar 4)Kr 30. The halogen with highest ionisation potential value 1) F 2) Cl 3) Br 4)I 31. In the long form of the periodic table, the elements having lower ionisation potential are present in 1) Group IA 2) Group IVA 3) Group VIIA 4) Zero group 32. As we go from left to right in a period, the effective nuclear charge 1) increases 2) decreases 3) remains the same 4) can’t be said 33. The alkali metal that has highest ionisation potential value is 1) Li 2)Na 3)K 4)Cs 34. The largest ion in the following: _ 1) Cl– 2) S2 3)Na + 4) all same 35. The first ionisation energy of lithium will be 1) greater than beryllium 2) lesser than beryllium 3) equal to sodium 4) equal to fluorine



2

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9th Class Chemistry

Chemical Bonding

By the end of this chapter, you will understand • • • • • • •

Reasons for nonreactivity of noble gases What is a chemical bond Electron theory of valency Electron dot structure of atoms - Lewis’ symbols Ionic bond and Its formation Conditions favourable an ionic bond formation Properties of Ionic compound

4

• • • • • •

Covalent bond and its formation Types of covalent bonds and their formation Properties of covalent compounds Polar covalent bond and its formation Co-ordinate covalent bond Deviation from octet rule  Lewis Dot Structures of Some Molecules

1. Reasons for Non-Reactivity of noble gases Electronic configuration of Noble Gases Noble gases are monoatomic in nature. They do not form molecules either with their own atoms or react chemically with any other element. Why? In order to answer the above, question, let us study electronic configuration of the noble gases as illustrated in a given table: Noble Gas Helium Neon Argon Krypton Xenon Radon

Symbol He Ne Ar Kr Xe Rn

Atomic number 2 10 18 36 54 86

Electronic configuration K L M N O P 2 2, 8 2, 8, 8 2, 8, 18, 8 2, 8, 18, 18, 8 2, 8, 18, 32, 18, 8

Notice that with the exception of helium, which has two electrons in its valence shell, all other noble gases have eight electrons in their valency shells. Thus, we can say that helium has duplet configuration in its valence shell, whereas all other noble gases have octet configuration in their valence shells.

As all noble gases are chemically inactive, following conclusions can be drawn i) ii)

Duplet configuration of valence shell makes an element inactive. Octet configuration of valence shell makes an element inactive. The above conclusions were drawn by Kossel and Lewis independently in 1916. They further stated that a duplet or an octet configuration of electrons in the valence shell is most stable and any atom having this configuration will be in the minimum state of energy.



1.

Chapter -5

Learning Outcomes

Why do atoms combine? We know that, noble gases do not enter into any chemical action due to their stable duplet (or) octet configuration in their outermost shell. Hence, they are chemically inactive. This assumption was made by Kossel and Lewis independently in 1916. With the exception of noble gases, atoms of all elements have 1-7 electrons in their outermost shell. Thus, the electronic configuration of outermost shell of these elements is incomplete. Hence, they are unstable and have a tendency to attain stability by attaining stable configuration. In other words, atoms of the same element or different elements combine with one another, so that they attain duplet (or) octet configuration. Consider sodium (2, 8, 1) and chlorine (2, 8, 7) atoms. Both are unstable in their atomic state. When sodium and chlorine are brought together, Sodium loses an electron, which is gained by chlorine atom. As a result Na+ (2, 8) and Cl– (2, 8, 8) are formed, thus making themselves stable. Further, stability is co-related to energy. A system with less energy is more stable. During a chemical reaction, the reactants lose energy and the resulting product has less energy. As the product has less energy, it is more stable. So, during a chemical reaction, stability can also be achieved by lowering the energy. We can thus conclude that, the cause of a chemical reaction is to attain stability. How do atoms acquire stable octet configuration? Atoms can complete the valence shell by acquiring octet configuration in two ways. By transfer of one or more electrons, from one atom to another. Generally, electropositive elements lose electrons and electronegative elements gain electrons.

110 2. By sharing one or more electrons between two or more atoms. Thus, we can conclude that, atoms tend to acquire 8 electrons in their outermost shell (except hydrogen, lithium and beryllium which tend to acquire 2 electrons), in order to attain stable state. This is called ‘octet rule’.

Formative Worksheet Which inert gas has a duplet configuration in its valence shell? a) Helium b) Neon c) Argon d) Krypton 2. Assertion: Duplet configuration implies that a given atom has 2 electrons in its valence shell. Reason : Elements with octet configuration in their valence shell are stable. a) Both assertion and reason are correct but reason is the correct explanation for the assertion. b) Both assertion and reason are correct, but reason is not the correct explanation for the assertion. c) Assertion is correct but reason is incorrect. d) Assertion is incorrect and reason is correct. 3. Chemical reactivity of an elements depends on: a) Outer shell electronic configuration. b) Reactivity of the nucleus. c) Core electrons. d) None. 4. Assertion :Less energy species are more stable. Reason : When energy is less, the velocity of the vibrating particles decreases, thereby increases the stability. a) Both assertion and reason are correct but reason is the correct explanation for the assertion. b) Both assertion and reason are correct, but reason is not the correct explanation for the assertion. c) Assertion is correct but reason is incorrect. d) Assertion and reason both are incorrect. 5. In each of the following sets, identify the more stable species. i) Al or Al3+, ii) N or N3–, iii) H or H2 6. Statement A: Pure metal is more stable than its ore. Statement B: Ore of metal is more stable than pure metals. a) ‘A’ is correct, ‘B’ is incorrect. b) ‘B’ is correct, ‘A’ is incorrect. c) ‘A’ and ‘B’, both are correct. d) ‘A’ and ‘B’, both are incorrect. www.betoppers.com

9th Class Chemistry 7.

8.

1.

9.

10.

11.

12.

13.

14.

Which one of the following statements is correct? a) Products of endothermic reactions are more stable. b) Products of exothermic reactions are more stable. c) Products of both exothermic and endothermic reactions are equally stable. d) None of the above. An element A is tetravalent and another element B is divalent. The formula of the compound formed by the combination of these elements is: Two atoms X and Y have 5 and 7 valence electrons. The formula of the compound formed by their combination is: Two elements X and Y have the following electron configurations, X= 1s2 2s2 2p6 3s2 3p6 4s2 and Y = 1s2 2s2 2p6 3s2 3p5. The formula of the compound formed by the combination of X and Y is: Electronic configuration of an element ‘A’ is 1s2 2s2 2p6 3s1 and electronic configuration of another element ‘B’ is 1s2 2s2 2p6 3s2 3p4 . The possible compound that can be formed between A and B is: In a short period, as atomic number increases, the valency of elements with respect to hydrogen: a) Increases b) Decreases c) Remains constant d) First increases and then decreases Cation is isoelectronic with anion in: a) Sodium chloride b) Potassium Bromide c) Lithium fluoride d) Rubidium bromide An element ‘A’ belongs to IIA group and another element ‘B’ belongs to VIA group. The compound formed between ‘A’ and ‘B’ contains: a) A++, B– ions b) A++, B– – ions –– ++ c) A , B ions d) A+, B– – ions

Conceptive Worksheet 1.

2.

3.

It was found that atoms having atomic numbers of 2, 10, 18, 36, 54, 86 are very stable and do not show any chemical reactivity, these elements were found to be gases and are called: a) Inert gases b) Diatomic gases c) Monoatomic gases d) Noble gases Which of the following element(s) do not form molecules? a) Helium b) Oxygen c) Nitrogen d) Argon “The duplet and octet configuration of electrons in the valence shell is most stable and any atom having this configuration will be in minimum state of energy”. This statement was given by

Chemical Bonding 4.

5.

6.

7.

8. 9.

10.

11.

12. 13.

14.

To attain a ______ participating atoms redistribute their electrons to get a electronic configuration of the nearest noble gas either octet or duplet. a) State of maximum energy b) State of minimum energy c) Stability d) None When two atoms approach each other and a bond is formed, potential energy: a) Increases. b) Decreases. c) Sometimes increases, sometimes decreases. d) None of the above. Atoms attain the octet configuration, (i) by transfer of one or more electrons from one atom to another. (ii) by sharing of one or more electrons between two or more atoms. a) ‘i’ is correct b) ‘ii’ is correct c) Both ‘i’ and ‘ii’ are correct d) None of the above The chemical stability is more for [Excepting inert gases] a) the parent atom b) their ions c) Both (a) and (b) d) None The maximum valency of sulphur is: a) 4 b) 6 c) 8 d) 7 Metal ‘M’ forms a peroxide of the type MO2 . Valency of the metal with respect to oxygen: a) 0 b) 1 c) 2 d) 4 Valency of the metal atom with respect to oxygen is maximum in: a) Mn2O7 b) OsO4 c) MnO2 d) CrO3 An atom ‘A’ has 2K, 8L and 3M electrons. Another atom ‘B’ has 2K and 6L electrons. The formula of the compound formed between A and B is: a) AB b) A2B3 c) A3 B2 d) AB2 Which of the following exhibits variable valency? a) Na b) H c) Al d) S In a short period, as the atomic number increases, the valency of elements with respect to oxygen: a) Decreases b) Remains constant c) First increases and then decreases d) Increases Which of the following has pseudo inert gas configuration: a) Na+ b) Cu+ c) K+ d) S– –

111

2. What is Chemical Bond?

a)

b)

During a chemical reaction, atoms come closer and are held together by a force of attraction to form molecules. This force of attraction is called a chemical bond. Chemical bonds are responsible for the existence of molecules. What happens when atoms are brought closer? Atoms experience two kinds of forces when they are brought closer. Attractive force: This is due to the attraction between the positively charged nuclei of atoms and the electrons of adjacent atoms. Repulsive force: This is due to the repulsion between the negatively charged electrons of the adjacent atoms. Repulsion also occurs between the positively charged nuclei of adjacent atoms. Initially, the attractive force is greater than the repulsive force. As a result, atoms start to come closer under the influence of this attractive force. In the process, they lose energy and at the same time gain stability. attractive force 

e + e

repulsive force

attractive force

e + e

Now, let’s see what happens when atoms come too close. After a certain point, even as the distance between the atoms becomes less, the repulsive force begins to gain the upper hand. In order to overcome the power of the repulsive force, energy from an external source is required. Hence, the energy of the system starts to increase.

3. Electron theory of valency Kossel and Lewi’s approach of bonding

1.

A number of attempts were made to explain formation of chemical bonds in terms of electrons, but it was only in 1916, Kossel and Lewis’ succeeded independently in giving a satisfactory explanation. They proposed a theory, based on electronic concept of atoms, known as electron theory of valency. The main postulates of this theory are: The secret of stability of atoms: Atoms with eight electrons in the outermost shell (two in the case of Hydrogen, Helium, Lithium and Beryllium) are chemically more stable. www.betoppers.com

9th Class Chemistry

112 2.

3.

4.

Cause of chemical reaction The cause for chemical reaction is to attain stability. An atom achieves this by acquiring the octet configuration (inert gas configuration) in its outermost shell. Type of electrons taking part in a chemical reaction (or) chemical bonding The electrons present in the outermost shell of an atom are responsible for chemical reaction. The outermost shell is called valence shell and hence, the electrons present in it are called valence electrons. The number of electrons taking part in a chemical reaction is called valency of that element. Attainment of nearest inert gas configuration The atoms of various elements achieve the nearest inert gas configuration, either by transfer (losing or gaining) or by sharing of electrons with another atom. This transfer or sharing of electrons results in the development of an attractive force between the atoms, which holds the atoms together by a bond.

Examples: S.No Element

1 2 3

Lithium Carbon Nitrogen

4

Chlorine

5

Calcium

4. Electron Dot structure of atomsLewis symbols In the formation of any molecule or formula unit, only the electrons present in the outermost shell are shown. The reason for not showing the inner shell electrons is that, they are well protected and do not involve in chemical reaction. Therefore, valence electrons are considered for the formation of the chemical bonds. G.N. Lewis’ introduced simple symbols called Lewis’ symbols to denote the valence electrons in an atom. Lewis’ symbols The symbol of the element, surrounded by the valence electrons of its atom, represented in the form of dots around it, is known as Lewis’ symbol or electron dot symbol.

Symbol

Atomic Number

E.G.

Li C

3 6

2, 1 2, 4

Li  :C:

N

7

2, 5

:N:

Cl

17

2, 8, 7

: Cl :

7

Ca

20

2, 8, 8, 2

:Ca

2

The number of dots present around the symbol, gives the number of electrons present in the outermost shell i.e., number of valence electrons. (ii) The number of electrons present in the outermost shell is the common valency of the element. The common valency of the element is equal to the number of dots around the symbol (if the dots are < 4, then the valency is equal to the number of dots and if the number of dots > 5, then the valency is 8 – number of dots.) For example: Li, Be, B and C have valencies 1, 2, 3 and 4 respectively and N, O and F have 3, 2, 1 respectively (i.e. 8 – number of dots). www.betoppers.com



Number of electrons in valence shell 1 4 5



Significance of Lewis’ Symbols (i)

Lewis’ symbol

 

Formative Worksheet 15. Assertion(A): If the magnitude of attractive forces is more than those of repulsive forces, then potential energy of the system increase. Reason (R): With the decrease in potential energy, system gains stability and a chemical bond is formed. a) Both assertion and reason are correct and reason is the correct explanation of assertion b) Both assertion and reason are correct but reason is not the correct explanation of as sertion c) Assertion is correct and reason is incorrect d) Assertion is incorrect and reason is correct

Chemical Bonding

113

16. The common or group valency is equal to: a) No. of valence electrons till group number 4. b) 8 - no. of valence electrons after group number 4. c) Only no. of electrons present in the valence shell. d) All 17. X, Y, Z are the three elements which belongs to IVA, VA, VIA groups respectively. Which of the following are correct Lewis’ dot representations of X, Y, Z? (X) (Y) (Z) a)

X

Y

Z

b)

X

Y

Z

c)

X

Y

Z

X Z Y d) 18. The Lewis’ symbols of three unknown elements are as follows :

R

A

(ii) Y

(iii) Z

Write their respective general valencies and the groups to which they belong. 21. The Lewis’ symbols of three unknown elements are as follows (i) A

15. The force of attraction that holds the atoms or ions or molecules together is known as: a) Chemical bond b) Gravitational pull c) Nuclear pull d) Magnetic pull 16. Atoms experience the following when they are brought closer. a) Attractive forces b) repulsive forces c) both (a) and (b) d) none 17. The dots in Lewis symbols represents: a) Valence electrons in an atom b) Low energy state of an atom c) Octet and duplet of an atom d) Nearest noble gas configuration 18. A, B, C are the elements of group IA, IIA, IIIA respectively. Identify the correct Lewis dot diagram of A, B and C in the following: (A) (B) (C) a) b)

M

Predict the general valency of each element and groups to which they belong. 19. Draw the Lewis’ symbols of (i) Oxygen (ii) Fluorine (iii) Aluminium 20. The Lewis’ symbols of three unknown elements are as following. (i) X

Conceptive Worksheet

(ii) B

(iii) C

Write the formula of the different compounds formed by their combination. a) A3B2, AC, A2, B2, BC, B2C, BC2, C2 b) A2B3, AC2, B2, B­2C, B2C2, C2, BC2, AC c) A2B4, A2C2, B2, BC, BC2, C2, B2C d) A3B2, AC, B2, BC, B2C, BC2, C2

A

A

B

C

B

C

c)

A

B

C

d)

A

B

C

19. G, D are the two elements which belongs to VIIA and Zero group respectively. Choose the correct Lewis’ dot diagram in the following. (G) (D) a)

G

D

b)

G

c)

G

D

d)

G

D

D

20. “Highly electronegative halogens & highly electropositive alkali metals are separated by noble gases”. This fact in relation to chemical bonding was given by 21. Inert nature of noble gases can be explained by _______ theory. 22. Electrons involved in bond formation are commonly called as________.

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9th Class Chemistry

114

5. Ionic bond and its formation Ionic bond and its formation Electron transfer and the formation of ionic bonds Electron transfer from one atom to another results in the formation of charged species, called ions. On losing electron(s), an atom has more protons than electrons. So it forms a positively charged ion, called a cation. On gaining electron(s), an atom has more electrons than protons, so it forms a negatively charged ion, called an anion. Elements which lose electrons are called electropositive elements and those which gain electrons are called electronegative elements. Atoms can lose or gain, no more than 3 electrons. Atoms having up to 3 valence electrons e.g., Na (2, 8, 1), Ca (2, 8, 8, 2) and Al (2, 8, 3) can lose these electrons to achieve the octet. Na (2, 8, 1)-1 electron = Na+ (2, 8) Ca (2, 8, 8, 2) –2 electrons = Ca2+ (2,8,8) Al (2, 8, 3) - 3 electrons = A13+ (2, 8) On the other hand, atoms having 5 to 7 valence electrons e.g., N (2, 5), O (2, 6) and Cl (2, 8, 7) can gain electrons to achieve the octet. N (2, 5) + 3 electrons  N3– (2, 8) O (2, 6) + 2 electrons  O2– (2, 8) Cl (2, 8, 7)+1 electron  Cl– (2,8,8) The cations and anions formed as a result of electron transfer are drawn towards each other due to the electrostatic force (coulomb force) of attraction. They form an ionic bond or an electrovalent bond. Note: The bond between two elements is ionic if the EN difference between them is greater than 1.7 The number of electrons transferred during an ionic bond formation is known as an electrovalency. Compounds containing ionic bonds are called ionic compounds. Examples of ionic compounds are NaCl(Na+Cl–), CaO(Ca2+O2 ), MgO(Mg2+O2–) and MgCl2 (Cl –Mg++ Cl–).

6. Conditions favourable for an ionic bond formation During ionic bond formation, the donor atom gets converted to a positive ion and the acceptor atom gets converted to a negative ion. For a strong ionic bond formation, let us see how the donor and acceptor atoms should behave.

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

1)

2)

3)



1)

2)

3)

Features of donor atoms Donor atoms lose electrons with greater ease if the following conditions are satisfied: Less ionisation potential (I.P.) If I.P. is less, less energy is required to remove the electrons. Hence, the electrons can be lost easily by the donor atom. Eg: Groups IA and II A have less I.P., hence, they are good donors. More size If the size is more, then the nuclear pull on the outer electrons is less. Therefore, electrons can be lost easily. Eg: IA group atoms have more size, hence, they are good donors. Less cation charge Charge on the cation should be less. A cation with more charge is difficult to form.For example, the formation of Ca++ occurs in two steps. Step 1 : Ca – le–  Ca+ Step 2 : Ca+ – le– Ca++ The extra charge on Ca+ results in extra nuclear pull. This makes the removal of the second electron from Ca+ more difficult. So, Ca++ is more difficult to form than Ca+. Features of acceptor atoms Acceptor atoms gain electrons more easily in the following conditions: High electron affinity and electronegativity If E. A and EN is more for an acceptor atom, then the atom has greater tendency to gain electrons. More energy is released when an electron is gained by it, resuming in greater stability of acceptor atom. Eg: Since groups VIA and VII A have greater electronegativity and affinity, they are regarded as good acceptors. Less size When the size of the atom is small, the nuclear pull on the electrons is more. Therefore, electrons can be gained easily. Eg: Group VII A has small atoms, hence, the elements of this group are good acceptors. Less anion charge The charge on anions should be less. An anion with more charge is difficult to form. For example, the formation of ‘O2–’ occurs in two steps. Step 1 : O  1e   O  Step 2 : O  1e  O

Chemical Bonding

115

The extra negative charge on O– results in repulsion of the second electron that is added. To overcome this repulsion, additional energy is required, thus making it more difficult for O– to gain an additional electron. So, O– – is more difficult to form than O–. So we can summarise: a)

b)

For donor atoms to lose electrons easily, they should possess less I.P., big size and less charge on cations. IA group has the above mentioned characters and therefore is the best donor group. For acceptor atoms to gain electrons easily, they should possess more EA, less size and less negative charge on anions. Group VII A has these features and so is regarded as the best acceptor group. Therefore, IA and VII A get along well with each other, forming a strong ionic bond.

Formative Worksheet 1) 1s

2

2

2

2

2

6

2) 1s 2s 2p

2

2

3) 1s 2s 2p

5

4) 1s 2s 2p

The tendency to form electrovalent bond is greatest in a) 1

b) 2

c) 3

d) 4

23. An element ‘X’ is strongly electropositive an element ‘Y’ is strongly electronegative both are univalent. The compound tied would be: +

–

a) X Y

-

+

b) X Y

28. Which of the following has strongest and weakest ionic bonds? LiI, NaI, KI, CsI

Conceptive Worksheet 23. The electronegativities of two elements are 1.0 and 3.5. Bond formed between the would be: 24. In a NaCl crystal, cations and anions held together by: 25. Which one of the following has an electrovalent linkage? a) CH4

b) MgCl2

c) SiCl4

d) BF3

26. Which of the following is easily formed? a) Calcium chloride

22. The electronic structure of four elements a, b, c and d are: 2

27. Arrange the following in the increasing order of their donating ability: Ca, Mg, K, Cs, Al, B?

c) X – Y d) X  Y

24. Which of the following is least ionic? a) CaF2 b) CaBr2 c) CaCl2 d) CaI2

b) Calcium bromide c) Potassium chloride d)Potassium bromide 27. Among the compounds NaCl, KCl and CsCl, the one with greatest ionic character is: 28. When an element of low ionisation potential combines with an element of very high electron affinity: 29. Which of the following is more ionic? a) Si3N4

b) AlN

c) BN

30. Arrange the following is the increasing ionic nature : MCl, MCl2, and MCl3. a) MCl < MCl2 < MCl3

25. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3 .

b) MCl2 < MCl < MCl3

26. Assertion : AlF3 is more readily formed than Al2O3

d) MCl < MCl3 < MCl2

Reason : As the charge on F is less than that in O, it can be formed easily. a) Both assertion and reason are correct and reason is the correct explanation of assertion b) Both assertion and reason are correct but reason is not the correct explanation of assertion c) Assertion is correct and reason is incorrect d) Assertion is incorrect and reason is correct

d) Ca3N2

c) MCl3< MCl2 < MCl 31. Match the following: Column-I

Column-II

i) More EA

p) Donor

ii) Less IP

q) Acceptor

iii) Less charge on cation iv) Greater size v) Less size vi) Less charge on anion www.betoppers.com

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116

7. Properties of Ionic compounds 1.

Physical state Generally, ionic solids are relatively hard. It is because of the close packing due to strong inter-ionic force of attraction present between oppositely charged ions. Brittleness of ionic compounds Even though ionic compounds like rock salt are hard solids, they break quite easily when dropped on floor.  ...... + ......  ...... + ......  ...... + Reason: The behaviour of ionic compounds is much like a glass, which  ...... ......  ...... ......  ......    breaks into many pieces on falling. ...... ...... ...... ...... ......    + + + This brittleness is because of shift in alignment of its ions. Normally, the allignment is such that the oppositely charged ions are next to each other as shown in the adjacent figure.

 Line of force

+



Allignment on impact

Structure of ionic solids Unit cell There is a basic unit in an ionic crystal, which when repeated three-dimensionally, gives complete crystal. This basic unit is called the unit cell. Melting and boiling points Ionic compounds possess high melting and boiling points. Reason: Melting and boiling points of ionic compounds involves breaking of the lattice structure and setting the ions free. In a lattice, there are strong electrostatic forces between oppositely charged ions. To break these strong electrostatic forces, considerable amount of energy is required. Hence, the melting points and boiling points of ionic compounds are high. Solubility Ionic compounds are soluble in water. Reason: Dissolving an ionic solid involves the setting of opposite ions free from the lattice into the solvent. This can happen when the strong electrostatic force of attraction between the opposite ions is weakened. Therefore, solvents having oppositely charged ions, called polar solvents should be used. The best polar solvent is water. Therefore, all ionic compounds are dissolved in water.

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+

 

4.

+



+

3.

+



2.

Normal alignment

+

Owing to the impact on falling, the allignment is disturbed, such that the ions with similar charge come next to each other. Since, like charges repel each other, the crystal breaks along the line of force.

Force of attraction

5.

6.

7.

Electrical conductivity Even though ionic solids consist of opposite ions, they are bad conductors of electricity. In ionic solids, a strong electrostatic force of attraction, making the ions immobile, holds the oppositely charged ions together. Hence, conductivity is not possible. However, in their fused or aqueous state, ionic compounds are good conductors of electricity owing to the presence of mobile ions. For instance, NaCl in its fused state or in its aqueous solution, has free Na+ and Cl– ions. The mobility of Na+ and Cl– results in conduction. High reactivity Ionic compounds react instantaneously in fused state. This is because of easy formation of free ions, rapid union of these ions in solutions, form new compounds. For example, the reaction between NaCl and AgNO3 is very rapid in solution state, resulting in the formation of AgCl, a precipitate and NaNO3. Directional properties A given ion in the ionic crystal is surrounded by a uniform electric field around it. Therefore, the electrostatic bonding force acts equally on the ion in all the directions. As there is no specific direction for the electrostatic bonding force, the ionic bond is a non-directional bond.

Chemical Bonding 8.

Isomorphism Crystals of different ionic compounds having similar arrangement of ions as well as geometry are known as isomorphs, and the phenomenon is known as isomorphism. Eg: ZnSO4. 7H2O & FeSO4. 7H2O A crystal of an isomorph, if placed in a saturated solution of other isomorphs, grows in size. The valency of elements forming isomorphous compounds is same.

Formative Worksheet 29. Assertion (A) : Ionic compound tend to be non-volatile. Reason (R): Inter molecular forces in these compounds are weak. The correct answer is: a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 30. Which one has greater melting point: Ag2O or BaO? a) Ag2 O b) BaO c) both d) none 31. Sodium sulphate is soluble in water, where as barium sulphate is sparingly soluble because, a) The hydration energy of sodium sulphate is less than its lattice energy. b) The lattice energy of barium sulphate is more than its hydration energy. c) The hydration energy and lattice energy of sodium sulphate are equal. d) None of these. 32. In which of the following solvents, should KCl be soluble at 25°C? (D = Dielectric constant value) a) C6H6(D = 0) b) CH3COOCH3 (D = 2) c) CH3OH (D = 32) d) CCl4 (D = 0) 33. If Na+ ion is larger than Mg2+ ion and S2– ion is larger than Cl– ion, which of the following will be least soluble in water? a) NaCl b) Na2S c) MgCl2 d) MgS 34. Assertion (A): Ionic compounds possess high melting and boiling points. Reasons (R) : Ionic compounds involves breaking of the lattice structure. In a lattice, there are strong electrostatic forces between oppositely charged ions.

117 The correct answer is: a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 35. An aqueous solution of silver nitrate gives a white precipitate with: a) C2H5Cl b) CHCl3 c) HCl d) None of the above 36. Which of the following has different crystal lattice from the remaining substances? a) KCl b) CsBr c) MgO d) CaS

Conceptive Worksheet 32. Which of the following true for ionic compounds? a) They are hard solids b) They can be broken down into pieces very easily c) They are soluble in non-polar solvents d) all the above 33. Which of the following boils at higher temperature? a) CCl4 b) CO2 c) C6 H12 O6 d) KCl 34. Many ionic crystals dissolve in water because, a) Water is an amphoteric solvent. b) Water is a high boiling liquid, which has no taste and no odour. c) The process is accompanied by a positive heat of solution. d) Water decreases the inter ionic attraction in the crystal lattice due to solvation. 35. Most of the ionic substances: a) Are non-electrolytes in molten state b) Have directional character c) Are soluble in polar solvents like water d) Conduct electricity in solid state 36. Which one of the following is more soluble in water? a) AgF b) AgI c) AgCl d) AgBr 37. An ionic solid is a poor conductor of electricity because, a) Ions do not conduct electricity. b) Charge on the ions is uniformly distributed. c) Ions occupy fixed positions in solids. d) Ions have uniform field of influence around them. 38. Molten sodium chloride conducts electricity due to the presence of: a) Free electrons b) Free ions c) Free molecules d) Free atoms www.betoppers.com

9th Class Chemistry

118

8. Covalent bond and its formation As the electronegativity difference between two hydrogen atoms, in a hydrogen molecule is zero, transfer of electrons is not possible. Hence, the two hydrogen atoms combine due to the electrostatic force of attraction, developed by the sharing of electrons. Definition: A bond formed by the equal contribution and equal sharing of electrons between two atoms or more atoms is known as covalent bond (co-sharing, valence  valence electron). Since, the formation of a covalent bond results in the formation of a molecule, it is also called molecular bond. G.N. Lewis did the study of covalent bond. He explained covalent bond formation by the electron dot structure called Lewis Structure. When is the bond between two atoms covalent? When non-metals come together, the tendency to donate or accept the electrons is not possible due to the less electronegativity (EN) difference. Thus, in order to acquire stable configuration (an octet or duplet) of a noble gas, sharing takes place between them, resulting in formation of covalent bond. Generally, if the EN difference between two nonmetals is less than 1.7, a covalent bond is formed between them due to their combination. Representation of covalent bond The covalent bond between a pair of two atoms is represented by a small line[ – ]. For example, H2 can be represented as H–H. Covalency: The number of electron pairs shared between two atoms of the same element or different elements during the formation of a molecule is known as covalency. Eg: Covalency of hydrogen molecule is equal to 1 and that oxygen molecule is 2. Bond pairs and lone pairs Have a look at the Lewis dot structure of oxygen molecules.

There are six electron pairs (1), (2), (3), (4), (5) and (6). Out of them, the electron pairs (5) and (6) are www.betoppers.com

involved in bonding and they are called bond pairs. The remaining electron pairs (1), (2), (3) and (4) present in the valence shell of the atom but not involved in the bonding are called lone pairs. Bond pair of electrons The shared pair of electrons, which result in the formation of a bond, is called the “bonded pair”. Lone pair of electrons The pair of electrons, present in the valence shell but not involved in the bonding is called the “nonbonded pair” or “lone pair.” Conditions favourable for covalent bond formation Covalent bonding is all about sharing of electrons. The ideal conditions necessary for atoms to share electrons are: 1) Atoms should be of small size. 2) They should have high electronegativity and ionisation potential. 3) The electronegativity and the ionisation potential of the combining atoms should be almost the same. Favourable conditions for the covalent bond formation are satisfied by the elements of VA, VIA, and VIIA groups. The electrons in the outermost shell (valence electrons) in the elements of VA, VIA and VIIA groups are 5, 6 and 7 respectively, and they can have stable octet configuration by sharing 3, 2 and 1 electron respectively. The molecules formed among the elements of VA, VIA and VIIA are mostly covalent molecules. Eg: SO2, PCl3, PBr3, SF6, SCl2, IF7, O2, N2, F2, Cl2, Br2, I2, etc. Note: Number of electron pairs shared between two atoms of the same element is equal to the number of electrons short of octet. Examples of covalent compounds F2, Cl2, I2, O2, N2, H2, HCl, H2O, NH3 etc. Note: The force of attraction present between the molecules of inert gases is the Vander Waal’s forces.

Formative Worksheet 37. In the formation of covalent bond, a) Transfer of electrons take place b) Electrons are shared by only one atom c) Sharing of electrons take place d) None of these 38. An element ‘Y’ has the ground state electronic configuration 2, 8, 8. The type of bond that exists between the atoms of ‘Y’ is: a) Ionic b) Covalent c) Metallic d) Vander Waal’s

Chemical Bonding 39. The covalency of nitrogen in N2 molecule is: a) 0 b) 2 c) 3 d) 5 40. The covalency of Hydrogen, Chlorine, Oxygen and Nitrogen in their elementary molecules respectively is: a) 1, 2, 3, 4 b) 1, 1, 2, 3 c) 2, 1, 3, 4 d) 1, 3, 2, 4 41. The correct order of increasing covalent character is: a) SiCl4 < AlCl3 < CaCl2 < KCl b) KCl < CaCl2 < AlCl3 < SiCl4 c) AlCl3 < CaCl2 < KCl < SiCl4 d) SiCl4 < KCl < CaCl2 < AlCl3 42. In which of the following compounds, highest covalent character is found? a) CaF2 b) CaCl2 c) CaBr2 d) CaI2 43. Arrange the following in the order of increasing covalent character : NaCl, MgCl2 and AlCl3. 44. A compound has the formula H2Y (Y = Non-metal) State the following. a) The outer electronic configuration of Y b) The valency of Y c) The bonding present in H2Y d) The formula of the compound formed 40 between calcium  20 Ca  and Y.

119 43. Which of the following is incorrect about covalent bond? a) It is formed by sharing of the electrons b) It is stronger than ionic bond c) Its study was done by Kossel d) When two non-metals are combined, then a covalent bond is formed 44. Among the alkaline earth metals, the element forming predominantly covalent compound is: a) Be b) Mg c) Sr d) Ca 45. Arrange the following in the order of increasing covalent character : LiF, LiCl, LiBr and LiI.

9. Types of covalent bonds and their formation

1.

Conceptive Worksheet 39. When two atoms of chlorine combine to form one molecule of chlorine gas, the energy of the molecule is: a) Greater than that of separate atoms b) Equal to that of separate atoms c) Lower than that of separate atoms d) None of these 40. Silicon has 4 electrons in the outermost orbit. In forming the bonds: a) It gains electrons b) It loses electrons c) It shares electrons d) None of the above 41. Which of the following is a covalent compound? a) H2 b) CaO c) KCl d) Na2S 42. Which of the following substance has covalent bonding? a) Sodium chloride b) Solid neon c) Copper d) BeCl2

Covalent bonds are classified into different types based on I) The type of atoms involved in bonding II) The number of electron pairs involved in sharing III) The type of sharing Covalent bonds based on the type of atoms involved in bonding Based on the types of atoms involved in bonding, covalent bonds are classified into homogeneous and heterogeneous covalent bonds. Homogeneous covalent bond It is a covalent bond formed between the atoms of similar type. Examples: a) Formation of hydrogen molecule

(or) H – H (hydrogen molecule) b) Formation of chlorine :eu m lceol (or) Cl – Cl (chlorine molecule) c) Formation of oxygen molecule:

2.

(or) O = O (oxygen molecule) Heterogeneous covalent bond It is a covalent bond formed between the atoms of different types.

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d)

9th Class Chemistry Formation of methane molecule – CH4

e)

Formation of carbon tetrachloride –

120 Examples: a) Formation of Hydrogen Chloride HCl

b)

(or) H – Cl (Hydrogen chloride) After formation of a covalent bond, hydrogen has stable duplet configuration, and chlorine has stable octet configuration. Formation of water molecule – H2 O

C l

4

f) c)

Formation of Ammonia molecule – NH3

(Carbontetrachloride) Formation of Carbondioxide molecule – CO2

(or) O = C = O (Carbondioxide) II.

Covalent bonds based on the number of electron pairs shared. S.No. 1.

2

3.

Type of covalent bond Single covalent bond

Double covalent bond

Triple covalent bond

Definition

Representation

The covalent bond formed by sharing of 1 pair of electrons The covalent bond formed by sharing of 2 pairs of electrons The covalent bond formed by sharing of 3 pairs of electrons

Formative Worksheet The maximum number of covalent bonds by which the two atoms can be bonded to each other is: a) Four b) Two c) Three d) No fixed number 46. The molecule, which contains maximum number of electrons, is: a) CH4 b) CO2 c) NH3 d) BCl3 47. The molecule which contains only bonded pairs or no lone pair of electrons on the central atom is: a) H2O b) NH3 c) BeCl2 d) BrF5 48. Compound having maximum number of bonded pairs of electrons in its molecule is: a) Ethane b) Ammonia c) Sulphur hexafluoride d)Bromine Pentafluoride www.betoppers.com

49.

 H H ×

Examples H 2, Cl2, F2 etc

H-H O2    × ×× O O    × ××

O=O N2 × × ×

N N

× ×

N N

Match the following : a) NH3

45.

b) H2O c) O2 d) CCl4

i) 4 bond pairs and no lone pairs on the central atom ii) 2 bond pairs and 2 lone pairs iii) 3 bond pairs and 1 lone pair iv) 2 bond pairs and 4 lone pairs

50. Match the following: Molecule No.of Bond Pairs No.of Lone Pairs p) HCl i) 2 k) 1 q) CO2 ii) 3 l) 2 r) H2O iii) 1 m) 4 s) NH3 iv) 4 n) 3

Chemical Bonding p q r s a) iii, n iv, m i, l ii, k b) iii, m i, m ii, l iv, k c) ii, l iii, n iv, k ii, m d) iv, n iii, m ii, l i, k 51. Molecule which contains 4 bonded pairs and 2 lone pairs of electrons on the central atom is: a) XeF2 b) CO2 c) XeF4 d) SF6

Conceptive Worksheet 46. Match the following: Column-I Column-II i) CH 4 p) Homogeneous ii) H 2 q) Heterogeneous iii) Cl2 iv) O 2 v) CCl4 vi) H 2 O p q a) i, ii, iii iv, v, vi b) ii, iii, iv i, v, vi c) i, iii, v ii, iv, vi d) none of these 47. The number of electron pairs involved in the bond formation in hydrogen cyanide molecule is: a) Two b) Three c) Four d) Five 48. The number of electron pairs present on central atom in SF6 molecule are: a) 4 b) 6 c) 8 d) 7 49. The number of electrons shared in the formation of: i) Single bond ii) Double bond iii) Triple bond (i) (ii) (iii) a) 1 2 3 b) 2 4 6 c) 6 2 4 d) 6 4 2 50. In which of the following, double bond is present? a) CH4 b) CCl4 c) H2 O d) None 51. In which of the following molecule(s), multiple bond is present? a) CO2 b) O2 c) N2 d) HCN 52. Only triple bond is present in: a) N2O b) CO2 c) HCN d) N2 53. Molecule having maximum number of lone pairs of electrons on central atom is: a) PH3 b) H2S c) CH4 d) BrF5

121

10. Properties of covalent compounds Physical state Generally, covalent compounds are gases, liquids or soft solids at room temperature. Reason : Covalent compounds are composed of molecules. There exists a intermolecular force of attraction known as Vander waal’s force of attraction between these molecules’. These Vander waal’s forces are weak and hence covalent compounds exist as soft solids, liquids and gases. If the Vander waal’s force of attraction is more, then the covalent compound exists as a soft solid, if they are moderate, they exist as liquids and if the forces are negligible, they exist as gases. Halogens like F2, Cl2 , Br2 , I2 all are covalent in nature. But F2 and Cl2 are gases, Br2 is a liquid and I2, a solid. Because, we have seen earlier that the covalent compounds exist as molecules. Between these covalent molecules, a weak intermolecular force of attraction known as Vander waal ‘s force of attraction exists. The Vander waal’s force of attraction increases with the molecular weights. As we move down the halogen group, molecular weight increases, increasing the Vander waal’s force of attraction. This results in a change of state gradually, from gas to liquid and to solid. Melting and boiling Points Covalent compounds possess low melting and boiling points. Reason: Melting and boiling involves the breaking of intermolecular force of attraction between the molecules. As the intermolecular forces are weak between the molecules, less energy is required to break them. Hence, melting and boiling points of covalent compounds are low. Covalent solids are soft, with a low melting point. But diamond (carbon), even though a covalent compound, is the hardest substance known to man. Also among all the elements, diamond (carbon) has the highest melting point (3900°C). Because, each carbon atom in diamond is strongly bonded with the four adjacent carbon atoms by covalent bonds. So, we have a network of strongly bonded carbon atoms. This results in a closely-packed threedimensional structure. This makes diamond the hardest of all the substances. As more energy is needed to set the carbon atom free from this network of strongly bonded carbon atoms, the melting and boiling points are very high for diamond.

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9th Class Chemistry

122 Solubility: Covalent compounds are soluble in nonpolar solvents. Reason: Like dissolves like. Covalent compounds have molecules without the opposite polarity and hence they are non-polar compounds. Therefore, covalent compounds are soluble in non-polar solvents like kerosene, benzene, ether, carbon disulphide, carbon tetrachloride, etc. Covalent compounds, being non-polar, are insoluble in polar solvents like water. But some covalent compounds like alcohol, urea and sugar are soluble in water. Because, this is due to the attraction between covalent molecules like alcohol, urea, sugar and the water molecule. This attraction is called “Hydrogen bonding”. The attractive force of Hydrogen bond pulls out-the urea molecule from its solid phase into the solvent (water) and the urea gets dissolved.[Note: We shall learn about Hbonding in detail, in future classes]. Conductivity Charge carriers are required for substances to conduct electricity. Ionic solutions have charge carriers in the form of mobile ions. Metals have charge carriers in the form of electrons. But covalent compounds do not have any charge carriers and hence they are bad conductors of electricity. Covalent compounds do not conduct electricity. But graphite is a good conductor of electricity even though it is a covalent compound. Graphite has a

structure comprising layers. Each layer consists of a network of ‘hexagonal carbon rings’. Each carbon atom in the ring is covalently bonded to three adjacent carbon atoms. Therefore, out of the four valence electrons of carbon, three get trapped in the covalent bond, leaving the fourth electron free. This free electron in graphite is responsible for the conductivity of graphite. Speed of reaction Reactions of covalent compounds are slow. Reason: The reactions of covalent compounds involve, firstly, the breaking of the existing bonds in the molecules and then the formation of new bonds. Breaking of bonds requires extra energy, and until sufficient energy is available, the reaction is not initiated. Further, to initiate the reaction, the reacting molecules should collide. And among all the collisions, only a small fraction is fruitful. Since a covalent reaction involves a number of operations, it is slow. Directional property Covalent bonds are directional. The direction is along the inter-nuclear axis (line joining the nuclei of two reacting atoms). This directional property leads to different arrangement of the atoms along the line of direction. The result being that different molecules are formed with the same structure. This phenomenon is called Isomerism and the molecules are called Isomers.

Comparison between ionic and covalent compounds

Ionic compound 1. Crystalline solids at room temperature. 2. High melting and boiling points. 3. Hard and brittle.

Covalent compounds Gases, liquids or soft solids under ordinary conditions. Low melting and boiling points with the exception of giant molecules. Soft and waxy with the exception of giant molecules. Usually insoluble in water and in polar solvents. Soluble in non-polar solvents. Bad conductors of electricity with a few exceptions having layer lattice structure.

4. Highly soluble in water and polar solvents. Insoluble in non polar solvents 5. In solid state, bad conductors of electricity. Good conductors in molten state and in solutions. 6. Undergo ionic reactions. Rate of reactions are Undergo molecular reactions. Rate of very high. Reactions are fast and instantaneous. reactions are low. Reactions are slow.

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Chemical Bonding

123

Formative Worksheet 52. Substance X has a melting point of 1500°C. It conducts current at room temperature. Substance Y is gas at room temperature. It does not conduct current. Substance Z is soluble in water and the solution conducts current. However, in solid state Z does not conducts current but melts at 1020°C to form a conducting liquid. Deduce whether X, Y, Z. i) have giant or molecular structure. ii) are bonded covalently or ionically. X a) Giant structure

Y

Z

Covalent

Ionic

Giant structure

Ionic

Covalent b) Covalent

Covalent c) Giant structure

Ionic

Covalent

Giant structure

Covalent

Ionic d) Ionic

Ionic

53. Which of the following statements about LiCl and NaCl is wrong? a) LiCl has lower melting point that NaCl. b) LiCl dissolves more in organic solvents whereas NaCl does not. c) LiCl would ionise in water more than NaCl. d) Fused LiCl would be less conducting than fused NaCl. 54. CCl4 is insoluble in water because, a) H2O is non-polar. b) CCl4 is non-polar. c) They do not form inter molecular H–bonding. d) They do not form intra molecular H–bonding. 55. Which of the following when dissolved in water forms a non conducting solution? a) Green vitriol b) Chile or Indian salt petre c) Alcohol d) Potash alum 56. Assertion (A): Graphite is a good electrical conductor. Reason (R): The free electrons in graphite conducts electricity. a) Both assertion and reason are correct and rea son is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct.

57. Which of the following statements is incorrect? a) Sodium hydride is ionic. b) Beryllium chloride is covalent. c) CCl4 gives a white ppt. with AgNO3 solution. d) Bonds in NaCl are non-directional. 58. Which of the following statements is incorrect? i) Covalent bond is highly directional. ii) Ionic compounds do not exhibit space is isomerism. iii) Molecules react quickly than ions. a) Both ii and iii b) Both i and ii c) Both i and iii d) Only iii 59. The following are some statements about the characteristics of covalent compounds: i) The compound formed by the combination of a metal and non-metal must be a covalent compound. ii) Most of the covalent compounds are gases at room temperature. iii) All organic compounds are covalent compounds. The correct combination is a) all are correct. b) only i and ii are correct. c) only ii and iii are correct. d) all are incorrect. 60. Generally covalent compounds are insoluble in polar solvents. But, covalent substances like urea, glucose, alcohol, etc of soluble in water. This is because of a) high electronegativity. b) urea, glucose, alcohol are polar in nature. c) water is non polar in nature. d) Formation of hydrogen bond.

Conceptive Worksheet 54. Assertion (A): Chlorine is a gas where as Bromine is a liquid. Reason (R): Vander waal forces of attraction are more in bromine than in chlorine. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 55. Which of the following species possesses giant network covalent structure? a) Solid CO2 b) Ice c) SiO2 d) NaCl

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9th Class Chemistry

124 56. Assertion (A): Melting point of diamond is very high. Reason (R) : Ionic bonds are present in diamond. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 57. Which of the following is not a property of covalent compounds? a) They have low melting points. b) They are not electrical conductors. c) They exhibit space isomerism. d) They undergo chemical reactions quickly. 58. Which of the following is not a characterstic property of a covalent compound? a) It has low melting point and boiling point. b) It is formed between two atoms having no or very small electronegativity difference. c) They have no definite geometry. d) They are generally insoluble in water 59. Which elements forms a net work of covalent bonds in the solid state? a) Li b) O2 c) Cl2 d) C

11. Polar covalent bond and its formation Polar covalent bond We have seen that covalent bond is all about sharing of electrons. Consider a covalent compound AB formed from A and B. If both A and B exercise the same amount of influence on the shared electrons, then A and B are said to share the electron pair equally. This is seen in H2, where both atoms A and B belong to the same element. Such compounds are called non-polar covalent compounds. It is interesting to see what happens when one of the atoms in the covalent bond influences the shared electron pair, more than the other. If this happens, then the electron pair is said to be shared unequally between A and B. During the unequal sharing, the atom which exercises more influence on the electron pair gets www.betoppers.com

the partial negative (  ) charge and the other atom gets the partial positive (  ) charge. For example, in H2O molecule the electronegativity of “O” is more, and its influence is greater on the shared pair. This results in a partial negative charge on oxygen and partial positive charge on hydrogen and electron pair is unequally shared.

 H

 O

 H

The covalent bond formed due to unequal sharing of electrons is called a Polar covalent bond and the molecule is called a Polar molecule. Examples of polar molecules are HF, NH3, HCl, etc. Properties of polar covalent compounds Physical state: They are generally gases or liquids or soft solids. Molecules HCl H 2S HF H 2O State Gas Gas Liquid Liquid

Reason: The molecules are acted upon by weak intermolecular forces which is comparatively more than vander Waal‘s force, yet fairly small that the molecules cannot be held firmly. They also exist in the gaseous, liquid or soft-solid state for the same reasons, as described in nonpolar covalent compounds. Melting point: They have generally low melting and boiling points. Molecules

H 2O

Melting point 0 °C Boiling point 100°C

H Cl

NH3

-113 °C -77.7 °C 85 °C - 35.5 °C

Reason: The molecules are held by weak intermolecular forces. Thus, a very small amount of energy is required to break the bonds between molecules. Thus, these compounds have low melting and boiling points. Solubility: Polar covalent compounds have free ions. Thus, they are soluble in polar covalent solvents like water and in doing so, form ions. They are also soluble in non-polar covalent liquids. Conduction of electric current: Polar covalent compounds have free ions and hence, dissociate in water. Thus, they are good conductors of electricity.

Chemical Bonding

125

Comparison between non-polar and polar covalent compounds

Non-polar covalent compounds

Polar covalent compounds

1. Nature of sharing The shared pair of electrons are The shared pair of electrons are unequally distributed between the equally distributed between the atoms. atoms. 2. Force of attraction As the sharing of electrons is equal, the atoms attract each other with the equal force.

As the sharing of electrons is unequal, the force of attraction is more towards the more electronegative element. 3. Symmetry and charge

Since there is no electronegative or less electronegative difference between the combining atoms, a non -polar molecule is symmetrical and electrically neutral.

As there exists considerable electronegative difference between the combining atoms, the polar molecules are non symmetrical and charged. The atom which attracts the electrons more towards itself acquires partial negative charge and the other atom acquires

4. Combining atoms Non-polar covalent bond is generally formed between two atoms (i) having zero E.N. difference. Eg: Atoms of the same element H 2 , C l2 , O 2 , N 2 etc. (ii) having less E.N. difference. Eg: CH 4 , CCl 4 , BeCl 2 , etc.

Polar covalent bond is generally formed between (i) dissimilar atoms such as in H 2O, NH 3. (ii) atoms having different electronegativity and atomic radii, such as in H -Cl, H -F, H -Br, H -S -H.

Formative Worksheet 61. A polar covalent bond shows 50% ionic character. Then, the electronegativity difference of the atoms that form the bonds. a) More than 1.7 b) Equal to 1.7 c) Less than 1.7 and greater than 1.5 d) Much greater than 1.7

62. Assertion (A):

The covalent HCl molecule is polar in nature. Reason (R): Because the electronegativity of hydrogen is less (2.1eV) than that of chlorine (3.0 eV) a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. www.betoppers.com

126 63. The electronegativity values of C, H, O and N are 2.5, 1, 3.0 and 3.5 respectively. The most polar bond is a) S - H b) O - H c) N - H d) C - H 64. Which of the following molecules is non polar but it contains polar bonds in it?. a) BCl3 b) H2 c) NH3 d) CHCl3 65. Assertion (A): Though flourine is more electronegative than silicon, SiF4 is non-polar. Reason (R): The four fluorine atoms have thesame force of attraction on silicon. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 66. Assertion (A): HF is the compound most likely to contain H+ ions. Reason (R): Since it has the greater difference in electronegativity between H and F. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 67. Assertion (A): Polar molecules have higher melting and boiling points than non polar compounds of similar nature and molecular size. Reason (R): In addition to Vander waal’s attractive forces, there is also dipole dipole attraction. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct.

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9th Class Chemistry

Conceptive Worksheet 60. A chemical bond in which the electron cloud isunevenly shared is known as a/an _____ bond a) Ionic b) Covalent c) Co-ordinate d) Polar covalent 61. If the electron pair forming a bond between two atoms A and B is not the centre, then the bond is: a) Single bond b) Polar covalent bond c) Non-polar bond d) Double bond 62. Hydrogen chloride molecule contains: a) Polar covalent bond b) Double bond c) Co-ordinate bond d) Electrovalent bond 63. The compound(s) containing polar covalent bonds is(are): a) HCl b) NH3 c) H2 O d) HF 64. Which one of the following compounds is non-polar? a) NH3 b) HCl c) CCl4 d) H2O 65. Which of the following bond has the most polar character? a) C - O b) C - Br c) C - F d) C - S 66. A non polar molecule is ___ and ____ therefore, no separation of charge takes place in its molecule. a) Symmetrical and electrically neutral b) Unsymmetrical and electrically neutral c) May be symmetrical or unsymmetrical d) Symmetrical and not electrically neutral 67. The two compounds that are covalent when taken pure but produce ions when dissolved in water. a) NaCl, NaBr b) H2O, Cl2 c) HCl, H2O d) HF, NH3

12. Co-ordinate covalent bond The bond in which the shared pair of electrons is contributed by only one atom but, is shared by both atoms or molecules is known as co-ordinate covalent or dative bond. Sidgwick and Powell proposed the co-ordinate covalent bond. Eg : NH3  BF3, NH4, NH4 , H3O+ Formation of NH3  BF3 : The central atom of boron (B) is covalently bonded with three “F” atoms and has six electrons in its outermost shell. This configuration is known as sextet configuration. For completing the octet configuration, boron is falling short of two electrons.

Chemical Bonding

127

H

The central atom of ammonia is nitrogen. It has one lone pair in its outermost shell. In BF3 molecule, the central atom ‘B’ needs two electrons for the octet, and in the second molecule NH3, the central atom ‘N’ has one lone pair, which is lying idle. Nitrogen shares its lone pair with Boron as shown below, resulting in a special covalent bond. This is known as the co-ordinate covalent bond or the dative bond. In this type of covalent bond, the shared pair of electrons is contributed by one of the participating atoms alone, but is shared by both of them.

H

H

F

N

B

H

F

F

H

H

F

N

B

H

F

H

H

Hydrogen ion needs two electrons to attain stability. Therefore, when ammonia (NH3) combines with hydrogen ion, the lone pair of ammonia is shared with hydrogen ion results in the formation of dative bond. The formation of dative bond can be shown as follows:

H H

F

(i) Carbon monoxide (CO):

C O

(ii) Ozone (O3): (iii) Sulphur dioxide (SO2) :

O = OO O = S O

(iv) Sulphur trioxide (SO3) :

O

S

O

O

The central atom of water is oxygen. It has two ii) lone pairs of electrons. The hydrogen ion needs two electrons to attain stability. When H2O and H+ are combined, oxygen iii) shares its one of the lone pairs with H+, resulting in the formation of dative bond. The formation of hydronium ion is shown iv) below: 

H H

H

Some examples of molecules containing co­ordinate covalent bonds:

i)

H

N H

The atom which contributes the shared pair is called the donor atom, and the other atom which makes use of it is known as the acceptor atom. The coordinate covalent bond is represented by an arrow (  ) directed from the donor to the acceptor.. According to Lewis, a base is one, which donates a lone pair of electrons and an acid is one which accepts the lone pair of electrons. So, in a coordinate covalent bond donor atom called as Lewis base and accepted atom called as Lewis acid. Formation of hydronium ion (H3O+) from water molecule and hydrogen ion: Electron dot structure of H2O:

O

N

v)

Formation of ammonium ion ( NH+4 ): Ammonium ion is formed by the combination of vi) ammonia (NH 3 ) and hydrogen ion (H +). In ammonia, the central atom is nitrogen and has one vii) idle lone pair of electrons as shown below:

Properties of coordinate covalent compounds The properties of coordinate compounds are intermediate between the properties of electrovalent compounds and covalent compounds. The main properties are described below: Physical state: These exist as gases, liquids and solids under ordinary conditions. Melting and boiling points: Their melting and boiling points are higher than pure covalent compounds and lower than pure ionic compounds. Solubility: These are sparingly soluble in polar solvents like water but readily soluble in non-polar (organic) solvents. Stability: These are as stable as the covalent compounds. The addition compounds are, however, not very stable. It is also a strong bond because the paired electrons cannot be separated easily. Conductivity: Like covalent compounds, these are also bad conductors of electricity. The solutions or fused mass do not allow the passage of electricity. Molecular reactions: These undergo molecular reactions. The reactions are slow. Isomerism: The bond is rigid and directional, thus, coordinate compounds show isomerism.

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9th Class Chemistry

128 viii) Dielectric constant: The compounds containing coordinate bond possess high values of dielectric constants. Special type bond: Hydrogen bond: Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of the other molecule. Examples : HF, H2O, NH3, C2H5OH, etc. Note: Hydrogen bonding will be discussed in detail in future classes.

Formative Worksheet 68. In which type of bond formation, can a proton participate? a) Hydrogen bond b) Electrovalent c) Dative d) Covalent 69. Which of the following combination is best explained by the coordinate covalent bond ? a) H2 + I2 b) Mg + 1/2 O2 c) Cl + Cl d) H+ + H2O 70. Dative bonds are present in: a) NH4Cl b) SO2 c) H2S d) HRr 71. A. CuSO4.5H2O I. Ionic, covalent and dative bonds B. K4[Fe(CN)6]

ii. Covalent and hydrogen bonds

C. Ice

iii. Ionic and covalent bonds

D. Solid NaOH

iv. Ionic, covalent, dative and hydrogen bonds

72. In which of the following molecules, both the donor and acceptor are molecules is: a) Ammonia Boron trifloride b) Hydronium ion c) Carbonmonoxide d) Ammonium ion 73. “The mechanism of co-ordinate linkage may be suggested as a combination of electrovalent and covalent linkage.” Explain: 74. Explain the structure of: a) Hydronium ion b) Ammonium ion

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Conceptive Worksheet 68. In which of the following molecules, the shared pair of electrons is contributed by only one individual atom or molecule? a) NH3 b) CO c) H2 O d) CO2 69. NH3 and BF3 form an adduct readily because, they form a) A coordinate bond b) A covalent bond c) An ionic bond d) A hydrogen bond 70. Which of the following contains a co-ordinate bond? a) Water b) Ammonia c) Ammonium ion d) Ethylene 71. Dative bond is present in the molecule of a) NH3 b) CO2 c) N2 O d) PCl5 72. The bond formed between a Lewis acid and a Lewis base is a) Ionic bond b) Covalent bond c) Dative bond d) Hydrogen bond 73. Substances which can contribute pairs of electrons to form a co-ordinate covalent bonds are called a) Lewis acids b) Lewis bases c) non-polar d) amphiprotic 74. Co-ordinate covalent bond is HNO3 is present: a) Between Nitrogen atom and Hydrogen atom. b) between 2 Oxygen atoms. c) between nitrogen atom and oxygen atom. d) None of the above. 75. In N2O5, the donor atom(s) present is/are: a) Nitrogen atom b) Oxygen atom c) Two Nitrogen atoms d) Two Oxygen atoms

13. Deviation from octet rule

i)

There are several stable molecules known, in which the octet rule is violated, i.e., atoms in these molecules have electrons in the valence shell either short of octet or more than octet. Some important examples are: BeCl2 molecule: BeCl2, (beryllium chloride) is a stable molecule. Here, each beryllium atom contributes two of its valence electrons forming two single covalent bonds with two chlorine atoms i.e., after the bond-bond formation, Be has four electrons in its outer shel1.

Chemical Bonding ii)

129

BF 3 molecule: In BF 3 , each boron atom contributes three of its valence electrons forming three single covalent bonds with three fluorine atoms. After the bond formation, “B” has six electrons in the outer shell.

iii) PCl 5 molecule: Phosphorus atom has five electrons in the valency shell. It forms five single covalent bonds with five chlorine atoms utilising all the valence electrons. By doing so, it attains 10 electrons in the outer shell.

iv) SF6 molecule: Sulphur atom has six electrons in the valency shell. It forms six single covalent bonds with six fluorine atoms utilising all the valence electrons and thereby attains 12 electrons in the outer shell. F

F F

F F v)

F

IF7molecule: Iodine forms seven single covalent bonds with seven fluorine atoms utilising 7 valence electrons. The iodine atom attains 14 electrons in outermost shell. F F

F

I

F F

F F

Method of writing Lewis dot structures of the molecules: Although Octet rule and Lewis structures do not present a complete picture of covalent bonding, they do help to explain the bonding scheme in many compounds and account for the properties and reaction of molecules. For this reason, we should practice writing Lewis’ structures of the compounds.

The basic steps involved in drawing the structures are as follows: Step l: Write the rough skeletal structure of the compound using chemical symbols, placing the bonded atoms one on another. In general, the least electronegative atom occupies the central positions and other atoms occupy the surrounding positions. It is to be noted that the positions of hydrogen and fluorine are always terminal in Lewis’ structure. For example, the skeletal structure of CO2 is OCO, that of CH4 is H HC H, NH3 is , HN H etc., H

H

Step 2: Count the total number of valence electrons present in the given molecule. The counting of valence electrons should be done as follows: (i) For a neutral molecule, the number is equal to the sum of valence electrons of all the atoms present in a molecule. Example: Number of valence electrons present in H2O = 2 × no. of valence electrons of hydrogen + No. of valence electrons of oxygen. = (2 × l) + (l × 6) = 8 (ii)For polyatomic anions, add the number of negative charges to the total. Example: Total number of valence electrons present in SO24 : = (Total no. of valence electrons present in SO4) + (number of negative charges). = (No.of valence electrons present in S) + 4(no. of valence electrons present in O) + total number of negative charges. = 6 + 4(6) + 2 =6 + 24 + 2 = 32. (iii) For polyatomic cations, subtract the number of positive charges from the total. Example: Total number of valence electrons present in NH+4 = (Total no. of valence electrons present in NH4) – (Number of positive charges) = (No. of valence electrons present in N) + 4(no. of valence electrons present in H) –Total number of positive charges. = 5 + 4 (1) – 1 = 5 + 4 – 1 = 8 Step 3: Draw a single bond (electron pair) between the central atom and each of the surrounding atoms. Determine the total number of electrons shared between the central atom and the combining atoms. This is equal to double the number of single bonds present between the central atom and surrounding atoms. www.betoppers.com

9th Class Chemistry

130 Step 4: Distribute the remaining electrons as unshared pairs so that each atom (except hydrogen) has 8 electrons if possible. Step 5: After completing the above steps, if central atom has fewer electrons, than octet, we convert the single bonds to double bonds or triple bonds by using lone pairs from the surrounding atoms to complete the octet of the central atom.

Formative Worksheet 75. For each of the following molecules, identify the number of electrons around the central atom. (i) PF5 (ii) SF6 (iii) H2SO4 (i) (ii) (iii) a) 5 6 6 b) 10 12 12 c) 10 12 6 d) 10 6 12 76. Which of the following represents right Lewis structure of NF3? a)

F N

F

b)

F N F

F N F F

O H O P O H O H

d) Its bond structure can be represented as

O  H–OPO–H

O O C O

O

d) None

H

77. Which of the following is the right Lewis representation of carbonate ion? a)

a) A is true, B is false b) A is false, B is true c) Both A and B are true d) Neither A nor B is true 79. Which of the following is true for phosphoric acid? a) The total number of valence electrons in it are 32 b) It has two coordinate covalent bonds c) The correct Lewis structure is

F

F c)

O H O S O H O

2–

80. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are wrongly shown. Write the correct Lewis structure for acetic acid.

H

H

O

C

C

O

H

2–

b)

O O C O

H

81. Draw Lewis structures for the followings : (a) PCl3, (b) HCN, 2–

O

c)

O

C O

d) All 78. A) The Lewis dot structure of nitric acid is

H O N O O B) The Lewis dot structure of sulphuric acid is www.betoppers.com

(c) ClO 3 ,

(d) NO+.

82. Draw dot and cross diagrams for the following molecules. Then use electron repulsion theory to predict their shapes. (i) CCl4, (ii) H2S, (iii) NF3, (iv) NO, (v) AlCl3

Chemical Bonding

131

Conceptive Worksheet 76. Assertion(A) : Molecules satisfying octet rule are only stable.

14. Lewis Dot Structures of Some Molecules (A) Molecules containing ionic bonds only

Reason (R) : In a additional to octet configuration, stability also depends on symmetry of a molecule and energy. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 77. The octet rule is not valid for the molecule a) CH4

b) Cl2

c) N2

d) SCl4

78. A) The number of electrons around the central atom in BF3 is 3. B) The number of electrons around the centre atom in AlCl3 is 6. a) A is true, B is false b) A is false, B is true c) Both A and B are true d) Neither A nor B is true 79. Which one is an electron deficient compound? a) ICl

b) NH3

c) BCl3

(B) Molecules Containing covalent bods only

d) PCl3

80. The total number of valence electrons in SO42: a) 30

b) 32

c) 12

d) 14

81. In poly atomic molecules, the central position is occupied by _______ and the peripheral positions are occupied by ___________ respectively. a) Less electropositive, least electronegative. b) Less electronegative, least electropositive. c) Less electronegative, more electronegative. d) None 82. The number of dative bonds in NaOH is: a) 3

b) 2

c) 1

d) 0

83. Write the Lewis dot structure of (a) PH3 and (b) BCl3. Is the octet rule obeyed in these structure?

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132

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9th Class Chemistry

Chemical Bonding

(C) Compounds containing both ionic & covalent bond s

133

(E) Compounds containing electrovalents, covalent and co-ordinate bonds

(D) Molecules containing covalent & co-ordinate bonds

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9th Class Chemistry

134

Additional Information

i)

ii)

Shapes of unit cells Broadly speaking, the shapes of a unit cell are of the following types: Simple or primitive: In this type, atoms, ions or molecules represented by points, are present only at the corners of the unit cell.

Face-centred: In this type, points are present at the corners as well as at the centres of each of the six faces.

iii) Body-centred: In this type, points are present at the corners and an additional point is present at the centre of the unit cell.

iv) End face-centred: In this type, points are present at the corners and at the centres of the two opposite faces.

We have many Na+ and Cl– ions arranged in 3-D pattern. So “ NaCl” is actually (Na +)n (Cl–)n or (Na+Cl–)n. Such a regular arrangement of oppositely charged ions is called lattice. In an ionic compound, we have opposite ions arranged in 3-D pattern.

Let us understand the structure of ionic solids by considering the example of a common ionic www.betoppers.com

compound, the common salt (NaCl). Face-centred cubic (FCC): Unit cell of NaCl has an FCC structure. In this structure, if one type of ions are present at the eight corners of the cube, the opposite ions are present at the centres of the six faces.

The cube consists of eight corners (A, B, C, D, E, F, G H). It has six faces. The top and bottom faces are ABCD & EFGH with their centres as T and U, the left, right faces are ABFE and DCGH with their centres as P and R and the front, and back faces are, DHE & BCGF with their centres as S and Q. Co-ordination number : The number of oppositely charged ions surrounding a positive or negative ion in a crystal lattice is called co-ordination number. Let us have look at the NaCl structure by considering many unit cells, as shown below. It is seen that each black dot (Cl–) is surrounded by six small circles (Na+). Similarly, each small circle (Na +) is surrounded by six black dots (Cl–).

Sodium Chloride Structure So in a NaCl lattice, each ion is surrounded by six opposite ions. Thus in NaCl the coordination number is 6. Body - centred cubic (BCC) Unit of CsCl has the BCC structure. If one type of ion (Cl–) is present at eight corners of the cube the opposite ion (Cs+) is present at the centre of the cube and vice-versa.

Chemical Bonding

135 surrounded by water molecules. This process involves hydration of Cl– ions.

(a)

(b) Unit cell of CsCl In figure (a) ‘Cs’ is surrounded by eight oppositely charged Cl – ions and in Fig (b) Cl– is surrounded by eight oppositely charged Cs+. As each ion is surrounded by eight opposite ions, the co-ordination number CsCl is eight. Mechanism for solubility Best solvent suited for dissolving ionic compounds – The universal solvent: They are solvents whose molecules have opposite charges, such molecules are called polar molecules and the solvent is called a polar solvent. Water is a polar solvent. This is because the water molecule (H2O) has partial positive charge on the hydrogen atoms ( ) and partial negative charge on the

The polar nature of a solvent depends upon the magnitude of the opposite charges on the solvent molecules. The more the magnitude of opposite charges, the more the polar nature of the solvent molecules. The polar nature of a solvent can be understood with the help of a constant called dielectric constant. The greater the dielectric constant, the more the polar nature of the solvent. Among all the solvents, water has the maximum dielectric constant value of 80. It means that, water has the capacity to reduce 80 times the electrostatic force of attraction among the opposite ions of an ionic compound. So the opposite ions of an ionic compounds are set free in water easily. Hence, water is the best solvent for ionic compounds. Similarly, the positive ion (Na+) in NaCl crystal gets hydrated as follows:

oxygen atom () .

Molecules, which do not have opposite charges on them, are called non-polar molecules and the solvent is a non-polar solvent. Dissolution of NaCl in water: When NaCl, an ionic solid is dropped into water, the positively charged hydrogen atoms of water come closer to the negative ions (Cl–) at the crystal surface. The resulting attraction pulls the negative ions (Cl–) into the solvent.

The energy released during the hydration process is called the hydration energy. Lattice energy and hydration energy and its influence on solubility: During the dissolution of an ionic compound in a polar solvent, two types of energy operations occur: 1) During the first operation, the ions from the crystal are separated and shifted into the solvent. This requires energy, and the energy that is absorbed is called lattice energy. 2) During the second operation, the ions are shifted into the solvent and get hydrated. This gives out energy, and the energy so released is called Hydration Energy. Solubility of an ionic compound in a solvent is more if the hydration energy is more than the lattice energy.

When the negative ion enters the solvent, it gets www.betoppers.com

9th Class Chemistry

136

Summative Worksheet 1.

2.

3. 4.

5.

6.

7.

8.

9.

An electrovalent compound does not exhibit space isomerism due to: a) Presence of ions b) High melting point c) Strong electrostatic force between constituent ions d) Non-directional nature of electrovalent bond The coordination numbers of cation and anion in NaCl crystal are respectively. a) 8, 6 b) 8, 8 c) 6, 6 d) 6, 8 Which is a covalent compound? a) RbF b) MgCl2 c) CaC2 d) NH3 The bond between two identical non-metal atoms has a pair of electrons: a) Unequally shared between the two b) Transferred fully from one atom to another c) With identical spin d) Equally shared between them Which of the following have more number of shared electrons with respect to two carbon atoms? a) Benzene b) Acetylene c) Ethane d) Ethylene A covalent bond is likely to be formed between two elements which: a) Have similar electronegativities b) Have low ionisation energies c) Have low melting points d) Form ions with a small charge Arrange the following in the order of increasing polarising power : Na+, Ca2+, Mg2+ and Al3+ a) Al+3 < Ca+2 < Mg+2 < Na+ b) Al+3 < Mg+2< Ca+2 < Na+ c) Na+ < Ca+2 < Mg+2 < Al+3 d) None In which of the following molecule, the central atom has three lone pairs of electrons: a) Ammonia b) Xenon difluoride c) Chlorine trifluoride d) Hydrogen sulphide Match the following: p) NH3 (i) Single q) N2 (ii) Double r) O2 (iii) Triple s) H2 p q r s a) i iii ii i b) ii i iii ii c) iii ii i ii d) i i ii iii

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10. Among the species NO2, CO2, NO2 & CN the one containing an unpaired electron is b) NO2 c) CN– d) NO2 Covalent compounds are soluble in a) polar solvents b) non-polar solvents c) concentrated acids d) all solvents Which of the following properties would suggest that a compound under investigation is covalent? a) It conducts electricity on melting. b) It is a non-electrolyte. c) It has a high melting point. d) It is a compound of a metal and a non-metal. Which of the following bonds have directional character? a) onic bond b) metallic bond c) covalent bond d) both covalent and metallic bonds Assertion (A): All covalent compounds are bad conductors of electricity. Reason (R): HCl in aqueous solution is a good conductor of electricity. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. The bond present in iodine monochloride is: a) Coordinate bond b) Electrovalent bond c) Covalent bond d) Metallic bond The compound, which contains both ionic and covalent bonds, is a) CH4 b) C2H2 c) KCN d) KCl The type of bonds present in ammonium chloride are a) Only ionic and dative b) Only covalent and electrovalent c) Only covalent and coordinate d) Ionic, covalent and coordinate The bonds present in N2O5 are: a) Ionic b) Covalent c) co-ordinate covalent d) Hydrogen The bonds present in [Cu(NH3)4] SO4 between copper and ammonia are a) Ionic b) Covalent c) Co-ordinate d) Hydrogen a) CO2 11.

12.

13.

14.

15.

16.

17.

18.

19.

Chemical Bonding 20. The types of bonds present in CuSO4.5H2O are: a) Electrovalent and covalent b) Electrovalent, covalent and co-ordinate c) Covalent and co-ordinate and co-ordinate d) Electrovalent

137 8.

9.

HOTS Worksheet 1.

2.

3.

4.

5.

6.

7.

An atom X has 2, 8, 7 electrons in its shell. It combines with Y having 1 electron in its valence shell. a) What type of bond will be formed between X and Y? b) Write the formula of the compound formed? An atom P has 2, 8, 7 electrons in its orbits. It combines with Q having one electron in its valence orbit. a) What type of bond will be formed between P and Q? b) Write the formula of the compound formed. Number of lone pairs of electrons in 9 gms. of water are: [N = Avogadro Number] a) 2N b) N/2 c) N d) N/4 The total number of valency electrons in

PO34 ion is: a) 32 b) 16 c) 28 d) 30 Match the following: Substance Nature A) Graphite i) Has ten electrons around its central atom B) AlF3 ii) Covalent liquid C) CCl4 iii) Ionic compound D) PCl5 iv) Covalent Polymer The correct matching is: A B C D a) iv ii i iii b) iv iii ii i c) iii ii i iv d) i iii iv ii a) Ammonium hydroxide b) Zinc sulphate c) Water d) Carbon tetrachloride e) Phosphorus trichloride Some of the first 20 chemical elements are H, C, N, O, F, Ne, Na, Mg, Al, Si, P, S and Cl, a) Choose any four of these elements and give the formula of a chloride of each of them. Indicate the nature of bonding (electrovalent or covalent) in each case. b) Choose any four of these elements and write the name and formula of a hydride compound of each with hydrogen.

10. 11.

12.

13.

On the basis of the knowledge of electronic configuration of elements, predict the nature of bonds in each of the following molecules. N2, CO2, HCl, MgCl2, CCl4, HCN Predict the type of bond (Ionic or covalent) likely to be formed when pairs of elements having following atomic numbers combine : a) 3 and 7 b) 6 and 9 c) 11 and 16 d) 8 and 16 The molecule that deviates from octet rule is a) NaCl b) BF3 c) MgO d) NCl3 Which of the following molecules are the examples of incomplete octet of the central atom? a) LiCl b) BeH2 c) BCl3 d) PCl5 Which of the following molecule does not obey the octet rule and also has lone pair on the central atom? a) CCl4 b) PCl3 c) NH3 d) SCl4 Octet rule is not followed in: a) SF6 b) PF5 c) BeCl2 d) All the three

14. The total number of valence electrons in NH4 ion a) 8 b) 9 c) 10 d) 11 15. Which of following is true for N2O ? a) It is also called laughing gas b) Its rate of diffusion is same as carbon dioxide and propane (C3H8) c) Its bond structure is N  N  O d) Its electron dot formula is

N N O

IIT JEE Worksheet 1.

2.

3.

4.

Given electronic configuration of four elements E1, E2, E3 and E4 are respectively 1s2, 1s2 2s2 2p2, 1s2 2s2 2p5 and 1s2 2s2 2p6 . The element which is capable of forming ionic as well as covalent bonds is: a) E1 b) E2 c) E3 d) E4 Which of the following has both polar and non-polar covalent bonds? a) S8 b) H2S c) H2 d) H2O2 The bonds present in K4[Fe(CN)6] are a) All ionic b) All covalent c) Ionic & covalent d) Ionic, covalent & co-ordinate covalent The type of bonds present in Na[Ag(CN)2] are: a) Ionic and covalent b) Covalent and dative c) Ionic and dative d) Ionic, covalent and dative www.betoppers.com

9th Class Chemistry

138 5.

Both ionic and covalent bonds are present in: a) KCl

6.

b) IF5

c) K2O

9.

List - 1

d) KOH

List - 2

q) Conductor of electricity 2) Ice

p) NaCl

1) Covalent bond

r) Good ionic solid

q) CH 4

2) Ionic bond

r) NH+4

3) Metallic bond

s) Cu metal

4) Covalent and dative bond

3) Diamond

s) Hydrogen bonded solid 4) Graphite 5) Zeolite a) p  3, q  2, r  1, s  5 b) p  2, q  3, r  2, s  4

a) p  2, q  1, r  4, s  3

c) p  3, q  4, r  1, s  2

b) p  2, q  4, r  1, s  3

d) p  5, q  4, r  1, s  2

c) p  2, q  3, r  1, s  4

10.

d) p  1, q  2, r  3, s  4 List - 2

p) C2 H 6

1) Ionic bond

q) Iron wire

2) Metallic bond

r) H 2 O

3) Coordinate, covalent bonds

s) H3 O

+

ii) The compound having monatomic cation and monatomic anion contains only ionic bond. iii) The compound having dative bond must possess covalent bond also. The correct combination is:

4) Covalent bond

a) All are correct

5) Coordinate

b) Only i and ii are correct

a) p  2, q  3, r  4, s  1

c) Only ii and iii are correct

b) p  3, q  2, r  4, s  4 c) p  4, q  2, r  4, s  3 d) p  1, q  5, r  2, s  3 Match the following:[L – II] List - 1

List - 2

The following are some statements about the type of chemical bond present in a given compound. i) All metallic hydroxides contains both ionic and covalent bond.

Match the following:[L – II] List - 1

8.

List - 2

p) Highly covalent material 1)Potassium chloride

Match the following: List - 1

7.

Match the following:[L – II]

d) Only i and iii are correct 11.

X,Y and Z are three substances. i) X does not conduct electricity in solid or liquid state. ii) Y conducts both in fused and solution state.

p) [Ni(NH3 )6]Cl2

1) Covalent bond

iii) Z conducts only in solution state.

q) Solid CO2

2) Covalent bond, Vander

What type of compounds X, Y and Z are in terms of linkage?

waals r) HF vapour s) C6 H 6

X

Z

a)

Non-polar covalent

Ionic

4) Hydrogen bond, covalent bond

Polar covalent

b)

No n-polar covalent

Polar covalent

Ionic

c)

No n-polar covalent

Ionic

Polar co valent

d)

Ionic

Non-polar covalent

Polar co valent

5) Metallic bond a) p  4, q  3, r  2, s  1 b) p  1, q  2, r  3, s  5 c) p  1, q  3, r  4, s  2 d) p  3, q  2, r  4, s  1

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Y

3) Ionic, Covalent and dative bond

Chemical Bonding 12.

139

Four elements coded A, B, C and D form a series of compounds e.g. AB, B2, CB3, DB2 and DB3. If the atomic number of these elements is not necessarily in the order are 13, 19, 26 and 35. What is the nature of bonding in each of them? CB3

DB2

DB 3

a)

Ionic

AB

Covalent

Covalent

Ionic

Ionic

b)

Covalent

Covalent

Ionic

Ionic

Covalent

c)

Ionic

Covalent

Ionic

Covalent

Ionic

d)

Ionic

Ionic

Covalent

Covalent

Covalent

13.

B2

State the kind of bond formed when atoms are: i) having zero electronegativity difference. ii) having small electronegativity difference. iii) having large electronegativity difference. i

ii

iii

a)

100%, non polar

More covalent, less ionic

More ionic less covalent

b)

More covalent, less ionic

100 % non polar

More ionic, less covalent

c)

More covalent, less ionic

More ionic, less ionic

100 % non polar

d)

Only covalent

Only ionic

100 % non polar

14. Consider the hypothetical elements X, Y and Z as:

X,

Y, Z

i) To which group in the periodic table these belong. ii) Write the electron dot structure for simplest compound of each with H2. iii) Write electron dot formulae for the ions formed when X, Y react with Na. 15. Choose the kind of chemical bonding from the ionic bond, covalent bond, both ionic and covalent bonds present in the following compounds. a) Magnesium oxide b) Ammonium nitrate c) Sulphuric acid

d) Carbon dioxide

e) Potassium chloride

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140

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9th Class Chemistry

Chapter -6

Gaseous State

Learning Outcomes

By the end of this chapter, you will understand • • • • • •

Comparative properties of different states of matter Unique properties of gases Measurable properties of gases Measurement of pressure Boyle’s law Charles’ Law

• • • •

Gay-Lussac’s law Avogadro’s Law Ideal gas equation Graham’s Law

1. Comparative properties of different states of matter The distinguishing features of the three states of matter - solids, liquids and gases can best be understood from the following table.

Solids

Liquids

Gases

1.

Solids have definite shape and definite volume.

Liquids have no definite shape, but have a definite volume.

Gases have neither definite shape nor definite volume.

2.

Position of the molecules of solid is fixed.

Position of molecules of a liquid is not fixed. The molecules can change their position within the liquids.

Position of molecules of a gas is not fixed. The molecules can move freely in all directions.

3.

Solids cannot be compressed.

Liquids can be very slightly compressed.

Gases can be easily compressed.

4.

Solids cannot flow. They can be heaped.

Liquids flow from higher to lower level.

Gases can flow in all directions.

5.

Solids have very small intermolecular spaces.

Liquids have small intermolecular spaces, which is more than the solids.

Gases have very large intermolecular spaces.

6.

Solids have very strong intermolecular forces of attraction.

Liquids have less intermolecular forces of attraction than the solids.

Gases have negligible intermolecular forces of attraction.

7.

Solids have infinite number of free surfaces.

Liquids have only one free surface.

Gases have no free surface.

8.

Solids do not need a vessel to contain them.

Liquids need a vessel to contain them.

Gases need a vessel to contain them.

Note: The state of matter of a substance is determined by the temperature. The higher the temperature, the greater the randomness among molecules and hence the state of matter is decided. Another factor that determines the state of matter is pressure.

9th Class Chemistry

142

2. Unique properties of gases 1.

2.

3.

4.

5.

6.

No fixed volume or shape The force of attraction between the molecules of a gas is negligible. Thus, the molecules of a gas can fill the entire space of the container. Hence, the gas attains the shape and size of the containing vessel. Exert same pressure in all directions The molecules of a gas constantly bombard the sides of the containing vessel and hence, exert some pressure. High compressibility The decrease in volume of a given mass of gas either by increasing pressure or decreasing temperature is called compressibility. The high compressibility of the gases is due to the fact that they have large intermolecular spaces. On applying pressure, these molecules simply come close to each other, thereby decreasing the volume of a gas. High expansibility The volume of a given mass of a gas can be increased, either by decreasing pressure or by increasing temperature. This is called expansibility. When pressure on an enclosed gas is reduced, its molecules simply move apart, thereby increasing intermolecular spaces and hence, the volume increases. When an enclosed gas is heated, the kinetic energy of its molecules increases. Thus, the molecules move faster and farther from each other. This in turn results in the increase in volume. Low Density The gas occupies an extremely large volume as compared to solids and liquids. Thus, the mass per unit volume of a gas is very small as compared to the liquids and solids. This accounts for the low density of the gases. Exception: There also exists a gas which has more density than many solid metals. It is Xenon which is the most dense gas. Density of Xenon is 5.9 g/cc which is more than the density of many metals like Lithium, Sodium, Magnesium, Titanium, Aluminium, Bromine, Rubidium, Cesium, etc. High rate of diffusion The spontaneous intermixing of gases is called diffusion. The intermolecular spaces in a gas are very large. Thus, when two gases are brought in to contact with each other, their molecules just move into one another’s intermolecular space, thereby forming a homogeneous mixture.

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Formative Worksheet 1.

2.

3.

Which one of the following statements is not correct about the three states of matter i.e. solids, liquids and gases? 1) Molecules of a solid possess least energy, whereas those of a gas possess highest energy. 2) In general, the density of solids is highest energy, whereas that of gases is lowest. 3) Gases, like liquids, possess definite volume. 4) Molecules of a solid possess vibratory motion. Statement A: Gases have neither fixed volume nor fixed shape. Statement B: Gases neither exert pressure nor diffuse from one place to another. 1) Statement ‘A’ is false 2) Statement ‘B’ is true 3) Statement ‘A’ and ‘B’ are both false 4) Statement ‘A’ is true but ‘B’ is false Assertion: Gases are highly compressible. Reason : The force of attraction between the gas molecules is large and hence is compressible. 1) Both Assertion and reason are correct and reason is the correct explanation of assertion. 2) Both assertion and reason are correct and reason is not the correct explanation of assertion. 3) Assertion is correct, reason is incorrect. 4) Assertion is incorrect, reason is correct

Conceptive Worksheet 1. 2.

3. 4.

Name the word from which the term ‘gas’ was derived. Which of the following is not correct regarding gases? 1) Gases do not have definite shape and volume. 2) Volume of a gas is equal to volume of container confining the gas. 3) Confined gas exerts uniform pressure on the walls of its container in all directions. 4) None of the above. Which form of matter is highly compressible? Which of the following is true regarding the intermolecular force of attraction among the three different states of matter? 1) Solids > Liquids > Gases 2) Gases > Liquids > Solids 3) Solids = Liquids = Gases 4) Gases > Solids > Liquids

Gaseous State 5.

The reason for high rate of diffusion of gas molecules is : 1) The intermolecular spaces in a gas are very large. 2) The intermolecular force is very negligible. 3) Gas molecules move randomly in a straight line direction. 4) None of the above.

3. Measurable properties of gas The measurable properties of gases are volume, temperature, pressure and amount of gas.

1. Volume Gases always occupy the complete volume of the container on account of their high expansibility. Thus, the volume of a gas is always equal to the volume of the container. Units: a) 1 millilitre (1 ml) = 1 cm3 (1 cc) b) 1 litre (1l)= 1 cubic decimetre (dm3)

143 b) Kelvin temperature is measured in kelvin = K. [Details of Kelvin scale will be discussed in upcoming pages] However, rise or fall in temperature of 1 K = 1°C. c) Temperature in kelvin = 273 + temperature in °C, K = 273 + °C Note: Temperature in Kelvin scale is also called absolute temperature.

3. Pressure The gas molecules continuously bombard the walls of the container. This continuous bombardment results in a force on the walls of the container. The force exerted per unit area of the wall is the pressure of the gas.

Note: 1 litre = 1000 ml = 1000 cm3 = 10–3 m3 = 1dm3 Factors affecting volume: i) On compressing, the volume of a gas decreases. This indicates that volume is affected by pressure. ii) It is common observation that on heating, substances, including gases they expand. This indicates that volume is affected by temperature. iii) On blowing a balloon, it expands. This indicates that the volume of a gas increases with its amount of its number of molecules.

2. Temperature Temperature of a gas is expressed in terms of motion of gas molecules. Gas molecules are in continuous motion. A gas molecule moves along a straight line in any direction before it collides with another gas molecule. Millions of collisions occur in a second in a container containing a gas. Such a motion is called random motion. The Temperature is the index of random motion. The higher the temperature, the more the randomness. Let us take a look at the precise relationship between the motion of a gas molecule and its temperature. The amount of motion of a molecule is measured by the kinetic energy of a gas molecule. The average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. Units of temperature : a) Celsius temperature is measured in degrees Celsius = °C

So, we may say that the pressure of a gas is due to the bombardment of gas molecules on the walls of the container and is measured as force per unit area. Factors that decide the pressure of a gas:

Pressure of a gas is basically the force exerted per unit area of the wall of the container in which it is kept. The force depends on: i) The number of molecules striking a unit area of the wall per second (collision frequency) and ii) The velocity with which they strike. The absolute unit of pressure is atmosphere. The other units of pressure are bar, torr, cm of Hg and mm of Hg. 1 atm = 105 Pascal = 105 N/m2 = 106 dyne / cm2 = 76 cm of Hg = 760 mm of Hg = 760 torr.

4. Amount of gas The amount of gas is measured in terms of moles. 1 mole of any substance is its weight of equal to its gram molecular weight. For example, 1 mole of hydrogen gas weighs 2 grams. 1 mole of oxygen gas weighs 32 grams and so on. The formula to calculate no. of moles is:

Given weight of gas n = Gram molecular weight

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9th Class Chemistry

144

Formative Worksheet 4.

The radius of a spherical container is 10cm. What is the volume occupied by any amount of gas filled

5.

Conceptive Worksheet

7. 8. 9.

h

into it? 10 gm of a gas present in a cube of length ‘  ’ cm

is transferred to a cube of length 2 cm. What is the percentage change in the volume occupied by the gas? 6. A liquid of volume 1m3 is poured into litre flasks of ‘n’ in number. Find the minimum value of n, which can accommodate 1m3 of liquid. 7. The temperature variation at a place is 35°C. Find the same variation in Kelvin scale . 8. One atmosphere is numerically equal to approximately _____dynes cm–2 9. One atmosphere = x Newton/meter2 = y torr = z dyne/cm2. Find the values of x, y and z. 10. How many moles are present in 8 g of it? 11. What is the weight of 4 moles of nitrogen gas?

6.

Vacuum Connected to gas container

Which of the following is (are) the unit(s) of volume? 1) litre 2) millilitre 3) cubic metre 4) decimetre cube What is value of –2730C in Kelvin scale ? The value of 273oC in Kelvin scale is_________. Which one of the following is not the unit of pressure? 1) Newton 2) Torr 3) Pascal 4) Bar

4. Measurement of pressure The pressure of a gas is measured by a ‘Manometer ’ and the atmospheric pressure is measured by a ‘Barometer’. The practical unit of pr essure of a gas is atmospheres or centimetre of mercury column (cm of Hg) in a barometer. In SI system, the unit of pressure is Pascal (Newton /m2).

Manometer It consists of a J-shaped tube closed at the longer arm which has a vacuum at its end. The shorter arm is connected to the gas whose pressure is required to be measured. In the manometer the pressure of mercury is measured as the difference in the levels of the mercury of the two arms. If the difference is 76 cm at 0°C, then the pressure of the gas corresponds to one atmosphere. www.betoppers.com

Mercury

Manometer

Barometer A barometer is used to measure atmospheric pressure. It was devised by Evangelista Torricelli. Construction of a Barometer: Fill a 100-cm long tube of steel closed at one end completely with mercury. Invert this tube into a trough containing mercury. The mercury present at the closed end falls down leaving behind an empty space. This empty space is nothing but a vacuum, known as ‘Torricelli vacuum’. Vaccum

h

Air pressure

Barometer

The mercury column is supported by the force of atmospheric pressure. The height (h) of the mercury column indicates the atmospheric pressure. Note: i) If the atmospheric pressure is normal, then mercury level is equal to 76 cm. ii) If the atmospheric pressure is high, then mercury level will be more than 76cm. iii) If the atmospheric pressure is low, then the mercury level will be less than 76cm. Measurement of atmospheric pressure: Atmospheric pressure(P) can be obtained by a barometer, using the formula: P=h×d×g where, ‘h’is the height of liquid column, ‘d’ is the density of the barometric liquid and ‘g’is the acceleration due to gravity. If ‘h’ is expressed in meters, ‘d’ in Kg/m3 and ‘g”

Gaseous State in m/s2, the pressure obtained by this formula is in terms of Newton / m2 ’ or ‘dyne/cm2 ’. Pressure measured in this way is called Absolute pressure. In the formula ‘P = hdg’, the terms ‘d’ and ‘g’ for a given barometer are constant and the atmospheric pressure is proportional to the height of the Mercury column. So, in day - to - day usage, the atmospheric pressure is expressed in terms of the height of the Mercury column. The height of the Mercury level is an index of the atmospheric pressure. For normal atmospheric pressure, the height of the Mercury column is 760 mm or 76 cm. Uses of Mercury Barometer 1. Mercury bar ometers are used by Meteorologists, as the height of the mercury in the glass tube indicates the weather changes in the atmosphere in the following manner. Rise in the height of mercury level  Rise in the atmospheric pressure  Return of fair weather Rise in the height of mercury level  Rise in the atmospheric pressure  Return of fair weather No change in the height of mercury level  No change in the atmospheric pressure  No change in weather 2. It also indicates the change in altitudes. Implications of low atmospheric pressure: i. Artificial pressure in air crafts: Air-crafts travel at high altitudes. At such altitudes there will be 1017 molecules in one cubic centimetre of air whereas on the surface of the earth their presence is 1000 times more. Further, the weight of the air column also decreases with altitude. So, at a higher altitude the pressure becomes very much less and the passengers encounter dangers like bursting of body cells, bleeding of noses and ears and also an increase in the blood pressure. To avoid these problems an artificial pressure is created inside the plane. ii. Faster breathing on mountains: At the top of the mountains and high altitudes, there are a fewer number of molecules of air. Since we could inhale only fewer air molecules each time, we feel suffocated and tend to breathe faster to make up for the deficit. iii. Popping of ears: At the top of a tall mountain, the pressure is less. This causes our ears to pop in order to balance the pressure between the outer part and the inner part of our ears. Countering the absence of pressure in space. The Astronaut’s way : There is no external

145 pressure in space. Astronauts, while travelling through space, wear heavy space suits so as to create an external pressure. These space suits prevent the bursting of body cells and bleeding of nose and ears.

Formative Worksheet 12. An oil barometer of greater is height than a water barometer. True of false. 13. Explain what happens when a hole is made in the top portion of the inverted tube of a barometer. 14. What will be the height of mercury column in a barometer, it it is taken to the moon? 15. The column of mercury in a barometer is 76 cm Hg. Calculate the atmospheric pressure if the density of mercury = 13600 kgm–3. (Take g = 10 ms–2) 16. If the mercury in the barometer is replaced by water, what will be the resulting height of the water column? Density of water = 1000 kgm–3 density of mercury = 13600 kgm–3 17. Calculate the total pressure at the bottom of a lake of depth 5.1 m (atm. pressure = 105 P), density of water = 103 kg/m3,g = 10 m/s2. 18. At a given place a barometer records a pressure of 70 cm of mercury. If the mercury in the barometer is replaced by water, what would be the height of the water column? Take density of mercury = 13600 kg/m3, density of water = 1000 kg /m3

Conceptive Worksheet 10. a) The force exerted by atmospheric pressure on 1m2 area of the earth’s surface is approximately equal to the weight of two elephants. b) We are carrying an eternal load of 18 adults on our head. Substantiate the two statements made above. 11. Two million molecules per square inch beat against a window pane every second. This for ce is sufficient to power a steam engine or a turbojet. Such enormous force should shatter the glass into pieces. But this does not happen. Why? 12. Our body is accustomed to high atmospheric pressure on ground. If we encounter, a low atmospheric pressure as is the case on high altitude we will experience the bursting of body cells, bleeding of nose and a high blood pressure. But in space, where astronauts dare to travel, atmospheric pressure does not exist at all. How do they cope with such condition? www.betoppers.com

9th Class Chemistry

146 13. On mountain tops, we breathe more and our ears tend to pop. Why? 14. Name the instruments used to measure the pressure of atmosphere and gas. 15. Name the best suited liquid to use in a barometer. 16. At higher altitudes the passengers encounter dangers like bursting of body cells, bleeding of noses. Give reasons. 17. Assertion (A): At high altitudes, we breathe faster. Reason (R): At high altitudes, there are a fewer number of molecules of air. Since we could inhale only fewer air molecules each time, we feel suffocated and tend to breathe faster. 1) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. 2) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. 3) Assertion is correct and Reason is incorrect. 4) Assertion is incorrect and Reason is correct. 18. The weight of ¼ moles of a gas is 16 g. Identify the gas. 19. Which of the following contains maximum no. of moles? P : 4 g of He R : 48 g of O3 Q : 16 g of CH4

5. Boyle’s law At first, gas laws were propounded by Robert Boyle. Keeping the temperature (T) and amount of gas (m) as constants, he studied the variations of pressure and volume as follows: He took a ‘J’- shaped tube closed at one end and filled it with mercury. The gap between the closed end of the tube and mercury enclosed the volume of air (V). The level of mercury is the same in both arms of the tubes and hence the pressure acting on the enclosed air is equal to the atmospheric pressure (Pa). Patmospheric (Pa) Air

Patmospheric (Pa)

air. He, thus, pointed out the inverse relation between pressure and volume. Boyle’s law states that “the volume of a given mass of gas varies inversely with the pressure, provided the temperature remains constant”. Mathematically, V 

1 or PV = k P

(T= constant )  P1V1 = P2 V2 = P3 V3 = Pn Vn

GRAPHS Graph between P and V: The graph between P and V is a curve known as hyperbola.

P

V

The graphs drawn at constant temperature are known as isotherms. Graph between P and 1/V: The graph between P and 1/V is a straight line passing through the origin.

P

1/V

Graph between PV and V: The graph between PV and V is a straight line parallel to volume axis.

Air h

V

PV

J-Tube

V

Now, on adding more amount of mercury into the tube, he observed that volume of the enclosed air has decreased. This is because the addition of mercury results in more pressure on the enclosed www.betoppers.com

Graph between PV and P: The graph between PV and P is a straight line parallel to volume axis.

Gaseous State

147

PV

P

Modified Boyle’s law We can draw a relation between pressure and density of a given mass of gas at constant temperature as follows: We know PV = K (constant) as per Boyle’s law.

 P×



m P K = K  = = Constant d d m

22. A gas present in a cube of length ‘l’ is shifted to another cube of length 2 ?. What is the pressure of the gas in the 2nd case, if the pressure of the gas in 1st case is ‘P’ at same temperature? 23. A box of volume V having gas at pressure ‘P’ is connected to another empty box of volume 2V by a tube of negligible volume. Find the pressure exerted by the gas in the new box. 24. Which of the following is right for the following graph?

P

P1 P2 = d1 d 2 ( or ) P  d .

T3 T2 T1

The above equation is called Modified Boyle’s law.

V Some Problem Solving Tips and Understandings  Before substituting the terms in the formula, convert the terms into and also express them in the same system of unit. This is the rule for substitution.  Always express the answer in terms of given variables. This is the rule for expressing answers.  Volume is inversely proportional to pressure of the gas only when mass and temperature are kept constant.  If the terms in a problem are Pressure (P), Volume (V) and Temperature is constant, then Boyle’s Law is applicable.

Formative Worksheet 19. If 20cm3 gas at 1atm is expanded to 50cm3 at constant T, what is the final pressure? 20. A sample of a given mass of gas at a constant temperature occupies 95cm3 under a pressure of 9.962 × 104Nm–2. At the same temperature, what is its volume at a pressure of 10.13 × 104 Nm–2 21. A gas of volume 100cc. is kept in a vessel at pressure 104 Pa maintained at temperature of 24oC. If now the pressure is increased to 105 Pa, keeping the temperature constant, then find the volume of the gas

Hint: At constant volume, the pressure is directly proportional to its absolute temperature. 1) T1 = T2 = T3 2) T3 > T2 > T1 3) T1 > T2 > T3 4) T2 > T1> T3 25. A certain mass of a dry gas at 26°C and 760mm pressure has a density of 2.8gm/cc. What will be its density at 26°C and 740mm pressure?

Conceptive Worksheet 20. An air bubble rising up from the bottom of a pond. Which gas law does it demonstrate? 21. A gas occupied a volume of 250 ml at 700 mm Hg pressure and 250oC. What additional pressure is required to reduce the gas volume to its 4/5th value at the same temperature? 22. There is a gas in a sphere of radius ‘2r’ at a pressure ‘P’. If the same amount of gas is present in another sphere of radius ‘r’ at same temperature, find its pressure: 23. A certain mass of a dry gas at 26°C and 760mm pressure has a density of 2.8gm/cc. What will be its density at 26°C and 740mm pressure? 24. If the pressure of a gas becomes 5 times its original pressure, then the density of the gas changes from ‘d’ gm/lit to______ gm/lit.

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9th Class Chemistry

148

Kelvin Scale

6. Charles’ law This law was proposed by Charles and it gives the relation between the volume and absolute temperature. It states that “the volume of a fixed mass of a gas is directly proportional to its absolute temperature, provided the pressure remains constant”. Mathematically, V  T or

V K T

This scale of temperature was given by Lord Kelvin. The starting point of Kelvin scale is absolute zero i.e., –2730C, which corresponds to one Kelvin. The difference between any two successive points on the scale is same as that of centigrade scale.The Kelvin scale is also called absolute scale of temperature. T ( K ) = t (0C) + 273

Graph between V and T The graph between V and T is a straight line passing through origin as shown in the figure.

V1 V2 ( P= constant)  T  T 1 2 Charles' Law is based on his observation, according to which the volume of a given mass of a gas at

1 273 0 of its volume at 0 C for every degree of rise or fall of its temperature respectively. constant pressure increases or decreases by

t   Vt = V0 1+  where, Vt is the volume of the  273  gas at t0C and V0 is the volume of the gas 0°C. Based on Charles' observation, it was found that volume at –2730C should be expected to be zero. This temperature is called absolute zero. All the properties of the gases become zero at absolute zero.

Graph between Vt and t 70 60 50 40 30 20 10 50 O

50

100 150 200

273°C

The graph between Vt and t is a straight line as shown in the figure. From the graph it is clear that as we lower the temperature, the volume decreases. At a point –273 0 C and the volume of the gas corresponds to zero. www.betoppers.com

T

The graphs drawn at constant pressures are known as isobars.

Standard Temperature and Pressure As volume depends on temperature and pressure, they should invariably be mentioned during volume measurements. The standard temperature and pressure at which we measure the volume of gas is 0°C and 1 atmosphere of pressure. This temperature and pressure is called S.T.P ( Standard temperature and pressure) or N.T.P ( Normal temperature and pressure) Some Problem Solving Tips and Understandings

80

250 150

V

 During problem solving never to substitute the temperature in the centigrade scale. As a rule, convert the temperature in the centigrade scale into Kelvin Scale.  Do not write K to express temperature in the Kelvin scale.  Volume is directly pr oportional to absolute temperature only when pressure and mass of the gas are kept constant.  Cylinder with movable piston, rubber balloons, etc., are the examples of constant pressure system.  If the terms in a problem are Volume (V), Temperature (T) and Pressure is constant, then Charles’ Law is applicable.

 Whenever STP or NTP is mentioned in the problem, you need to immediately note the value of temperature as 0 C or 273K and the value of pressure as 1 atm or 760 mm of Hg as per the requirement.

Gaseous State

Formative Worksheet 26. A gas is enclosed in a vessel at standard temperature . At what temperature is the volume of enclosed gas 1/8 of its initial volume, keeping the pressure constant? 27. At what temperature is the volume of a gas is double the volume of a gas at room temperature (27°C)? 28. An open vessel contains 200 mg of air at 17°C. What weight percent of air would be expelled if the vessel is heated to 117°C? 29. The volume of a gas by 20% without changing the pressure. To what temperature the gas must be heated if the initial temperature of the gas is 27°C? 30. At constant pressure, what is the volume of air which escapes from a litre flask if the temperature changes from 27°C to 127°C? 31. A student forgot to add the reaction mixture to a round bottomed flask at 27°C but, instead he placed the flask on the flame. After a lapse of time, he realized his mistake, and, using a pyrometer, he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?

Conceptive Worksheet 25. 500ml of nitrogen at 27°C is cooled to –5°C at the same pressure. Calculate the new volume. 26. A cylinder has a movable piston. The piston is stationed at a height of x metres at 27°C. What should be the change in temperature such that the piston moves up by one-fifth of its original height? 27. At constant pressure, what is the volume of air escaped from a litre flask, if the temperature changes from 27°C to 127°C? 28. When measured at 27°C, the volume of oxygen gas was found to be 400ml. What would be the new volume at 127°C at constant pressure? 29. If the temperature of a gas is doubled from 10°C to 20°C, then the volume also doubles. True/False. 30. If the temperature of a gas at constant pressure is changed from 15°C to – 15°C the change in its density is______ times the initial density. 31. The density of nitrogen is 1.25kg/m3 and that of oxygen is 1.45kg/m3 at 27°C temperature. To what temperature should the oxygen be heated to make it as light as the nitrogen?

149

7. Gau Lussac’s law “Volume remaining constant, the pressure of a given mass of a gas increases or decreases by 1/273 of its pressure at 00 C for every 10 C rise or fall in temperature”. Mathematically, Pt = P0 +

 273 + t  P0 T ×t = P 0  or  = P0 273 273  273 

P P  T i.e., T = constant or P1 = P2 at constant T1 T2 volume. Statement: At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.

Graph between P and T The graph between P and T at constant volume is the straight line passing through origin. P

T

Note: The graphs plotted at constant volumes are called Isochors. Some Problem Solving Tips and Understandings  Pressure of a gas is directly proportional to absolute temper ature only when mass and volume of the gas are kept constant .  Gas cylinders, spheres, containers with fixed piston represent constant volume system.  If the terms in a problem are Pressure (P), Temperature (T) and Volume is constant, then Gay-Lussac's Law is applicable.

Formative Worksheet 32. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder indicates 12 atm. at 27°C. Due to a sudden fire in the building, its temperature starts rising. At what temperature will the cylinder explode? 33. An iron cylinder contains a gas at a pressure 10 atm at 300 K. The cylinder can withstand a pressure of 100 atm. The room in which the cylinder is placed catches fire. Predict whether the cylinder bursts off before it melts or not? (Melting point of the material of the cylinder is 2500 K). www.betoppers.com

9th Class Chemistry

150 34. A 10.0 litre container is filled with a gas to a pressure of 2.00 atm at 0°C. At what temperature will the pressure inside the container be 2.50 atm? 35. A car tyre initially has a volume of 10 litres. The tyre is inflated to a pressure of 3 atmosphere at 27°C with air. While driving, the temperature of the tyre increases to 47°C. What will be the pressure at this temperature? 36. A cylinder contains helium at a pressure of 250 kPa at 300 K. The cylinder can withstand a pressure of 1 × 106 Pa. The room in which cylinder is placed catches fire. Predict whether the cylinder will blow up or not before it melts. (M.P. of the cylinder = 1800 K).

Conceptive Worksheet 32. The graph plotted at constant volume is known as___________. 33. Which of the following graph represents Gay Lussac’s Law? 34. Which of the following true for Gay-Lussac’s Law?

P1 T1 t   1) Pt = PO 1+ 2) P = T   273  2 2 3) Both 4) None 35. Which of the following is a constant volume system? 1) A cylinder 2) A cylinder with a movable piston 3) Balloon 4) All

8. Avogadro’s law What happens to the size of a balloon when we blow air in to it? The size (volume) of the balloon increases. Volume of the balloon increases as mass (no. of moles )of air increases. Thus, we can say that volume is directly proportional to mass of a gas. This is called Avogadro’s Law. Under the equal conditions of temperature and pressure, the volume of gas is directly proportional to its mass or number of moles.

V1 n1 m1 = = Mathematically, V  n  m  V2 n 2 m 2 Note: Avogadro’s Law is applicable only when temperature and pressure are kept constant.

9. Ideal gas equation Now, let us summarize the gas equations involving volume…. www.betoppers.com

Gas law

Equation

Constant condition

Boyle’s law

V

1 P

m and T

Charles’ law

VT

m and P

Avogadro’s law

Vn

T and P

Observations: i) Boyle’s equation gives the variation of volume with respect to pressure, when mass and temperature are kept constant . ii) Charles’ equation gives the variation of volume with respect to absolute temperature, when mass and pressure are kept constant . iii) Avogadro’s equation gives the variation of volume with respect to no. of moles, when temperature and pressure are kept constant. If all these three vary simultaneously, can we find the variation in volume? Yes, we can, with the help of Ideal Gas Equation It helps us to find the variation in volume, when all the terms – P, T and n-vary simultaneously. Ideal gas equation is obtained by combining all the gas equations. By combining then, we get

nT  PV  nT  PV = nRT , P This equation is called Ideal gas equation., where 'R' is called universal gas constant. R  f {nature of gas, temperature, amount of gas & pressure of gas} V

Units and Values of R System SI CGS In calories

Value 8.314 Joule/ mole/ Kelvin 8.314  107 erg/ mole/ Kelvin 2 calories/ mole/ Kelvin

Note: 1 mole of any gas at S.T.P. occupies 22.4 litres. The density of the gas can be calculated by d  P = Pressure, M = Molecular weight, R = Universal gas constant T = Absolute temperature

PM RT

Gaseous State

Ideal and Real Gases It this world of humans an ideal individual is one who has achieved freedom from all bondage, entanglements, fear, insecurity etc,. In the world of gases too an ideal gas is one which has achieved freedom from the bondage of intermolecular forces under all circumstances. Such an ideal gas satisfies the following: 1. It obeys all the gas laws under all conditions of temperature and pressure. 2. It obeys the ideal gas equation and all gas laws under all conditions. The central theme around which the concept of ideal gas revolves is the zero intermolecular force of attraction. Let us check the conditions where the intermolecular forces of attraction come into picture: Consider a very high pressure and low temperature. These conditions make the volume very less. As a result, the gas molecules come very close to each other. This brings the intermolecular forces of attraction into picture and the central theme of the ideal gas gets diluted. Practically speaking, there are no ideal gases and there exists no gas with zero intermolecular forces of attraction at all temperatures and pressure. As there are no perfectly ideal gases, all gases are said to be real gases. Real Gases Real gases are those gases which do not obey all the gas laws under all conditions of temperature and pressure. Which is then the most idealistic real gas? We have noticed that all gases are real gases. Let us check which real gas is closer to the ideal gas behaviour.An ideal gas is characterised by zero intermolecular forces of attraction. Molecular force of attraction depends on the molecular weight of the gas molecules. The less the molecular weight, less the molecular force. The gas that is closest to an ideal gas has the least intermolecular force of attraction.  Less molecular weight. The gas with the least molecular weight is hydrogen. Hence, the behaviour of Hydrogen is closer to the ideal gas behaviour than any other. Can we convert a real gas to an ideal gas? Of course! Yes. To convert a real gas into an ideal gas, the intermolecular force should vanish or become negligible. Let us study the conditions which make this possible. The intermolecular force weakens as the intermolecular distance increases. This happens

151 when the volume of a gas is more. Volume of a gas is more at high temperatures and at low pressure. So a real gas approaches an ideal gas at a high temperature and low pressure. At high temperature Real gas

APPROACHES

Ideal gas

At low temperature

Some Problem Solving Tips and Understandings  Volume of a gas is directly proportional to the mass or number of moles only when pressure and temperature are kept constant.  If the terms in a problem are Volume (V), Mass (m) or number of Moles (n) and pressure, temperature are constants, then Avogadro’s Law is applicable.

Formative Worksheet 37. A rubber balloon of volume 6 litres contains 6 gm of N2 . Find the mass of nitrogen to be added to the balloon to expand its volume to 10 litres under the same conditions of temperature and pressure. 38. Calculate the volume occupied by 11 gm of CO2 at STP. 39. A container containing hydrogen gas is heated such that its volume increases by 40% and pressure decreases to 80% of its original value. If the original temperature was –13°C, calculate the temperature to which the gas was heated. 40. The radius and temperature of an air bubble at the top of a pond become twice when present at the bottom. If the pressure at the bottom is ‘P’, what will be the pressure at the top? 41. We know that volume of gas depends on temperature and pressure but can we make the volume of gas unchanged in spite of change in temperature and pressure? 42. A certain quantity of hydrogen gas occupies a volume of 300ml at a certain temperature and pressure. What volume would half of this mass of hydrogen occupy at triple the absolute temperature if the pressure were 1/9th of the original gas? 43. If the mass of gas is halved and pressure is doubled, then how should temperature be changed such that volume remains unchanged? www.betoppers.com

9th Class Chemistry

152 44. A cube of length l has a gas enclosed at NTP. It is shifted to another cube where the temperature is 546K and pressure is 0.5 atm. From this information can you find the side of the second cube? 45. A gas is shifted from a container of volume V litres to another container of volume 4V litres. The temperature of the gas in first container is 273°C and in the second is 546°C. Find the pressure in the first container if the pressure in the second is P.

Conceptive Worksheet 36. How is Ideal gas obtained ? 37. Which of the following is correct? 1) PV = nRT

2) V =

nRT P

RT 4) All PV What does ‘R’ represent in ideal gas equation? Write the different values of R. What is the volume occupied by 1 mole of any gas at STP ? The density of a gas can be calculated by the formula: 3) n =

38. 39. 40. 41.

1) d =

PM RT

43. 44.

45. 46.

RT PM

3 )

PR 4) All MT If the absolute temperature of a gas is doubled and the pressure is reduced to one-fourth, the volume of the gas will ________. At S.T.P., the density of nitrogen monoxide is _____. At 0°C and one atm pressure, a gas occupies 100 cc. If the pressure is increased to one and a halftimes and temperature is increased by one-third of absolute temperature, then final volume of the gas will be 8.2 L of an ideal gas weighs 9.0 g at 300 K and 1 atm pressure. The molecular mass of gas is An electronic vacuum tube was sealed off during an experiment at a pressure of 8.2 × 10–10atm at 27°C. The volume of the tube was 30 dm3. The number of gas molecules remaining in the tube are: (Hint : 1 mole of any gas contains 6.023×1023) d=

42.

2) d =

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10. Graham’s law Diffusion The gas molecules being placed far apart from each other have very less (almost negligible) inter molecular forces acting between them. Due to this, the gas molecules have a greater degree of freedom of movement. This makes them occupy the complete space within a container. The same reason also makes the gas molecules occupy the intermolecular gaps of other gases. This makes the non-reacting gases to have a tendency to intermingle with the other gases without losing their own chemical identity. The gas molecules just occupy the intermolecular spaces and form a homogenous mixture. This property of the gases is called Diffusion. Diffusion: The process of spontaneous intermixing of two or more gases in varying proportions to form a homogeneous mixture irrespective of the gravitational force is called diffusion. Conditions for diffusion i) The gases should be non-reactive. ii) The gases in the following pairs do not diffuse with each other as they react with each other. a) Carbon monoxide and Chlorine gas (CO & Cl2), b) Nitrogen monoxide and Oxygen gas (NO & O2), c) Hydrogen gas and Chlorine gas (H2 & Cl2). Examples: a) In laboratories, it is common that if any gas with a high aroma is prepared in one corner of the lab, its aroma can be sensed immediately in an other corner due to diffusion. b) In the Bhopal gas tragedy of 1984, many people succumbed to the gas leak and died while some others suffered by the havoc. The gas escaped from the factory at night and diffused all over the city, thus causing the tragedy. c) Diffusion is also seen in liquids. When a few drops of ink are added to water in a glass beaker, beautiful continuously changing patterns are observed, due to the diffusion of ink particles. After some time we get a homogeneous solution. Applications of diffusion 1. Diffusion helps in the detection of Marsh gas in coal mines and thus helps in avoiding explosions. (Marsh gas is Methane, an explosive. In the presence of O2 it proves to be an explosive).

Gaseous State 2.

The difference in rates of diffusion of gases is used in the separation of isotopes. 235 Eg: Isotopes of Uranium  92 U,

3. 4.

5.

153

238 92

U  are separated

by this principle. Poisonous and foul-smelling gases present in the atmosphere get diluted by diffusion. Molecular weights and densities of unknown gases can be calculated by comparing the rates of diffusion. The phenomenon of diffusion is useful in the separation of gaseous mixtures. Effusion Likewise, gases exhibit another property called effusion which can be defined as follows: Definition: The passage of gases through a small aperture from a region of high pressure to a region of low pressure is called ‘Effusion’. Example: When you switch on the knob of a gasstove in a kitchen, the gas starts escaping through the holes of the burner. This is an example of effusion. The same rules apply for both diffusion and effusion. Rates of diffusion and rate of effusion (r) The volume of gas escaping per second is called the rate of diffusion or the rate of effusion of the gas.

V --------- (1), where ‘V’ t is the volume of the gas escaping in a time of ‘t’ seconds. It can be written as: r 

An activity Keep two cotton plugs, one dipped in conc. Hydrochloric acid, and the other in liquid ammonia at the two ends of an open test tube. Close the open ends with two corks. What do you observe after sometime?

Ammonia and hydrochloric acid vapourise and react to form a whitish ring of ammonium chloride. The ring is not formed at the centre; it is observed near the HCl end of the cotton plug.

Why does this happen?

We know that Ammonia (NH3) is lighter than hydrochloric acid (HCl). The lighter ammonia diffuses faster than the heavier hydrochloric acid. This makes ammonia move a longer distance than HCl and hence a ring is formed away from the Ammonia cotton, nearer to the hydrochloric acid cotton end. Conclusion: Lighter gases diffuse more rapidly than heavier gases. This conclusion is standardized as Graham’s Law of diffusion. Statement: "It states that rates of diffusion or effusion of gases, under similar conditions of temperature and pressure, are inversely proportional to the square root of their densities." Thus, if r1 and r2 are the rates of diffusion or effusion of two gases with densities d1 and d2 respectively, the law may be mathematically represented as :

r1 d2  r2 d1 .................. ( 1 ) Note: i) Diffusion is the spontaneous intermixing of gases. ii) Escaping of gas through a small aperture or orifice from a region of high pressure to that low pressure is known as effusion. We also know that at constant temperature and pressure, the density of a gas is directly proportional to its molecular weight. i.e. d  M 

d1 M1 ( or )  d 2 M2

d 2 M 2 ........(2)  d1 M1 Combining equations (1) and

(2) we get

r1 d2 M2   r2 d1 M1 Note: Rate of diffusion is also equal to the volume of the gas escaping in unit time. i.e., r =

r1 V1 / t1 V  r  V /t  t 2 2 2

r1 V /t = 1 1 = r2 V2 / t 2

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Formative Worksheet 46. If the rate of diffusion of two gases is in the ratio of 1 : 4, find the ratio of their molecular weights. 47. At a definite pressure and temperature 100ml of hydrogen diffused in 20 minutes. How long will 40ml of oxygen take to diffuse under similar conditions? 48. Do the following pairs of gases has same rate of diffusion under identical conditions of temperature and pressure? 1) N2O, CO2 2) CO2, C3H8 3) C3H8, N2 O 4) None 49. Hydrogen chloride gas is sent into a 100 meter tube from one end and NH3 gas is sent into the tube from the other end. At what distance a white ring of NH4Cl is formed ? 50. One litre of oxygen gas takes 1 hour to diffuse out of a vessel. How long will it take to diffuse one litre of Hydrogen gas through the same vessel under the same conditions of temperature and pressure?

Conceptive Worksheet 47. What are the relative rates of diffusion of hydrogen and oxygen gases? 48. At a definite pressure and temperature 100ml of hydrogen diffused in 20 minutes. How long will 40ml of oxygen take to diffuse under similar conditions? 49. A gas of unknown identity effuses at the rate of 83.3 mL s–1 in an effusion apparatus in which carbon dioxide effuses at the rate of 102 mLs–1. Calculate molecular mass of the unknown gas. 50. What is the ratio of rates of diffusion of SO2 and O2? 51. The ratio of rates of diffusion of SO2, O2 and CH4 is______. 52. 50 ml of hydrogen diffuses out through a small hole from a vessel in 20 minutes. The time needed for 40 ml of oxygen to diffuse out is________. 53. The relative rate of diffusion of a gas (Mol. wt. = 128) as compared to oxygen is: 54. If 4ml of oxygen diffuses through a very narrow hole, how much hydrogen would have diffused under identical conditions? 55. 20 L of SO2 diffuses through a porous partition in 60 seconds. What volume of oxygen will diffuse under similar conditions in 30 seconds?

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Summative Worksheet 1.

2.

3.

4.

A gas occupies a volume of 1500 cc at a pressure of 720 mm of mercury. Find the pressure at which its volume becomes 1000 cc at constant temperature. At constant temperature, a gas is at a pressure of 1080 mm of Hg. At what pressure will its volume decrease by 40%? A gas present in a spherical container of radius 10 cm is shifted to another spherical container of radius 20 cm. If the pressure exerted by the gas in first container is 760 mm of Hg, find the pressure in the second container at constant temperature. Find the no. of cylinders of volume 20 cc required to fill it at a pressure of 1 atm from a cylinder of volume 20 litres containing a gas at a pressure of 100 atm. Assume that the shifting took place at same temperature.

279 cc of a gas at 870 C is cooled to standard temperature at constant pressure. Calculate the volume of gas at standard temperature. 6. A gas is enclosed in a vessel with movable piston at standard temperature. At what temperature does the volume of the gas increase by 4 times the initial? 7. A gas is present in a container of volume ‘V’ litres. It is connected to three more empty containers of volumes V , 2V and 3V respectively with a tube of negligible volume. What is the pressure exerted by the gas after connection, if the initial pressure is 120 cm of Hg? 8. 4 gm of hydrogen is added to a balloon containing 8 gm of hydrogen. Find the percentage change in volume of the gas, if temperature and pressure were to be constant. 9. Oxygen is present in 1litre flask at a pressure of 7.6 × 10–10mm of Hg. Calculate the no. of molecules in the flask at 0°C. 10. When 2 g of a gas A is introduced into an evacuated flask kept at 25 °C, the pressure is found to be 1 atm. If 3 g of another gas B is added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behaviour, calculate the ratio of their molecular weights. 11. If the temperature difference in centigrade scale is 10, its equivalent value in kelvin is__________. 12. The height of water column in a water barometer is 600 mm. What will be the height of mercury column in a mercury barometer to measure the same pressure? 5.

Gaseous State

155

HOTS Worksheet

13. The height of mercury in the long tube of a manometer is l1 cm. When it is connected to a container containing hydrogen, the height of

14.

15.

16.

17.

18.

19.

20. 21.

22.

23.

24. 25.

mercury column was found to be l2 . What is the pressure exerted by the hydrogen gas in Pascal? A sample of nitrogen gas occupies a volume of 58.0 cm3 under the existing barometric pressure. If the pressure is increased by 125 torr, and volume is reduced to 46.6 cm 3 , what is the prevailing barometric pressure? What percentage of a sample of nitrogen must be allowed to escape if its temperature, pressure and volume are changed from 220 0C, 3 atm and 1.65 litres, to 400C, 0.7 atm and 1 litre respectively? To what temperature must an ideal gas be heated to double its pressure, if the initial volume of gas at750C is decreased by 15%? A good vacuum pump will bring a tight system down to 10 –10 mm of Hg. How many gas molecules remain at this pressure and at a temperature of 300K? A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends. If the length of the tube is 100cm, find the distance from the lighter gas end, where a white smoky ring of NH4Clis formed. Three grams of helium gas diffuses from a container in 15 minutes. The mass of sulphur dioxide diffusing from the same container over same time interval will be ____. The rate of diffusion of nitrogen is half that of an unknown gas ‘G’. Identify ‘G’. The density of phosphorus vapour at 310°C and 775 torr is 2.64 gdm–3 . What is the molecular formula of phosphorus? A gaseous mixture of helium and oxygen is found to have a density of 0.518 gdm–3 at 25°C and 720 torr. What is the percent by mass of helium in this mixture? If the pressure of a gas has become one fifth of its initial, absolute temperature is increased by 3 times and mass is decreased by 25%. Find the percentage change in volume. Find the volume occupied by 1 gas molecule of S.T.P. If the mass of 6023 molecules of an unknown gas is 10 × 10–22g, find the molecular weight of the gas at S.T.P.

1.

2.

3.

4. 5.

6.

7.

8.

9. 10.

11.

12.

13.

14.

15.

How high would a column of air be required to cause the barometer to read 700mm Hg, if the atmosphere is of uniform density 1.2 kg/m3 and density of mercury is 1.36 × 104 kg/m3? A certain mass of 2.5 litres of oxygen exerts a pressure of 740mm. What volume will the same mass of oxygen have under a pressure of 780mm? A sphere of radius ‘r’ contains a gas at a pressure of 10 atm. If it is connected to another sphere of radius ‘2r’, then the pressure inside the new setup is______. Calculate the pressure required to compress isothermally a 105L sample of air at 1 atm to 35L. Calculate the resulting temperature change, if a 52L sample of gas at 25°C is isobarically expanded to 104L. What will happen to the pressure of a 5-litres sample of a gas at 5 atmosphere, if it is heated from 250K to 300K and the volume is held constant ? The mass of nitrogen is increased from 7g to 21g. Find the % increase in volume at same conditions of temperature and pressure. Calculate the number of moles of hydrogen contained in 18L of gas at 27°C and 70 cm pressure. Given that R = 0.0821 litre-atom mole –1 K –1 . Further, if the mass of hydrogen taken as above is found to be 1.350g, calculate the molecular mass of hydrogen. 380ml of a gas at 27°C and 800mm of Hg weighed 0.455g. Identify the gas. The density of a gas is found to be 1.56g/L at 745 mm pressure at 650 C. Calculate the molecular weight of the gas. The density of a gas at 270C and 1 atmosphere is 'd', pressure remaining constant. At what temperature does the density become 0.75d? At a definite pressure and temperature 100ml of hydrogen diffused in 20 minutes. How long will 40ml of oxygen take to diffuse under similar conditions? Hydrogen chloride gas is sent into a 100-meter tube from one end and NH3 gas is sent into the tube from the other end. At what distance is a white ring of NH4Cl formed ? What is the molecular weight of a substance, each molecule of which contains 9 carbon atoms, 13 hydrogen atoms and 2.33 × 10 –23 g of other components ? If % change in volume is 10. What is the % change in pressure of a given mass of gas at constant temperature? www.betoppers.com

9th Class Chemistry

156 16. A mixture of CO and CO2 is found to have density of 1.5 g/L at 300K and 700mm. What is the composition of the mixture ? 17. An evacuated glass vessel weighs 50.0g when empty, 148.0g when filled with a liquid of density 0.98 g/mL and 50.5g when filled with an ideal gas at 760mm Hg at 300 K. Determine the molecular weight of the gas. 18. The total pressure of a mixture of H2 and O2 is 1.00bar. The mixture is allowed to react to form H2O, which is completely removed to leave only pure H2 at a pressure of 0.35 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same conditions of temperature and volume, calculate the composition of the original mixture. 19. The respiration of a suspension of yeast cells was measured by observing the decrease in pressure of gas above cell suspension, using a manometer, the fluid of which had density of 1.034g/cm3 and the gas was confined to a constant volume of 16.0cm3. The entire apparatus was immersed in a thermostat at 370C. There was a drop of fluid level by 37mm in 30min in the open side of the manometer. Calculate the rate of oxygen consumption by the yeast cells in mm3 of O2 at S.T.P. per hour. 20. A cylinder contains 100g of an ideal gas (mol.wt = 40) at 2 atm pressure and 27 0 C. There was decrease in volume of cylinder due to dent. But the valve attached to the cylinder cannot keep the pressure greater than 2 atm. Thus, 10g of gas is leaked out. i) Calculate volume of cylinder before and after dent. ii) Had the valve been very strong, what would have been the pressure after dent ? Note : Temperature remained constant during this process. 21. Two flasks of equal volume are joined by a narrow tube of negligible volume. Initially, both flasks are at 300 K containing 0.6 mole of O2 gas at 0.5 atm pressure. One of the flasks is then placed in a thermostat at 600 K. Calculate the final pressure and the number of moles of O2 gas in each flask. 22. A meteorological balloon had a radius 1.0m, when released at sea level at 200 C and expanded to a radius of 3.0m at an altitude when the temperature was – 200C. What is the pressure inside the balloon at that temperature ? 23. Pressure of 1g of an ideal gas A at 27°C is found to be 2 bar. When 2g another ideal gas B is introduced in the same flask at same temperature, the pressure www.betoppers.com

becomes 3 bar. Find the relationship between their molecular masses. 24. A 1000ml sample of a gas at -730C and 2atm is heated to 1270C and the pressure is reduced to 0.5 atmosphere. What will be the final volume ?

IIT JEE Worksheet 1.

Volume of 0.5 mole of a gas at 1 atmosphere pressure and 2730C temperature is 1) 22.4 litres 2) 11.2 litres 3) 44.8 litres 4) 5.6 litres 2. A sample of gas has a volume of 0.2 lit measured at 1 atm pressure and 00C. At the same pressure, but at 2730C its volume will become 1) 0.1 litre 2) 0.4 litre 3) 0.8 litre 4) 0.6 litre 3. If pressure and absolute temperature of a given mass of a gas are doubled, the new volume will be ______ initial volume. 1) same as 2) double 3) four times 4) half 4. Which of the following pairs among the gaseous species will diffuse through a porous plug with the same rates of diffusion? 1) CO, NO2 2) NO2, CO2 3) NH3, PH3 4) NO, C2H6 5. The molecular weight of a gas which diffuses through a porous plug at 1/6 of the speed of hydrogen under identical conditions is _________. 6. Which one of the following indicates the value of the gas constant R? 1) 1.987 cal/deg/mol 2) 8.3 cal/deg/mole 3) 0.0821 lit/deg/mole 4) 1.987 joules/deg/mole 7. 2 gm of H2 or 32 gm of O2 occupy a volume of ______cm3 at N.T.P. 8. A gas ‘A’ diffuses at a rate which is twice that of another gas ‘B’. The ratio of molecular weights of A and B is 1) 1.00 2) 0.75 3) 1.5 4) 0.25 9. The magnitude of R in cc-atmosphere/mole/K is ____________. 10. At S.T.P the volume of 0.5 g of hydrogen gas is _________ litres. 11. A certain mass of a gas occupies a volume of 10 litres at a pressure of 1 atmosphere and 300 K temperature. Which of the following changes does not produce a change in volume of the gas? 1) pressure double and temperature halved 2) doubling both pressure and temperature 3) pressure halved and absolute temperature doubled 4) pressure unchanged and absolute temperature halved.

Gaseous State 12. The value of R in joule per mole per degree kelvin is _________ . 13. The molar volume of an ideal gas at one atmosphere pressure and 2730C temperature is _______ litres. 14. At constant pressure, the volume of a fixed mass of an ideal gas is directly proportional to 1) absolute temperature 2) degree centigrade3 ) degree Fahrenheit 4) none of these 15. 2 litres of H2 and 11.2 litres of Cl2 are mixed and exploded. The composition by volume of the resulting mixture will be 1) 0.8 litre hydrogen and 22.4 litres HCl 2) 24 litres of HCl 3) 0.8 litre of Cl2 and 20.8 litres of HCl 4) 22.4 litres of HCl 16. A steel vessel of capacity 22.4 litres contains 2 g of H 2 and 8 g of O 2 and 22 g of CO 2 at 0 0 C temperature. The total pressure of the gases is ________ atmospheres. 17. At what temperature would the volume of a given mass of gas, at constant pressure, be twice its volume at 00C? 1) 1000C 2) 5460C 3) 3730C 4) 2730C 18. Which of the following expressions at constant pressure represents Charles’ Law?

1 1 2) V  2 3) V  T 4) V= d T T 19. A gas of volume 100 cc is kept in a vessel at pressure 104 P maintained at temperature 240C. If the pressure is increased to 10 5 P, keeping the temperature constant, then the volume of the gas becomes 1) 10 cc 2) 100 cc 3) 1 cc 4) 1000 cc 20. A gas is initially at 1 atm pressure. To compress it 1) V 

to

1 th of the initial volume, pressure to be applied 4

is

1 atm 4 21. The density of neon will be highest at 1) S.T.P 2) 00C, 2atm 0 3) 273 C, 1atm 4) 2730C, 2 atm 22. In a closed flask of 5 litres, 1.0 gm H2 is heated from 300 to 600 K. Which statement is not correct? 1) pressure of the gas increases 2) the rate of collision increases 3) the number of moles of gas increases 4) the energy of gas molecules increases 1) 1 atm 2) 2 atm 3) 4 atm 4)

157 23. A gas mixture contains O2 and N2 in the ratio of 1: 4 by weight. Then the ratio of their number of molecules in the mixture is __________ 1) 1 : 4 2) 7 : 32 3) 3 : 16 4) 3 :32 24. At constant temperature and pressure, equal volumes of gases contain the same number of molecules. This is known as : 1) Hund’s rule 2) Avogadro’s hypothesis 3) Gay Lussac’s Law 4) Charles’ Law 25. If equal weights of oxygen (atomic weight 16) and nitrogen (atomic weight 14) are placed in separate containers of equal volume at the same temperature, which one of the following statements is true? 1) both flasks contain the same number of molecules 2) more molecules are present in the oxygen flask 3) the nitrogen has a greater average of kinetic energy per molecule 4) pressure in the nitrogen flask is greater than the one in oxygen flask 26. 1 cc (N.T.P.) of a hydrocarbon vapour is as heavy as 4 cc of oxygen. What quantity of the hydrocarbon will occupy a volume of 2.24 litres? 1) 6.4 g 2) 2.2 g 3) 12.8 g 4) 64 g 27. Weight of 112 ml of oxygen at N.T.P. on liquefaction would be 1) 0.32 g 2) 0.64 g 3) 0.16 g 4) 0.96 g 28. The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is 1) 64.0 2) 32.0 3) 4.0 4) 8.0 29. 50 ml of gas A diffuse through a membrane in the same time as for the diffusion of 40 ml of a gas B under identical pressure - temperature conditions. If the molecular weight of A is 64, that of B would be ________. 1) 100 2) 250 3) 200 4) 80 30. A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, the white ammonium chloride ring first formed will be 1) at the centre of the tube 2) near the hydrogen chloride 3) near the ammonia bottle 4) throughout the length of the tube



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9th Class Chemistry

B y t h e e n d o f t h i s c h a pt e r , y o u w i l l u n d e r s t a n d      

1.

Introduction Laws of Chemical Combination Dalton’s Atomic Theory Atoms and Molecules Atomic Weight Atomic Mass Unit

     

Molecular Weight Gram atomic Weight Gram Molecular Weight Avogadro’s Number Mole Percentage Composition

Introduction In our previous classes and chapters, we had the glimpses of microscopic part of the chemical world. We have known the details of the constituents of the microscope world electrons, protons, neutrons, atoms, molecules, ions etc. Now, there are some basic questions to be answered. How can we find the number of atoms / molecules / ions, etc. present in the given amount of substance? What is the weight of the given number of atoms, molecules, ions, etc. ? And many more such questions related to the reaction. All the above questions are very important in chemistry and for chemists. Practically, they form the foundation for many branches of chemistry and various other fields. Counting the number of atoms is impossible, weighing the number of atoms or molecules or ions is impractical and weighing the products formed in a reaction every time is a tedious job. To simplify all these intricate and challenging tasks, you need to comprehend the following chapter . This chapter helps you to study the relation between the microscopic the macroscopic worlds.

2.

Laws of chemical combination

1. 2. 3.

There are three important laws of chemical combination. These are: Law of conservation of mass (or matter), Law of constant proportions, and Law of multiple proportions. The laws of chemical combination are the experimental laws which have been formulated

Chapter - 71

Mole Concept

Learning Outcomes

by scientists after performing a large number of experiments involving various types of chemical reactions. These experimental laws ultimately led to the idea of ‘atoms’ being the “smallest unit” of matter. In fact, the laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter.

Law of Conservation of Mass In the 18th century, scientists noticed that if they carried out a chemical reaction in a closed container, then there was no change of mass. This preservation of mass in a chemical reaction led to the formulation of the law of conservation of mass (or law of conservation of matter). Law of conservation of mass was given by Lavoisier in 1774. According to the law of conservation of mass : Matter is neither created nor destroyed in a chemical reaction. The substances which combine together (or react) in a chemical reaction are known as ‘reactants’ whereas the new substances formed (or produced) as a result of chemical reaction are called ‘products’. The law of conservation of mass means that in a chemical reaction, the total mass of products is equal to the total mass of reactants. There is no change in mass during a chemical reaction. Since there is no gain or loss in mass in a chemical reaction, the mass remains conserved. Please note that the term ‘total mass’ of reactants and products includes solids, liquids and gases - including air that are a part of the reaction. Example When calcium carbonate is heated, a chemical reaction takes place to form calcium oxide and carbon dioxide. It has been found by experiments that if 100 grams of calcium carbonate are decomposed completely then 56 grams of calcium

9th Class Chemistry

160 oxide and 44 grams of carbon dioxide are formed. This can be written as :

same masses of hydrogen and oxygen elements are obtained in every case. This experiment shows that water always consists of the same two elements, hydrogen and oxygen, combined together in the same constant proportion of 11 : 89 or 1 : 8 by mass. And this is the law of constant proportions.

heat  Calcium  Carbon Calcium  oxide Carbonate dioxide 100 g 56 g 44 g

In this example, calcium carbonate is the reactant and it has a mass of 100 g. Calcium oxide and carbon dioxide are the products and they have a total mass of 56 g + 44 g = 100 g. Now, since the total mass of products (100 g) is equal to the total mass of reactant (100 g), there is no change of mass during this chemical reaction. The mass remains the same or conserved. So, this example supports the law of conservation of mass.

Law of Multiple Proportions This law was discovered by John Dalton (1803). This law states that: When two elements combine with each other to form two or more than two compounds, the masses of one of the elements which combine with fixed mass of the other, bear a simple whole number ratio to one another.

Law of Constant Proportions The law of constant proportions was given by Proust in 1779. He analysed the chemical composition (type of elements present and percentage of elements present) of a large number of compounds and came to the conclusion that the proportion of each element in a compound is constant (or fixed). Based on these observations, Proust formulated the law of constant proportions. According to the law of constant proportions : A chemical compound always consists of the same elements combined together in the same proportion by mass. This law means that whatever be the source from which it is obtained (or the method by which it is prepared), a pure chemical compound is always made up of the same elements in the same mass percentage. Example Water is a compound which always consists of the same two elements, hydrogen and oxygen, combined together in the same constant proportion of 1 : 8 by mass (1 part by mass of hydrogen and 8 parts by mass of oxygen). Let us discuss this in a little more detail. We know that water is a compound. If we decompose 100 grams of pure water by passing electricity through it, then 11 grams of hydrogen and 89 grams of oxygen are obtained. Now, if we repeat this experiment by taking pure water from different sources (like river, sea, well, etc.), the www.betoppers.com

Formative Worksheet 1.

2.

3.

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heat  Calcium  Carbon Calcium  oxide Carbonate dioxide 100 g 56 g 44 g

From the reaction given above, the assumptions made are. (I) 100gms of calcium carbonate decomposes to give 56 gms of calcium oxide and 44 gms of carbon dioxide. (II) The total mass of reactants is 100 g. (III) The total mass of the products is 100 g. (IV) There is no mass change during the reaction. (A) All statements are correct (B) Statements I & II are correct (C) Statements III & IV are correct (D) Statements I & IV are correct Sodium carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water. In an experiment, 5.3 gm of sodium carbonate reacted with 6 gm of ethanoic acid to form 8.2 gm of sodium ethanoate, 2.2 gms of carbondioxide and 0.9 gm of water. This data is in accordance with law of conservation of mass. What is the total mass of reactants (or) products? (A) 9.3 g (B) 12.3 g (C) 11.3 g (D) 10.3 g x g of potassium chlorate on decomposing produced 1.92g of oxygen and 2.96g of potassium chloride. What is the value of x? (A) 2.44g (B) 4.88g (C) 7.32g (D) 9.76g Certain nonmetal X forms two oxides I and II. The mass percentage of oxygen in I(X4O6) is 43.7

Mole Concept

5.

6.

7.

8.

which is same as that of X in 2nd oxide. Find the formula of 2nd oxide. (A) X2O3 (B) X2O4 (C) X2O5 (D) X2O7 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g. In another experiment, 3. 1.179g of copper was dissolved in the nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. This observation is in the agreement of. (A) Law of constant composition or definite proportions. (B) Law of multiple proportions. (C) Law of conservation of mass. (D) Law of reciprocal proportions. Hydrogen and oxygen are known to form two compounds. The hydrogen content in one of these is 5.93%. While in other it is 11.2%. The ratio of mass of oxygen that combine with fixed mass of hydrogen respectively is (A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D) 4 :1 4. In an experiment, 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. What is mass of oxygen formed? What is the ratio of masses of copper and oxygen? (A) 0.634 g, 4 : 1 (B) 0.534 g 1 : 4 (C) 0.734 g 4 : 1 (D) 0.934 g 1 : 4 In an experiment 1.288 g of copper oxide was obtained from 1.03 g of copper and 0.258 g of oxygen. Calculate the ratio of copper and oxygen 5. in the sample? (A) 1 : 4 (B) 8 : 1 (C) 4 : 1 (D) 1 : 8

Conceptive Worksheet 1.

2.

The law which states that in a chemical reaction, the total mass of the products is equal to the total mass of the reactants. (A) Law of constant proportions (B) Law of conservation of energy (C) Law of conservation of mass 6. (D) Law of multiple proportions Assumption : The total mass of the reactants is equal to the total mass of the products. There is no change in mass during a chemical reaction. This is called. Law of conservation of mass. Reason: As the chemical reactions proceed by loss or gain of energy, there is no change in mass. (A) Assumption is correct, reason is correct, reason is correct explanation of assumption.

161 (B) Assumption and reason are correct, reason is not correct explanation of assumption (C) Assumption is correct, reason is incorrect (D) Assumption is incorrect, reason is correct Barium chloride reacts with sodium sulphate to form Barium sulphate and sodium chloride. Then, according to the law of conservation of mass. (A) the total mass of reactants Barium chloride, & sodium sulphate taken is greater than the total mass of products Barium sulphate, sodium chloride formed. (B) the total mass of reactants, Barium chloride and sodium sulphate is less than the total mass of products, Barium sulphate and sodium chloride. (C) The total mass of the reactants is equal to the total mass of products. (D) None of the above. Different kinds of matter show different properties due to the presence of: (A) Same kinds of atoms and molecules. (B) Same kind of atoms and different kinds of molecules. (C) Different kinds of atoms and molecules. (D) Different kinds of atoms and same kind of molecules. Statement I : The laws of chemical combination ultimalety led to the idea of ‘atoms’ being the “smallest unit of” matter. Statement II : The laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter. A) Statement I is correct, II is incorrect. B) Both I & II are correct. C) Statement I is incorrect and Statement II is correct. D) Both I & II are incorrect. Match the following : Column - I Column - II i) Ritcher p) Law of conservation of mass. ii) Joseph Proust q) Law of constant composition or definite proportions. iii) Antoine Lavoisier iv) John Dalton

r) Law of multiple proportions. s) Law of reciprocal proportions. www.betoppers.com

9th Class Chemistry

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8.

9.

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11.

12.

(A) i-s; ii - r; iii - q; iv -p (B) i-s; ii - q; iii - p; iv - r (C) i-p; ii - s; iii - q; iv - r (D) i-s; ii - q; iii - p; iv - r Assertion: A certain elements X, forms three binary compounds with chlorine containing 59.68%, 68.95% and 74.75% chlorine respectively. These 13. data illustrate the law of multiple proportions. Reason: According to law of multiple proportions the relative amounts of an element combining with some fixed amount of a second element in a series of compounds are the ratios of small whole numbers. (A) Both assertion and reason are correct and 3. reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. The law of definite proportions is not applicable to nitrogen oxide because (A) Nitrogen atomic weight is not constant (B) Nitrogen molecular weight is variable (C) Nitrogen equivalent weight is variable (D) Oxygen atomic weight is variable When 3 grams of Carbon is burnt in 8 grams of Oxygen, 11 grams of Carbondioxide is produced. What is the mass of the Carbondioxide formed when 3 grams of Carbon is burnt in 50 grams of Oxygen? (A) 33 g (B) 11 g (C) 22 g (D) 44 g Hydrogen and Oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of Oxygen gas would be required to react completely with 3 grams of hydrogen gas? (A) 8 g (B) 16 g (C) 24 g (D) 32 g Two samples of lead oxide were separately reduced to metallic lead by heating in current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from other oxide. The data illustrates (A) Law of reciprocal proportions (B) Law of constant proportions (C) Law of multiple proportions (D) Law of equivalent proportions In two different experiments, copper oxide was obtained from copper. The ratio of mass of copper and mass of oxygen found in the two experiments

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are same. They are in the ratio of 4 : 1. The experiments were in accordance with ___. (A) law of conervation of energy (B) Law of multiple proportions (C) Law of constant proportions (D) Law of conservation of mass Chemical compounds are formed to have constant composition irrespective of their (A) Method of formation (or) Source (B) Conditions for formation (C) Substances taken for formation (D) Time of formation

Daltons Atomic Theory The laws of chemical combination were the experimental laws. The only logical explanation of the laws of chemical combination is that matter (say, elements) must be made up of minute “unit particles”, which take part in chemical combination in fixed whole numbers. These “unit particles” of matter were called atoms. So, in an attempt to explain the laws of chemical combination, Dalton put forward his atomic theory of matter. The theory that “all matter is made up of very tiny invisible particles (atoms)” is called atomic theory of matter. Dalton put forward his atomic theory of matter in 1808. The various postulates (or assumptions) of Dalton’s atomic theory of matter are as follows : 1. All the matter is made up of very small particles called “atoms”. 2. Atoms cannot be divided. 3. Atoms can neither be created nor destroyed. 4. Atoms are of various kinds. There are as many kinds of atoms as are elements. 5. All the atoms of a given element are identical in every respect, having the same mass, size and chemical properties. 6. Atoms of different elements differ in mass, size and chemical properties. 7. Chemical combination between two (or more) elements consists in the joining together of atoms of these elements to form molecules of compounds. 8. The “number” and “kind” of atoms in a given compound is fixed. 9. During chemical combination, atoms of different elements combine in small whole numbers to form compounds.

Mole Concept

163

10.

Atoms of the same elements can combine in more than one ratio to form more than one compound. Dalton’s atomic theory was based on the laws of chemical combination. For example, the postulate of Dalton’s atomic theory that “atoms can neither be created nor destroyed” was the result of law of conservation of mass given by Lavoisier. And the postulates of Dalton’s atomic theory that “the elements consist of atoms having fixed mass, and that the number and kind of atoms in a given compound is fixed,” came from the law of constant proportions given by Proust. Dalton’s atomic theory was the first modern attempt to describe the behaviour of matter (or properties of matter) in terms of atoms. This theory was also used to explain the laws of chemical combination in terms of atoms. Dalton’s atomic theory provides a simple explanation for the laws of chemical combination. (a) The postulates of Dalton’s atomic theory that “the elements consist of atoms and that atoms can neither be created nor destroyed” can be used to explain the law of conservation of mass. (b) The postulates of Dalton’s atomic theory that “the elements consist of atoms having fixed mass, and that the number and kind of atoms of each element in a given compound is fixed” can be used to explain the law of constant proportions.

Explanation of the Conservation of Mass

Law

of

According to Dalton’s atomic theory, atoms can neither be created nor destroyed. Now, since an atom cannot be created or destroyed, therefore, the number of various types of atoms in the products of a chemical reaction is the same as the number of all those atoms in the reactants. The same number of various atoms in products and reactants will have the same mass. So, the total mass of products is equal to the total mass of reactants. The mass remains the same (or conserved) in a chemical reaction. And this is the law of conservation of mass. This explanation will become more clear from the following example of calcium carbonate.

Calcium carbonate (CaCO3 ) is made up of 1 calcium atom, 1 carbon atom and 3 oxygen atoms. The products of its decomposition, calcium oxide (CaO) and carbon dioxide (CO2), taken together, also contain 1 calcium atom, 1 carbon atom and 3 oxygen atoms. Now, since the number of various types of atoms in the products (CaO and CO2 ) and reactant (CaCO3) remains the same, therefore, the mass of products and reactants also remains the same in this reaction. There is no change in mass during the decomposition of calcium carbonate to form calcium oxide and carbon dioxide. The mass remains conserved.

Explanation of the Law of Constant Proportions According to Dalton’s atomic theory, every element consists of small particles called atoms, each having a fixed mass. It also says that atoms of different elements combine to form compounds, and that the “number” and “kind” of atoms of each element in a compound is fixed. Now, since the “number of atoms”, the “kind of atoms”, and the “mass of atoms” of each element in a given compound is fixed, therefore, a compound will always have the same elements combined together in the same proportion by mass. And this is the law of constant proportions. This explanation will become more clear from the following example of water. According to Dalton’s atomic theory, hydrogen element consists of hydrogen atoms (H), and oxygen element consists of oxygen atoms (O). It also says that two hydrogen atoms always combine with one oxygen atom to form water molecule (H20). Since a water molecule always contains the same number of hydrogen and oxygen atoms, each atom having a fixed mass, therefore, the masses of hydrogen and oxygen elements in water will be in constant proportion. And this is the law of constant proportions.

Drawbacks of Daltons’s Atomic Theory It is now known that some of the statements of Dalton’s atomic theory of matter are not exactly correct. Some of the drawbacks of the Dalton’s atomic theory of matter are given below : 1. One of the major drawbacks of Dalton’s atomic www.betoppers.com

9th Class Chemistry

164 theory of matter is that atoms were thought to be indivisible (which cannot be divided). We now know that under special circumstances, atoms can be further divided into still smaller particles called electrons, protons and neutrons. So, atoms are themselves made up of three particles : electrons, protons and neutrons. 2. Dalton’s atomic theory says that all the atoms of an element have exactly the same mass. It is, however, now known that atoms of the same element can have slightly different masses. 3. Dalton’s atomic theory said that atoms of different elements have different masses. It is, however, now known that even atoms of different elements can have the same mass.

4.

Units of size: Angstrom (A°) Shape: Different molecules have different shapes. A few examples are as follows: i) The shape of water molecule (H2O) is angular. ii) The shape of ammonia molecule (NH3) is pyramidal. Note: Do not get confused with the terms ‘atom’ and ‘molecule’. An atom is the smallest part of an element which takes part in chemical reaction and do not exist independently (except inert gases), whereas a molecule is the smallest part of an element or a compound which can exist independently under ordinary conditions. Thus, chlorine gas consists of molecules of Cl2 under ordinary conditions; but at very high temperatures, these molecules will split up to form chlorine atoms.

Atom & Molecules

Types of molecules

Atom John Dalton (1766 – 1844 A.D) was the first chemist to use the name ‘atom’. An atom is the smallest particle of an element which can take part in a chemical reaction and is considered the basic unit of matter. e.g. O, Cl are atoms. The atom is built up of a number of sub atomic particles.Example: Pure copper is composed of only one type of atoms i.e., copper atoms. Calcium is made up of only one type of atoms i.e., calcium atoms. Constituents: Atoms are made up of many subatomic particles. The important of them being electrons, protons and neutrons. Units of size: Angstrom (A° and 1A° = 10–10 m) Shape: Atoms are assumed to have spherical shape. Molecule It was the Italian chemist, Amedo Avogadro (1776 - 1856 A.D), who introduced the word ‘molecule’ The smallest particle of an element or a compound that can exist independently is called a molecule. Example: O2 is a oxygen molecule, O3 is a ozone molecule, NH3 is a ammonium molecule, CH4 is a methane molecule, etc. Constituents: Molecules are made up of atoms; atoms in turn are made up of sub-atomic particles. www.betoppers.com

The molecules like, H2, Cl2, F2 are formed by the combination of atoms of the same element and the molecules like H2O, N2O, NO2 are formed by the combination of atoms of different elements. Thus, molecules can be categorised as follows: Homogeneous molecules: Molecules that are formed by the combination of atoms of the same element are called homogeneous molecules. Eg: H2, Cl2, F2, O3, S6, S8 Heterogeneous molecules: Molecules that are formed by the combination of atoms of different elements are called heterogeneous molecules. Eg: H2O, NO2, N2O. If you observe the examples of homogeneous molecules, it is clear that they differ in the number of atoms. Based on the number of atoms present, homogeneous molecules are further classified into the following types. Number of atoms of homogeneous Termed as Examples molecules Only one Monoatomic He, Ne, Ar Two Diatomic Cl2, F2, N2 Three Triatomic O3 Four Tetra atomic P4

More than four

5.

Poly atomic

S6 , S8 , C60

Atomic weight The atomic weight or the relative atomic mass (RAM) of an element is defined as the number of times an atom of an element is heavier than the

Mole Concept

mass of

165

The number of atoms of a particular isotope present in 100 atoms of a natural sample of that element is called its relative abundance which always remains constant for a given element. Natural chlorine is a mixture of two isotopes with relative abundances 75% (Cl-35) and 25% (Cl37) approximately. Then, the atomic weight of

1 th of C-12 isotope’s atom. 12

Relative atomic mass of an element

 RAM  =

Mass of 1atom of that element 1 × (Mass of C -12 atom) 12

Atomic weight has no units.

75  35   25  37  chlorine is   35.5

Table showing the atomic weights of first 20 elements S.No.

Ele ment

Atomic Atomic number we ight 1. Hydrogen H 1 1 2. Helium He 2 2 3. Lithium Li 3 7 4. Beryllium Be 4 9 5. Boron B 5 11 6. Carbon C 6 12 7. Nitrogen N 7 14 8. Oxygen O 8 16 9. Fluorine F 9 19 10. Neon Ne 10 20 11. Sodium Na 11 23 12. Magnesium Mg 12 24 13. Aluminium Al 13 27 14. Silicon Si 14 28 15. Phosphorus P 15 31 16. Sulphur S 16 32 17. Chlorine Cl 17 35.5 18. Argon Ar 18 40 19. Potassium K 19 39 20. Calcium Ca 20 40 The relative atomic mass of an element indicates the number of times one atom of that element is

100 Assume an element ‘E’ containing three isotopes.

Symbol

th

1 of mass of C- 12 isotopes atom. 12 For example, the atomic weight of calcium is 40. This means that an atom of calcium is on average is 40 times the mass of 1/12 the mass of C- 12 isotope’s atom.

heavier than

Atomic weights of many elements are not whole numbers due to the presence of stable isotopes.

E A1 , z E A2 and z E A3 If x, y and z are the percentage abundance of these three isotopes respectively, then z

The average atomic weight of ‘E’ 

xA1  yA 2  zA3 xyz

Formative Worksheet 9.

10.

Given a sample of carbon in which the relative abundance of isotopes is 98.892% of C12 (atomic weight = 12) and 1.108% of C13 (atomic weight = 13.0034). What is the average atomic weight of carbon ? Natural copper is composed only of Cu 63 and Cu65. The first of these isotopes has an atomic weight of 62.929 and the second 64.928. If the atomic weight of copper is 63.54, find the relative abundance of the two isotopes.

Conceptive Worksheet 14.

Oxygen occurs in nature as a mixture of atoms of which 99.75% have atomic weight of 15.9949 and 0.25% have atomic weight of 17.9992. What is the average atomic weight of oxygen?

6.

Atomic mass unit (a.m.u) It is the smallest unit of mass and is used to measure the masses of atoms and subatomic particles. The 1 th 12 the mass of C-12 atom. The other names of a.m.u. are Aston, Dalton and Avogram. Note: 1 a.m.u. = 1.66  10–24 g or 1.66  10–27 kg.

mass of one a.m.u. is equal to the mass of

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9th Class Chemistry

166

Relation of a.m.u with Atomic weight 7.

Molecular weight

Consider an element with atomic weight ‘x’. It means that one atom of the element is ‘x’ times heavier than 1/12th of C -12 isotope’s atom. That is, mass ‘of one atom of element = x × mass of 1/ 12th of C-12 isotope’s atom = x a.m.u ( 1 a.m.u = Mass of 1/12th of C - 12 isotope’s atom) If the atomic weight of an element is ‘x’, then the weight of one atom of that element is ‘x’ a.m.u.

Molecular weight or relative molecular weight is defined as the number of times a molecule is th

heavier than

1 the mass of C-12 isotope’s atom. 12

Average mass of one molecule RMM = Weight of 1/12th of C-12 atom . Relative molecular mass or molecular weight has no units. The molecular weight of an element or compound indicates the number of times a molecule

Examples to understand further: 1) The atomic weight of nitrogen is 14. So, the mass of an atom of nitrogen is 14. a.m.u (its atomic weight expressed with a.m.u). 2) The atomic weight of chlorine is 35.5. So, the mass of an atom of chlorine is 35.5 a.m.u (its atomic weight expressed with a.m.u).

Table showing the weights of first 10 elements At. Element no

Symbol

At.wt (appr)

1

Hydrogen

H

1

2

Helium

He

2

3

Lithium

Li

7

4

Beryllium Be

9

5

Boron

B

11

6

Carbon

C

12

7

Nitrogen

N

14

8

Oxygen

O

16

9

Fluor ine

F

19

10

Neon

Ne

20

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Weight of single atom ( in amu and grams) 1a.m.u =1×1.66×10-24g 2 a.m.u =2×1.66×10-24g 7 a.m.u =7×1.66×10-24g 9 a.m.u =9×1.66×10-24g 11 a.m.u =11×1.66×10-24g 12 a.m.u =12×1.66×10-24g 14 a.m.u =14×1.66×10-24g 16 a.m.u =16×1.66×10-24g 19 a.m.u =19×1.66×10-24g 20 a.m.u =20×1.66×10-24g

th

is heavier than

1 the mass of C-12 isotope’ss 12

atom. For example, the molecular weight of calcium carbonate is 100, it implies that mass of one molecule of calcium carbonate is 100 times heavier th

than

1 the mass of C-12 isotope’s atom. 12

Relation beween molecular weight and amu Consider an element/compound with molecular weight ‘y’. It means that one molecule of the element/compound is ‘y’ times heavier than 1/ 12th of C -12 isotope’s atom. That is, Mass of one molecule of element/compound = y × mass of 1/12th of C-12 isotope’s atom = y a.m.u ( a.m.u -Mass of 1/12th of C-12 isotope’s atom) If the molecular weight of an element/compound is ‘y’, then the weight of one molecule of that element/compound is ‘y’ a.m.u. Examples to understand further: 1) The molecular weight of water is 18. So, the mass of a molecule of water is 18 a.m.u (its molecular weight expressed with a.m.u). 2) The molecular weight of Sulphur dioxide is 64. So, the mass of a molecule of Sulphur dioxide is 64 a.m.u (its molecular weight expressed with a.m.u).

Mole Concept

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Table showing molecular weights and weight of single of molecule of some compounds in a.m.u

8.

H2 N2

Molecular weight 2 28

Weight of single molecule (in a.m.u & grams) 2 a.m.u=2×1.66×10-2 4g 28 a.m.u=28×1.66×10-24 g

Oxygen Carbon dioxide

O2 CO2

32 44

32 a.m.u=32×1.66×10-24 g 44 a.m.u=44×1.66×10-24 g

5. 6.

Carbon monoxide Ammonia

CO NH3

28 17

28 a.m.u=28×1.66×10-24 g 17 a.m.u=17×1.66×10-24 g

7. 8. 9. 10.

Methane Nitrogen dioxide Water Hydrochloric acid

CH4 NO2 H2O HCl

16 46 18 36.5

16 a.m.u=16×1.66×10-24 g 46 a.m.u=46×1.66×10-24 g 18 a.m.u=18×1.66×10-24 g 36.5 a.m.u=2×1.66×10-24g

11. 12.

Sulphuric acid Nitric acid

H2SO4 HNO3

98 63

98 a.m.u=98×1.66×10-24 g 63 a.m.u=63×1.66×10-24 g

S.No.

Molecule

Symbol

1. 2.

Hydrogen Nitrogen

3. 4.

Gram atomic weight Atomic weight of an element expressed in grams is known as its gram atomic weight. For example, the atomic weight of hydrogen is 1.008. So, the gram-atomic weight of hydrogen is 1.008 g. Gram atomic weight of any substance is also called its gram atom. For example, 1 gram atom of carbon weighs 12 gram and 1 gram atom of nitrogen weighs 14 grams. Number of gram atoms

Given weight = Gram atomic weight . For example, the number of gram atoms in 5 g of hydrogen =5/1 = 5. Note: 1) Weight of x gram atoms = x  Gram atomic weight. 2) 1 gram atom or gram atomic weight of an element contain = 6.023  1023 atoms. 3) Number of atoms in a given substance ( given element) = Number of gram atoms  6.023  1023. 4) Number of atoms in 1 gram of an element =

9.

Gram molecular weight It is the molecular weight of an element or compound expressed in grams. For example, the molecular weight of hydrogen gas is 2. So, the gram molecular weight of hydrogen is 2 g. Gram molecular weight of a substance is also called its gram molecule or mole molecule. For example, the weight of 1 gram molecule or mole molecule of H2O is 18 grams and the weight of 1 gram molecule of N2O is 44 grams. Number of gram moles =

Given weight . Gram Molecular weight

Note: 1) Weight of x moles of any compound = x  Gram molecular weight. 2) Number of molecules in a given substance= Number of gram molecules  6.023  1023. 3) Weight of substance in grams = Number of gram molecules  GMW.. 4) Gram atomic mass of an element and Molar mass of an element are just the same. 5) Gram molecular weight of a substance and Molar mass of a substance are also just the same.

6.023×1023 . Atomic weight

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9th Class Chemistry

168 19.

What is the weight of one carbon dioxide molecule?

The ratio of weight of one atom of an element to its atomic weight is equal to 20. (A) 1 amu

(A) 44g (B) 44 amu (C) 44 kg (D) 44 mg The mass of one molecule of a compound is 1.66 × 10–22 g. The molecular weight of the compound is (A) 166 (B) 1.66 (C) 100 (D) 1000 Statement A: The number of atoms present in gram atomic weight of different elements are equal. (L - I) Statement B: The number of molecules present in gram molecular weight of different substances is equal. (A) ‘A’ is true, ‘B’ is false (B) ‘A’ is false, ‘B’ is true (C) Both ‘A’ and ‘B’ are true (D) Both ‘A’ and ‘B’ are false Find the number of gram molecules present in the following: i) 5g of Neon ii) 7 g of nitrogen (i) (ii) (A) 0.25 0.25 (B) 0.25 0.5 (C) 0.5 0.25 (D) 1 2 Find the number of gram molecules of hydrogen present in 1 gram molecule of methane gas. (A) 1 (B) 2 (C) 4 (D) 8 Gases like H2, O2 are called elementary gases, whereas gases like SO2, NO2 and CH4 are called compound gases. 1 gram mole of which compound gas weighs least? (A) N2 O (B) NO2 (C) SO2 (D) CH4

Formative Worksheet 11.

(B) mass of

12

13.

14.

15.

16.

17.

18.

1 th of C – 12 isotopic atom 12

(C) 12 amu (D) None The mass of an atom of an element ‘x’ is 39. The number of atoms of it present in gram atomic weight of it is_______. (A) 1 (B) 1.66 × 1024 23 (C) 6.023 × 10 (D) 96500 The approximate number of electrons that are required to make 1 smallest unit of mass is (A) 6.023 × 1023 (B) 1.66 × 1024 (C) 1852 (D) 2500 Gram atom of any element contains (A) 6.023 × 1023 atoms (B) 3.0115 × 1023atoms (C) 1.505 × 1023 atoms (D) 12.0 × 1023 atoms Find the number of gram atoms present in the following: (i) 4g of hydrogen (ii) 1 g of helium (i) (ii) (A) 2 1 (B) 0.25 4 (C) 1 4 (D) 4 0.25 The ratio of weights of hydrogen and helium is 1 : 2. Find the ratio of number of gram atoms. (A) 2 : 1 (B) 1 : 1 (C) 1 : 4 (D) 4 : 1 How many gram atoms of the lightest element weigh same as 1 gram atom of the heaviest element? (A) 1 (B) 235 (C) 238 (D) 100

21.

22.

23.

24.

Conceptive Worksheet 15.

1 amu is equal to the mass of

weight of one atom of an element = x. Its atomic weight

(A)

weight of one molecule of a compound = y. Its molecular weight

(B) 16.

Then, x : y is

1 12 (C) 1 : 2 (A) 1:

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(B) 2 : 1 (D) 1 : 1

1 th of C - 12 atom 12

1 th of O-16 atom 14 (C) 1g of H2 (D) 1.66 × 10–23 kg Atomic weight of an element is ‘x’. The weight of one atom of that element is (A) ‘x’ amu (B) x × 1.66 × 10–24g –27 (C) 1.66x × 10 kg (D) All

Mole Concept 17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

The mass of one atom of an element is 40 × 1.66 × 10–24g. The number of protons in its nucleus is (A) 40 (B) 20 (C) 10 (D) 5 The weight of Helium atom in grams is (A) 2 (B) 4 –24 (C) 6.64 × 10 (D) 1.66 × 10–24 Atomic weight of an element is x. It means that weight of one atom of that element is 1 xg (A) ‘x’ g (B) 12 (C) 12 × x g (D) 1.66x × 10–24g Which of the following is correct? (A) Molecular weight of oxygen is 32. (B) Gram molecular mass of sulphur (S 8) is 256 g. (C) The weight of one molecule of O3 is 48 amu. (D) All Among all the naturally occurring elements, which one can generate the maximum number of gram atoms from a given amount? (A) Hydrogen (B) Uranium (C) Calcium (D) Mercury Which of the following is the smallest particle of matter that exist independently? (A) Atom (B) Molecule (C) Element (D) Compound In which of the following, the smallest particles exists in diatomic form? (A) Hydrogen gas (B) Nitrogen gas (C) Oxygen gas (D) Flourine gas Match the following: Column - I Column - II i) Sodium p) Monoatomic ii) Helium q) Diatomic iii) Oxygen r) Triatomic iv) Ozone s) Poly atomic v) Sulphur (A) i, ii - p; iii - q; iv - r; v - s (B) iii, ii - p; i - q; iv - r; v - s (C) i,iii - p; iv - q; ii - r; v - s (D) i,iv - p; iii - q; ii - r; v - s The units of molecular mass (or) molecular weight is (A) Amu (B) Grams (C) Both ‘a’ and ‘b’ (D) None The weight of ammonia molecule in grams is (A) 17g (B) 17 × 10–3 –24 (C) 17 × 1.66 × 10 (D) 17 × 1.66 × 10–27

169 27.

28.

100 g of which gas contains of the maximum number of gram molecules? (A) SO2 (B) O2 (C) He (D) H2 Identify the gas whose 2 gram molecule weigh 32g. (A) He (B) O2 (C) CH4 (D) SO2

10. Avogadro number Number of atoms present in gram atomic weight of an element is called Avogadro’s number. Avogadro number = 6.023 ×1023.It is denoted by N or NA. Similarly, the number of molecules present in gram molecular weight of nay substance is equal to Avogadro number (6.023 × 1023). Thus, Avogadro number of atoms = 6.023 ×1023 atoms. Avogadro number of molecules = 6.023 ×1023 molecules. Avogadro number of ions = 6.023 ×1023 ions.

11. Mole In general, we buy groceries in several ways. If we buy eggs or bananas, we buy them by the dozens – dozens – that is by number, by counting them. Eggs or bananas are easy to count. So are the oranges or apples.But there are the other types of items, though countable, are more conveniently sold by mass. A dozen peanuts is too small a number to buy by counting and several hundreds are difficult to count. So we buy them by weighing them. In the first case, it is more convenient to buy the eggs or bananas by counting rather than weighing them. Unknowing the information about the weight of eggs doesn’t affect any thing. In the second case, it is more convenient to buy the peanuts by weighing, rather than counting them, Unknowing the information about the number of peanuts doesn’t affect any thing. Now, let’s turn our attention from this everyday physical world to the chemical world of Scientists (Chemical world). Unlike the above two cases, both knowing weight and number of particles (atoms/molecules/ions/ chemical units ) present in the compounds is very important. Since this is of vital importance, ignorance or such knowledge may at times even lead to devastation and destruction. In fact, life itself would come to a standstill if such facts were www.betoppers.com

9th Class Chemistry

170 not known. Thus, we require a standard of measurement which is all embracing dealing at once with macroscopic world and the microscopic world. In other words, such a system should give us a comprehensive view of the number of particles in any given substance. For instance, units like grams or kilograms, which we use in everyday life, would be inadequate when dealing with two different substances of the same weight containing the same number of particles. Hence the conventional units of measurement would not serve our purpose here. But our ever thoughtful scientists have come to our rescue. They have come out with a solution by identifying a unit which is not only easy to use but also appropriate to its purpose. It gives just what is required – A magnified view of the anatomy of microscopic world, acting as a bridge between the tangible and the intangible worlds. Can you guess what this unit is ? It is nothing but .... mole. Let’s have a detailed understanding of mole....

The weight of 1 mole of ‘C’ atoms = 12 grams = GAW of ‘C’ The weight of 1 mole of H2O molecules = 18 grams = GMW of H2O Conclusion: The weight of 1 mole of any substance is equal to its gram atomic weight or gram molecular weight. of an element: of atoms weights

1 Mole

of a compound of molecules

of a salt of formula units

G A W Gram atomic we ight

G M W Gram mole cular weight

G F W Gram formula weight

(or) a of hydrogen element of hydrogen atoms weights

1g

Understanding of mole In Latin, the term mole literally means heap, pile or mass. In represents a definite number as century which means 100. it is used as the bridge in chemistry between the atomic and macroscopic scales. For example:In banks they do not count the coins, but weight them, as they know that a fixed number of a particular coin will always have the same mass. Similarly, chemists count atoms and molecules by weighing them.One mole of each of the different substances contains the same number of elementary units. Mole is the mass of the substance containing particles equal to Avogadro number (6.023 ×1023). 1. mole of any substance = Weight of 6.023 ×1023 2. mole of any subtance contains 6.023 ×1023 particles

Mole and Weight relation What is the weight of 1 mole of carbon atoms and 1 mole of water molecules? Wt. of 1 mole of 1 ‘C’ atoms

Wt. of 1 mole of 1 H 2O molecules

= 6.023 × 1023 × wt, of ‘C’ atom = 6.023 × 1023 × wt. of H2O molecules = 6.023 × 1023 × 12 a.m.u = 6.023 × 1023 × wt. of 18 a.m.u. 23 24 = 6.023 × 10 × 12 × 1.66 × 10 g = 6.023 × 1023 × 18 × 1.66 × 1024 g = 12 grams = 18 grams

What do you observe? www.betoppers.com

of hydrogen gas (or) of hydrogen molecules

1g

1 Mole of NaCl (or) of NaCl formula units

58.5 g

Mole and Particles relation How many particles are present in 1 mole of any substance ? We know that 1 mole of any substance contains 6.023 × 1023 respective particles. Particles can be: Atoms, molecules, ions, etc., Let us understand it with examples …

Mole Concept

171 of hydrogen element contain

1 Mole

of hydrogen gas contain

of NaCl contain

6.023

10 23 H atoms

6.023

10 23 H 2 molecules

6.023

10 23 NaCl formula units

6.023 10 23 Particles (atoms/ions/molecules, etc.,

1 Mole Of a substance

Mole and Volume relation for a gas 1 mole of any gas at STP or NTP occupies 22.4 litres. This volume occupied any gas at STP is also called Gram Molar Volume (GMV)

1 Mole of any gas

at STP occupies

22.4 litres

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9th Class Chemistry

172

Few more relations Weight – Particles relation: ‘GAW’ or ‘GMW’ of any substance consists of ‘N’ (Avogadro number) particles or ‘N’ particles of any substance weighs its GAW or GMW respectively. For example, 32 grams of Oxygen (O2) gas consists of ‘N’ molecules or ‘N’ molecules of Oxygen gas weighs 32 grams. Weight – Volume relation (only for gases): ‘GAW’ or ‘GMW’ of any gas occupies 22.4 litres at S.T.P. or 22.4 litres of any gas weighs ‘GAW’ or ‘GMW’ at S.T.P. For example, 2 grams of Hydrogen (H2) gas occupies 22.4 litres at S.T.P or 22.4 litres of Hydrogen gas at S.T.P weighs 2 grams Volume – Particles relation (only for gases): 22.4 litres of any gas at S.T.P consists of 6.023 × 1023 molecules or 6.023 × 1023 molecules of any gas at S.T.P. occupies 22.1 litres of a gas. For example, 2 grams of Hydrogen (H2) gas occupies 22.4 litres at S.P.T. or 22.4 litres of Hydrogen gas at S.T.P. weighs 2 grams.

SUMMARY OF RELATIONS

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Mole Concept

173

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9th Class Chemistry

174

SOME OF IMPORTANT FORMULAE Formulae to calculate

Number of moles

Number of molecules

Volume occupied at S.T.P.

Na 

Number of atoms of an element

m  6.023  1023 GMW

VSTP  

m  22.4 GMW

N m  n  6.023  1023

 or  N m  n

m  6.023  1023 GMW

Given weight  m  Gram mol weight  GMW 

Some more important relations: 1.

5. 6. 7.

8. 9.

Number of molecule in 1 gram of a substance 6.023  1023 = Molecular weight

11.

No.of atoms 

Weight of x gram atoms = x  Gram atomic weight Weight of x moles of any compound = x  Gram 15. molecular weight 1 gram atom or gram atomic weight of an element contains 6.023  1023 atoms. 1 gram molecule or gram molecular weight of a substance contains 6.023  1023 molecules. Number of atoms in a given substance ( given element) = Number of gram atoms 16. (ng)  6.023  1023 Number of molecules in a given substance ( Nm) = Number of moles (n)  6.023  1023 Number of atoms in 1 gram of an element 6.023  1023 = Atomic weight

10.

14.

Number of moles (n)

Given weight = Gram Molecular weight 3. 4.

13.

Weight of substance in grams = Number of moles  GMW Number of atoms of an element per molecule can be calculated if MW and percentage mass of that element are given by using the formula.

No of gram atoms or mole atoms

Given weight = Gram atomic weight . 2.

12.

Weight of an element in grams = Number of gram atoms  GAW W

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MW  Percentage mass At.wt  100

[ Note: Number of atoms is always is a whole number] No. of atoms present in given amount of substance (Na) = No. of molecules (Nm) × No. of atoms present in 1 molecule of the substance.

= No. of moles (n) × NA × No. of atoms present in 1 molecule of the substance. No. of subatomic particles (electrons / protons, etc) present in given amount of substance (Np) = No. of molecules (Nm) × No. of subatomic particles present in 1 molecule of the substance. = No. of moles (n) × NA × No. of subatomic particles present in 1 molecule of the substance.

Mole Concept

Formative Worksheet 25. 26. 27. 28.

29.

30.

31. 32. 33. 34.

35.

36.

37.

38.

39.

How many moles are 5 grams of calcium ? (Atomic mass of calcium = 40 u). What is the mass of 4 moles of aluminium atoms? (Atomic mass of A1 = 27 u) Calculate the number of atoms in 0.2 mole of sodium (Na). If 1 g of carbon contains x atoms, what will be the number of atoms in 1 g of magnesium?(C = 12 u, Mg = 24 u) How many grams of neon will have the same number of atoms as 4 grams of calcium ? (Atomic masses : Ne = 20 u, Ca = 40 u) The mass of a single atom of an element X is 2.65 × 10–23 g. What is its atomic mass ? What could this element be ? Convert 22g of carbon dioxide (CO2) into moles. (Atomic masses : C = 12 u ; O = 16 u) What is the mass of 0.5 mole of water (H2 0). (Atomic masses :H = lu ; O = 16u) What is the number of molecules in 0.25 moles of oxygen ? Which contains more molecules, 4 g of methane (CH4) or 4 g of oxygen (02) ? (Atomic masses:C = 12 u, H = 1 u, 0 = 16 u) If 1 g of sulphur dioxide contains x molecules, what will be the number of molecules in 1 g of methane ? (S = 32 u, O = 16 u, C = 12 u, H = 1u) How many grams of oxygen gas contain the same number of molecules as 16 grams of sulphur dioxide gas? (O = 16 u, S = 32 u) Calculate the number of aluminium ions present in 0.051 g of aluminium oxide (A12O3). (Atomic masses : A1 = 27 u; O = 16 u) How many moles and molecules of O2 are there in 64 g O2? What is the mass of one molecule of O2? (A) 3 moles, 12.04 × 1023 molecules, 6.3 × 10–23 g (B) 2 moles, 13.02 × 1023 molecules, 7.350 × 10–23 g (C) 4 moles, 11.04 × 1023 molecules, 5.152 × 10–23 g (D) 2 moles, 12.04 × 1023 molecules, 5.313 × 10–23 g From 200 mg of CO2, 1021 molecules are removed. How many grams and moles of CO2 are left? (A) 126.9 mg, 2.88 × 10–3 (B) 127.5 mg, 2.88 × 10–3 (C) 130 mg, 2.88 × 10–13 (D) 162.9 mg, 2.88 × 10–13

175 40.

41.

42.

43.

44.

45.

46. 47. 48. 49. 50. 51.

52. 53.

An alloy has Fe, Co and Mo equal to 71%, 12% and 17% respectively. How many cobalt atoms are there in a cylinder of radius 2.50 cm and a length of 10.0 cm? The density of alloy is 8.20 g/ mL. Atomic weight of cobalt = 58.9. (A) 0.198 × 1023 (B) 198 × 1023 23 (C) 19.8 × 10 (D) 1.98 × 1023 Calculate the number of Cl– and Ca2+ ions in 222 g anhydrous CaCl2. (A) 3N, 6N (B) 4N, 2N (C) 10N, 5N (D) 6N, 3N The dot at the end of this sentence has a mass of about one microgram. Assuming that black stuff is carbon, calculate approximate atoms of carbon needed to make such a dot. (A) 6 × 105 (B) 4 × 103 6 (C) 5 × 10 (D) 5 × 105 What is the molecular weight of a substance, each molecule of which contains 9 carbon atoms, 1 3 hydrogen atoms and 2.33 × 10 –23 g of other component? (A) 130 (B) 40 (C) 150 (D) 160 The total number of protons in 10 g of calcium carbonate is (N0 = 6.02 × 1023) (A) 1.5057 × 1024 (B) 2.0478 × 1024 24 (C) 3.0115 × 10 (D) 4.0956 × 1024 How many atoms are contained in one mole of sucrose (C 12 H22 O11 )? (A) 45 × 6.02 × 1023 atoms/mole (B) 5 × 6.62 × 1023 atoms/mole (C) 5 × 6.02 × 1023 atoms/mole (D) None of these 0.125 moles of any gas at S.T.P. occupies x litres. Find x. 2.8 litres of any gas at S.T.P make ________ moles. What is the volume occupied by 17.75 g of Cl2 at S.T.P. ? Calculate the mass of 500 cc of hydrogen at S.T.P. 2.24 litres of a gas at S.T.P. weigh 1.6 g. Identify the gas. In Victor – meyer’s method 0.2 g of a volatile compound on volatilization gave 56 ml of vapour at S.T.P. Find the molecular weight of the compound. Find the number of hydrogen molecules present in 1.12 litres of it at S.T.P. 1 molecules of any gas at S.T.P. occupies the same volume. True/False www.betoppers.com

9th Class Chemistry

176

Conceptive Worksheet 29. 30. 31. 32.

33. 34. 35. 36.

37. 38. 39.

40. 41.

42.

43. 44.

12. Percentage composition

How many moles are 9.033 × 1024 atoms of It is defined as the number of parts by weight of helium (He) ? different elements, present in hundred parts by Calculate the number of iron atoms in a piece of weight of the compound. iron weighing 2.8 g (Atomic mass of iron = 56u). It can be calculated by using the following formula. If one mole of carbon atoms weighs 12 grams, Percentage composition of an element in a what is mass in grams of 1 atom of carbon ? compound Which has more number of atoms, 100 grams of sodium or 100 grams of iron ? No. of atoms of × Atomic weight (Atomic masses : Na = 23 u ; Fe = 56 u) the elements (n) of the element (AW) = ×100 Convert 12.044 × 1022 molecules of sulphur dioxide Molecules weight of into moles. the compound (MW) What is the number of water molecules contained in a drop of water weighing 0.06 g ? [Note: Percentage composition is also the Calculate the mass of 3.011 × 1024 molecules of percentage by mass of atoms of an element present nitrogen gas (N2). (Atomic mass : N= 14 u) in mole of the compound.] The absolute mass of one molecule of a substance For example, percentage composition of calcium is 5.32 × 10–23 g. What is its molecular mass ? 1× 40 What could this substance be ? ×100 = 40%, that carbon is in CaCO3 = 100 Calculate the number of atoms present in 1 g He gas. 1 12  100 = 12% and that of oxygen is The density of O2 at NTP is 1.429 g/L. 100 Calculate the standard molar volume of gas. A compound contains 28% N and 72% of a 3×16 ×100 = 48% . metal by weight. Three atoms of metal combine 100 with two atoms of N. Find the atomic weight of Percentage composition can be also calculated for metal. a component in a compound. The number of oxygen atoms in 4.4 g of CO2 is % Composition of any component approximately is _____________ Weight of the component is compound The number of water molecules present in a  100 Total weight of the compound drop of water (volume 0.0018ml) at room temperature is_____________ For example, consider washing soda, Na 2CO3 The number of water molecules in 1 litre of 10H 2 O. Let us calculate the percentage water is (NA = Avogadro composition of water in washing soda. 286 g of number)_____________ washing soda contain 180 g of water. What is the volume occupied by 0.25 moles of 180 laughing gas at S.T.P?  100 = 62.93 % composition of water = 286 Find the number of moles in each of the cases, assuming S.T.P. condition. ormative orksheet (a) 2.24 litres of marsh gas (b) 1.12 litres of Tear gas (CCl3NO2) 54. Haemoglobin contains 0.25% iron by weight. The (c) 11.2 litres of N2 molecular weight of Haemoglobin is 89600. In which one of the following cases the number of Calculate the number of iron atoms per molecule hydrogen atoms is more ? of haemoglobin. (A) 3 (B) 4 Two moles of HC1 or One mole of NH3 (C) 5 (D) 2 55. Glucose (C6H12O6) is the most important nutrient in the living cell for generating chemical potential energy. What is the percentage composition of each element in glucose?

F

45.

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W

Mole Concept (A) C = 53.3%

H = 40% O = 6.7% (C) C = 40%

H = 6.7% O = 53.3% 56.

57.

58.

177 (B) C = 6.7%

H = 40% O = 53.3%

47.

48.

(D) C = 40%

H = 53.3% O = 6.7%

Iron, as Fe2+, is an essential nutrient. Pregnant women often take 325 mg ferrous sulphate (FeSO4) tablets as a dietary supplement. Yet iron tablets are poisonous, causing death in children. As little as 550 mg Fe2+ can be fatal to a 8 kg child. How many 325 mg tablets would it take to constitute a lethal dose to a 8 kg child? (A) 7 (B) 5 (C) 3 (D) 4 What weight of Fe2O3 can be obtained from 5 tons Fe? [Fe = 56; O = 16] (A) 6.124 tons (B) 5.412 tons (C) 7.142 tons (D) 8.214 tons Calcium is one of the most important elements in the diet because it is a structural component of bones, teeth, and soft tissues and is essential in many of the body’s metabolic processes. 99 percent of the total calcium is stored in bones and teeth. The deficiency of calcium leads to weakening of bones, teeth and soft tissues. So, children are often given extra calcium in the form of calcium tablets and other compounds like I) Calcium carbonate, CaCO3 II) Calcium lactate, Ca(C3H5O3)2 III) Calcium gluconate, Ca(C6H11O7)2 IV) Calcium citrate, Ca3(C6H5O7)2 Calculate the percentage of calcium in each of the above compounds. I II III IV A) 40 18.34 9.3 24.48 B) 9.3 40 24.48 18.34 C) 40 24.48 9.3 18.34 D) 24.48 9.3 18.34 40

49. 50.

Summative Worksheet 1.

2.

Calculate the percentage composition of: a) Potassium in potassium dichromate [K = 39, Cr = 52, O= 16] b) Phosphorus in calcium phosphate [Ca3(PO4)2] [Ca = 40, P - 31, 0 =16]

Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide. According to law of conservation of mass, if 5.6 gms of calcium oxide and 4.4 gms of carbon dioxide are formed, how much calcium carbonate was taken? (A) 11g (B) 10g (C) 15g (D) 12g 0.24 grams of sample of compound of Oxygen and Boron was found by analysis to contain 0.096 g of Boron and 0.144 g of Oxygen. Calculate the percentage composition of the compound by mass.

Boron  40% (A)

Oxygen  60%

Boron  30% (B)

Boron  44% (C) 3.

Conceptive Worksheet 46.

Calculate the mass of nitrogen supplied to the soil by 5 kg of urea. [CO(NH2)2] [N = 14, C = 12, O = 16, H = 1] Calculate the percentage composition of i) Silver in silver chloride ii) Iron in ferric oxide and iii) Potassium in potassium sulphate Calculate the percentage of hydrogen and oxygen in water. Insulin contains 3.4 % sulphur. Calculate minimum mol. wt. of insulin. (A) 614.27 (B) 941.17 (C) 841.27 (D) 714.17

4.

Oxygen  56%

Oxygen  70% Boron  42%

(D)

Oxygen  58%

The percentage of the three elements calcium, carbon and oxygen in a given sample of calcium carbonate is given is: Calcium = 40.0%; Carbon = 12.0% ; Oxygen = 48.0% If the law of constant composition is true, what weights of these elements will be present in 1.5 g of another sample of calcium carbonate? (A) 0.6 g, 0.12 g, 0.60 g (B) 0.7 g, 0.18 g, 0.80 g (C) 0.80 g, 0.20g, 0.72 g (D) 0.6 g, 0.18 g, 0.72 g Silver chloride was prepared in two days:(L-II) a) 0.5 g of a silver wire was dissolved in concentrated nitric acid to convert it into silver nitrate. Excess of hydrochloric acid was added to it. As a result white precipitate of silver chloride was formed. This was separated by filtration, washed, dried and then weighed. The weight of residue was formed to be 0.68 g. www.betoppers.com

9th Class Chemistry

178

5.

6.

7.

8.

9.

10.

11.

12.

b) In another experiment, 1 g of silver metal was heated with chlorine till it was completely converted to silver chloride. The weight of residue was found to be 1.33 g. Show that the above data is according to the law of constant composition. (A) 70.53%, 20.47% ; 72.22%, 22.81% (B) 73.53%, 26.47% ; 75.19%, 24.81% (C) 73.53%, 25.47% ; 74.19%, 24.81% (D) 72.53%, 25.57% ; 75.29%, 26.81% Sodium carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water. In an experiment, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid to form 8.2 g of sodium ethanoate, 2.2 g of carbon dioxide and 0.9 g of water. Show that this data verifies the law of conservation of mass.7 Calcium carbonate decomposes, on heating, to form calcium oxide and carbon dioxide. When 10 g of calcium carbonate is decomposed completely, then 5.6 g of calcium oxide is formed. Calculate the mass of carbon dioxide formed. Which law of chemical combination will you use in solving this problem? In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of copper. In another experiment, 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show that these figures verify the law of constant proportions. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas ? When 3 g of carbon is burnt in 8 g of oxygen, 11 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3 g of carbon is burnt in 50 g of oxygen ? Which law of chemical combination will govern your answer ? A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by mass. The ratio of masss of hydrogen and helium is 1 : 2. Find the ratio of number of gram atoms. (A) 2 : 1 (B) 1 : 1 (C) 1 : 4 (D) 4 : 1 How many gram atoms of the lightest element weigh same as 1 gram atom of the heaviest element? (A) 1 (B) 235 (C) 238 (D) 100

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13.

14.

15.

16.

17.

18.

19.

20.

weight of oneatom of an element = x. Its atomic weight weight of one moleculeof a compound = y. Its molecular weight Then, x : y is 1 (A) 1: (B) 2 : 1 12 (C) 1 : 2 (D) 1 : 1 What is the mass of one carbon dioxide molecule? (A) 44g (B) 44 amu (C) 44 kg (D) 44 mg The mass of one molecule of a compound is 1.66 × 10 –22 g. The molecular mass of the compound is (A) 166 (B) 1.66 (C) 100 (D) 1000 Find the number of gram molecules present in the following: (i) 5g of Neon (ii) 7 g of nitrogen (i) (ii) (A) 0.25 0.25 (B) 0.25 0.5 (C) 0.5 0.25 (D) 1 2 Find the number of gram molecules of hydrogen present in 1 gram molecule of methane gas. (A) 1 (B) 2 (C) 4 (D) 8 Gases like H2, O2 are called elementary gases, whereas gases like SO2, NO2 and CH4 are called compound gases. 1 gram mole of which compound gas weighs least? (A) N2 O (B) NO2 (C) SO2 (D) CH4 How many moles and molecules of O2 are there in 64 g O2? What is the mass of one molecule of O2? (A) 3 moles, 12.04 × 1023 molecules, 6.3 × 10–23 g (B) 2 moles, 13.02 × 1023 molecules, 7.350 × 10–23 g (C) 4 moles, 11.04 × 1023 molecules, 5.152 × 10–23 g (D) 2 moles, 12.04 × 1023 molecules, 5.313 × 10–23 g From 200 mg of CO2, 1021 molecules are removed. How many grams and moles of CO2 are left? (A) 126.9 mg, 2.88 × 10–3 (B) 127.5 mg, 2.88 × 10–3 (C) 130 mg, 2.88 × 10–13 (D) 162.9 mg, 2.88 × 10–13

Mole Concept 21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

An alloy has Fe, Co and Mo equal to 71%, 12% and 17% respectively. How many cobalt atoms are there in a cylinder of radius 2.50 cm and a length of 10.0 cm? The density of alloy is 8.20 g/ mL. Atomic mass of cobalt = 58.9. (A) 0.198 × 1023 (B) 198 × 1023 23 (C) 19.8 × 10 (D) 1.98 × 1023 Calculate the number of Cl– and Ca2+ ions in 222 g anhydrous CaCl2. (N = Avogadro number) (A) 3N, 6N (B) 4N, 2N (C) 10N, 5N (D) 6N, 3N The dot at the end of this sentence has a mass of about one microgram. Assuming that black stuff is carbon, calculate approximate atoms of carbon needed to make such a dot. (A) 6 × 1015 (B) 4 × 1013 16 (C) 5 × 10 (D) 5 × 1015 What is the molecular mass of a substance, each molecule of which contains 9 carbon atoms, 1 3 hydrogen atoms and 2.33 × 10 –23 g of other component? (A) 135 (B) 140 (C) 150 (D) 160 The total number of protons in 10 g of calcium carbonate is (N0 = 6.02 × 1023) (A) 1.5057 × 1024 (B) 2.0478 × 1024 24 (C) 3.0115 × 10 (D) 4.0956 × 1024 How many atoms are contained in one mole of sucrose (C 12 H22 O11 )? (A) 45 × 6.02 × 1023 atoms/mole (B) 5 × 6.62 × 1023 atoms/mole (C) 5 × 6.02 × 1023 atoms/mole (D) None of these A piece of copper weighs 0.635 g. How many atoms of copper does it contain ? (Atomic mass of copper = 63.5). (A) 6.02 × 1023 (B) 6.02 × 1022 21 (C) 6.02 × 10 (D) None of these Find the number of molecules present in 11 grams of laughing gas. (A) 6.02 × 1023 (B) 12.04 × 1023 (C) 1.51 × 1023 (D) 3.011 × 1023 11 g of gold costs Rs. 5, 000/-. With this information, can you find out the cost of 1 atom of gold in rupees? (A) 147 (B) 1.47×10–19 –23 (C) 1.47 × 10 (D) 1.47 × 10–21 560 cc of a gas weighs 1.1 g at S.T.P. Find the molecular mass of the gas. (A) 11 (B) 22 (C) 33 (D) 44

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HOTS Worksheet 1.

2.

3.

4.

5.

6.

7.

Gypsum is a hydrated calcium sulphate. A 1.0 g sample contains 0.791 g CaSO4. How many moles of CaSO4 are there in this sample? Assuming that the rest of the sample is water, how many moles of H2 O are there in the sample? What is the molecular formula of gypsum? A drug marijuana owes its activity to tetra hydro cannabinol, which contains 70% as many carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetra hydro cannabinol is 0.00318. Determine the molecular formula. In making transistors, one needs to control the concentration of impurity very carefully. Suppose you wanted to make a germanium transistor containing 1.0  1018 boron atoms per cm3 as impurity. If the density of germanium is 5.35 g/cc, what relative weights of germanium need to be mixed? A metal ‘M’ of atomic weight 59.94 has density of 7.42 g/cc. Calculate the volume occupied and radius of the atom of this metal assuming it to be a sphere. An hourly requirement of an astronaut can be satisfied by the energy released when 34.2 g of sucrose (C12H22O11) are burnt in his body. How many grams of oxygen would be needed in a space capsule to meet his requirement for one day. ( Hint: Combustion of sucrose is given by C12H22O11 + 12 O2  12CO2 + 11 H2O). Naturally occurring carbon consists of two isotopes, C 12 , C 13 . What are the percent abundances of the isotopes in a sample of carbon whose atomic weight is 12.01112? Find the % composition of each element present in Mg3(PO4)2 . Hint: % composition =

No.atoms  atomic weight of the element  100 . Molecular weight of the compound 8.

Haemoglobin is the oxygen carrying molecule of red blood cells, consisting of protein and non-protein substance The non protein substance is called heme. A sample of heme weighing 35.2 mg contains 3.19 mg of iron. If a heme molecule contains one atom of iron, what is the molecular weight of heme?

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180 9.

10.

11.

12.

13.

14.

15. 16.

17.

18. 19. 20.

Sulphur molecules exist under various conditions as S8, S6, S2 and S. (A) Is the mass of one mole of each of these molecules the same? (B) Is the number of molecules in one mole of each of these molecules the same? (C)Is the mass of sulphur is one mole of each of these molecules the same? (D)Is the number of atoms of sulphur in one mole of each of these molecules the same? (A) How many grams of sulphur are required to produce 100 moles of and 100 g of H2SO4 separately? (B) A compound contain 28 % N2 and 72 % of a metal by weight. Three atoms of metal combine with two atoms of N. Find the atomic weight of metal. (A) Calculate the number of atoms of oxygen present in 88 g of CO2. What would be weight of CO having the same number of oxygen atoms? (B) The dot at the end of this sentence has a mass of about one microgram. Assuming that the black stuff is carbon, calculate approximate atoms of carbon needed to make such a dot. Find the number of atoms and molecules present in each of the cases. (A) 6 g of hydrogen gas (B) 96 g of ozone (C) 64 g of sulphur (S8) Find the molecular mass from the following information: (A) 3.0115 × 1022 molecules of which weigh 2.4 g. (B) 1 lakh molecules of which weigh 19.92 × 10–18g. Find the volume occupied by the following gases at S.T.P. (A) 19 g of F2 (B) 5 g of Argon (C) 8.5 g of NH3 The volume occupied by 10 grams of CO2 is equal to the volume occupied by 10 g of O3 gas. True / False The volume occupied by the gram molecular mass of SO2 is equal to the gram molecular mass of SO3 under the same conditions of temperature and pressure. True/False Find the mass of the given volume of gases, at S.T.P. A) 5.6 ml of O2 B) 2.24 litres of CH4 C) 112 cc of brown gas (NO2) The mass of 100 ml of any gas at S.T.P is the same. True/False The volume of 0.7 g of a volatile compound at S.T.P is 140 ml. Find its molecular mass. The volume occupied by 1 billion molecule of CO2 gas is equal to the volume occupied by 1 billion molecules of CH4 gas. True or false.

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21. 22. 23. 24. 25.

26.

27. 28. 29. 30. 31.

Calculate the number of atoms of each element present in 22 grams of CO2. Calculate the total number of atoms of different elements in 342 grams of sugar. Calculate the total number of protons, electrons and neutrons present in 50 grams of CaCO3. Calculate the number of nucleons present in 24.5 g of king of chemicals. Calculate the following, present in 0.25 mole of SO2 . (A) Amount of SO2 (B) Number of molecules (C) Number of gram atoms of S (D) Number of gram atoms of O (E) Volume occupied by SO2 at S.T.P. (F) Total Number of protons of SO2 Calculate the following present in 1.505 × 1022 molecules of CO2 (A) Amount of CO2 (B) Number of moles (C) Number of gram atoms of C (D) Number of gram atoms of O (E) Volume occupied by CO2 at S.T.P. (F) Number of protons of CO2 How to calculate the number of atoms present in ‘x’ grams of a substance ? What is the mass of 2.8 litres of H2S at S.T.P.? Find the volume of 8 grams of SO2 at S.T.P. Find the mass of H2SO4 present in 2 moles of it. Find the number of moles of sodium atom present in 0.23 grams of sodium.

IIT JEE Worksheet 1.

Vapour density of a volatile substance is

4(CH 4  1) . Its molecular weight would be 2.

3.

4.

5.

(A) 8 (B) 2 (C) 64 (D) 128 The ratio of the rates of diffusion of a given element to that of helium is 4. What will be the molecular weight of the element? (A) 0.25 (B) 4 (C) 2 (D) 7 What is the volume (in litres of oxygen required at STP to completely convert 1.5 moles of sulphur into sulphur dioxide? (A) 11.2 (B) 22.4 (C) 33.6 (D) 44.8 Gas A with molecular weight 4, diffuses thrice as fast as the gas B, the molecular weight of B is (A) 24 (B) 12 (C) 36 (D) 18 Which of the following contains equal number of atoms as 12g of magnesium? (A) 12g of Carbon (B) 20g of Calcium (C) 24g of Carbon (D) 40g of Calcium

Mole Concept 6.

If atomic weight of carbon is 12 and Avogardo’s number is 6.0 × 1023, then the weight of one atom of carbon is (A)

7.

8.

9.

10.

11.

12.

13.

14. 15. 16.

181

12 g 6.0  1023

(B)

17.

6.0  1023 g 12

12 6.0  1022 g g (C) (D) 6.0  1022 12 180 g of water contains _____ number of moles. (A) 100 (B) 180 (C) 10 (D) 0.01 Four one litre flasks are separately filled with the gas hydrogen, helium, oxygen and ozone at the same room temperature and pressure. The ratio of total number of atoms of these gases present in the different flasks would be (A) 1 : 1 : 1 : 1 (B) 1 : 2 : 2 : 3 (C) 2 : 1 : 2 : 3 (D) 3 : 2 : 2 : 1 A flask of gaseous CCl4 was weighed at a certain temperature and pressure. The mass of CCl4 vapour would be about (A) The same as that of O2 (B) Five times that of O2 (C) One fifth as heavy as O2 (D) Twice as heavy as O2 20 litres of H2 gas weigh about (A) 12.2 g (B) 448g (C) 1.8g (D) 20g Vapour density of a gas is 22. That gas could be (A) Carbon dioxide (B) Nitrous oxide (C) Propane (D) Any of these A and B are two identical vessels. A contains 15g of ethane at 1atm and 298K. The vessel B contains 75g of a gas X2 at same temperature and pressure. The vapour density of X2 is (A) 75 (B) 150 (C) 37.5 (D) 45 How many moles of Helium gas occupy 22.4litres at 30°C and 1 atmosphere? (A) 0.11 (B) 0.90 (C) 1.0 (D) 1.11 22 The weight of 1 × 10 molecules of CuSO4 . 5H2O is _________. Equal volumes of all gases all contain equal number of moles, provided _____. The total number of electrons present in 18ml of water (density of water is 1 gm/cc is_________.

18.

At a given temperature and pressure, 2 volumes of A combine with 5 volumes of B to form 2 volumes of C and 1 volume of A combines with one volume of B to form 2 volumes of D. The formula of C is (A) AB5

(B) A5B2

(C) A2B5

(D) AB

CH 4  2O2  CO2  2 H 2O If 4gm of methane is to be completely burnt, the amount of O2 needed is

19.

20.

(A) 8gm

(B) 16gm

(C) 4gm

(D) 32gm

6 grams of magnesium atoms of atomic weight 24 react with excess of acid, the amount of hydrogen produced would be (A) 0.5gm

(B) 1.0gm

(C) 2gm

(D) 4gm

What quantity of lime stone ( CaCO3 ) on heating will give 56kg of CaO ?

21.

22.

(A) 1000kg

(B) 56kg

(C) 44kg

(D) 100kg

27g of ‘Al’ will react completely with ____ of oxygen. (A) 8gms

(B) 16gms

(C) 32gms

(D) 24gms

The number of water molecules in 1 litre of water is ( N = Avogadro’s number). (A) 18

(B) 18 × 1000

(C) N

(D) 55.55N

23.

3.5gm of Carbon monoxide at O°C and 760mm pressure contains ____ molecules.

24.

Which one of the following gas contains, the same number of molecules as 16g of Oxygen? (A) 16g O3

(B) 16g SO2

(C) 32g SO2

(D) all the above

25.

The number of moles of BaCO3 which contains 1.5moles of oxygen atoms is ____.

26.

The number of moles of water present in 90g of water is ____. www.betoppers.com

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Modern atomic weights of elements are based on _________ scale.

28.

The pair of gases containing the same number of molecules is (A) 22g of CO2 and 72g of N2 (B) 11g of CO2 and 28g of N2 (C) 44g of CO2and 7g of N2 (D) 11g of CO2 and 7g of N2

29.

30.

The vapour density of a gas is 11.2. The volume occupied by 11.2g of the gas at S.T.P. is (A) 2 litres

(B) 4 litres

(C) 11.2 litres

(D) 22.4 litres

Which of the following pairs of gases contain the same number of molecules? (A) 16g of O2 and 14g of N2 (B) 8g of O2 and 22g of CO2 (C) 28g of N 2 and 22g of CO2 (D) 32g of O2 and 32g of N2

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Chapter - 8

Solutions

Learning Outcomes By the end of this chapter, you will understand     

Solution Properties of a Solution Types of solutions Suspensions and Colloids Saturated, Unsaturated and Supersaturated Solutions

 Solubility of a Solution  Factors affecting solubility  Concentration of a Solution

1. Introduction

Some common examples of solutions are : Salt solution, Sugar solution, Vinegar, Metal alloys (such as Brass) and Air. Salt solution is a homogeneous mixture of two substances, salt and water, whereas sugar solution is a homogeneous mixture of two substances, sugar and water. Some more examples of the solutions are : Sea-water, Copper sulphate solution, Alcohol and water mixture, Petrol and oil mixture, Soda water, Soft drinks (like Coca Cola and Pepsi, etc.), and Lemonade (which is a sweetened drink made from lemon juice or lemon flavouring). The substances like salt, sugar, etc., which dissolve in water completely are said to be ‘soluble’ in water.

All extraordinarly problems have simple solutions provided there is a stable, calm and tranquil mind to think. It appears, the shrewd leader of the refugee community had it. The challenge ahead of him was to covey to Jadi Rana that their presence would add value to his kingdom without disturbing. He had to overcome this challenge with another handecap of language barrier but he came out successfully with a brilliant solution. He simply added little powdered sugar to the milk. Jadi Rana was too sharp to miss the message behind the gesture of sweetening the milk and he generously allowed the people to stay in his state and rest is history. Let’s hope all the problems of the world have such sugary solutions. In this chapter, we shall discuss about such sweet solutions, their types, strength of the solutions, etc.

2. Solution A solution is a homogeneous mixture of two or more substances. A homogeneous mixture means that the mixture is just the same throughout. The ‘substance which is dissolved’ in a liquid to make a solution is called ‘solute’, and the ‘liquid’ in which solute is dissolved is known as ‘solvent’. For example, salt solution is made by dissolving salt in water, so in salt solution, ‘salt’ is the ‘solute’ and ‘water’ is the ‘solvent’. Similarly, the substances like sugar, ammonium chloride, copper sulphate and urea, etc. , which are dissolved in water to make solutions are called ‘solutes’, whereas water is the ‘solvent’. Usually, the substance present in lesser amount in a solution is considered the solute, and the substance present in greater amount in a solution is considered the solvent. Please note that the solute particles are also called ‘dispersed particles’ and solvents are also known as ‘dispersion medium’

Tea

Soft drink

Alcohol and water

Note: Alloys are homogeneous mixtures of metals and cannot be separated into their components by physical methods. But still, an alloy is considered as a mixture because it shows the properties of its constituents and can have variable composition. For example, brass is a mixture of approximately 30% zinc and 70% copper.

3. Properties of a Solution 1.

Let’s perform the following activity and understand the properties of a solution. Shake some sugar with water in a beaker, the sugar seems to disappear in water and we get a transparent sugar solution.

9th Class Chemistry

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Try to observe the dissolved sugar particles using a microscope. The dissolved sugar particles cannot be seen even with a microscope, and the sugar does not settle down even on keeping the solution for quite some time. Allow the sugar solution to pass through a filter paper. The whole solution passes through the filter paper and no residue is left behind. Taste the sugar solution. The sweet taste of the sugar solution, however, shows that sugar is present in it.

Water

Sugar solution

Sugar

4. Types of Solutions

Sugar solution

Conclusions a) b)

c)

small that they cannot reflect light rays falling on them). From the above activity, we can now state the properties of a true solution which are as follows: 1. A solution is a homogeneous mixture. 2. The size of solute particles in a solution is extremely small. It is less than 1 nm in diameter (1nanometre = 10–9 metre). 3. The particles of a solution cannot be seen even with a microscope. 4. The particles of a solution pass through the filter paper. So, a solution cannot be separated by filtration. 5. The solutions are very stable. The particles of solute present in a solution do not separate out on keeping. 6. A true solution does not scatter light (This is because its particles are very, very small).

From these observations, we conclude that Sugar solution is a homogeneous mixture having the same composition throughout. Sugar solution is a true solution. In a true solution, the particles of the solute break up to such an extent that they disappear into the spaces between the solvent molecules. So, in a sugar solution, the sugar particles break up to such an extent that they disappear into the spaces between the water molecules. A sugar solution does not scatter a beam of light passing through it and render its path visible (because the sugar particles present in it are so

(a) Aqueous solution: A solution in which water is the solvent, is called aqueous solution. (b) Alcoholic solution: A solution in which alcohol is the solvent, is called alcoholic solution. (c) Dilute solution: A solution in which the amount of solute is relatively small compared to the amount of solvent in a given mass of its is called a dilute solution. (d) Concentrated solution: A solution in which the amount of solute is relatively large compared to the amount of solvent in a given mass of it is considered as concentrated solution.

Types of Solutions based on the physical state of solute and solvent Types of solution

Physical state of

Physical state of

solute

solvent

Solid in Solid

Solid

Solid

Solid in Liquid

Solid

Liquid

Solid in Gas Liquid in Solid

Solid Liquid

Gas Solid

Liquid in Liquid Liquid in Gas Gas in Solid

Liquid Liquid Gas

Liquid Gas Solid

Milk Fog H 2 in platinum

Gas Gas

Liquid Gas

Soda water Air

Gas in Liquid Gas in Gas

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Examples

Alloys Salt solution Smoke Cheese

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185

5. Suspensions and Colloids

4.

Separation: The solid particles reset in the suspension can be easily separated by ordinary filter papers. No special filter papars are needed for the purpose. So, a suspension can be separated by filtration. Unstablility: The suspensions are unstable. The particles of a suspension settle down after some time. The setteled particles of a suspension is known as a precipitate.

We have studied that homogeneous mixtures are regarded as solutions, or true solutions. The heterogeneous mixtures in which the components or constituents are present in more than one phase are of two types. These are known as suspensions 5. and colloidal solutions. Thus, we conclude that there are three types of mixture solutions. These are solutions (or true solutions), colloidal solutions and suspensions. They differ mainly in the size of the 6. Scattering of light: A suspension scatters a beam of light passing through it (because its particles are particles which are also responsible for the difference in their properties. In a true solution, the quite large). –9 size of the particles is less than 1 nm (1 nm = 10 Colloids m = 10–7 cm). In a colloidal solution, it is between 1 A colloid is a kind of solution in which the size of to 100 nm while in a suspension, the size of the solute particles is intermediate between those in true particles is more than 100 nm . solutions and those in suspensions. The size of solute

Solution

Suspensions

Colloidal solution

Suspension

particles in a colloid is bigger than that of a true solution but smaller than those of a suspension. Though colloids appear to be homogeneous to us but actually they are found to be heterogeneous when observed through a high power microscope. So, a colloid is not a true solution. Some of the examples of colloids (or colloidal solutions) are : Soap solution, Starch solution, Milk, Ink, Blood, Jelly and Solutions of synthetic detergents. Colloids are also known as colloidal solutions.

A suspension is a heterogeneous mixture in which Types of colloidal solutions the small particles of a solid are spread throughout We have stated earlier that the colloidal solutions a liquid without dissolving in it. are the heterogeneous mixtures. This means that Examples: the constituents are not present in a single phase. Chalk-water mixture, Muddy water, Milk of Actually there are two phases in a colloidal solution. magnesia, Sand particles suspended in water, and These are known as dispersed phase and dispersion Flour in water. Chalk-water mixture is a suspension medium. of fine chalk particles in water; muddy water is a The component present in smaller proportion is the suspension of soil particles in water; and milk of dispersed phase while the one present in greater magnesia is a suspension of magnesium hydroxide proportion is the dispersion medium. in water. Please note that solid particles and water Note: Please note that the dispersed phase in a remain separate in a suspension. The particles do colloidal solution is comparable with solute in a true not dissolve in water. solution. Similarly, the dispersion medium can be Properties of a suspension compared with the solvent. However, they differ in 1. Heterogeneous in nature: A suspension is a the sense that in a true solution, solute and solvent heterogeneous mixture. There are two pohases. are present in a single phase but in colloidal solution, The solid particles represent one phase while the they represent separate phases. In other words, a liquid in which these are suspended or distributed true solution is homogeneous while colloidal solution forms the other phases. is of heterogeneous nature. All this happens because 2. Size: The size of solute particles in a suspension is of the difference in the particle size. quite large. It is larger than 100 nm (10–7 m) in Just as in case of true solutions, the substances diameter. belonging to all the three states of matter can act 3. Visibility: The particles of a suspension can be as dispersed phase or dispersion medium depending seen easily seen with our naked eyes and also under upon their relative amounts or proportions. Thus, a microscope. www.betoppers.com

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nine different types of colloidal solutions are also possible. But there are actually eight and not nine as we notice in true solutions. When two different gases are mixed, they always form a homogeneous and not heterogeneous mixture. The different types of colloidal solutions along with a few examples are listed in the form of a table S. no

Dispersed

1. 2. 3. 4. 5. 6. 7. 8.

phase Gas Gas Liquid Liquid Liquid Solid Solid Solid

Dispersion Medium Liquid Solid Gas Liquid Solid Gas Liquid Solid

Name of colloidal Solutions Foam Solid foam Aerosol Emulsion Gel Aerosol Sols Solid sols

Examples Soap leather, whipped cream, soda water Pumice stone, rubber, bread Mist, fog, cloud, insecticide spray Milk, cod liver oil, tonics in liquid form Jelly, butter, cheese, boot polish, curd Smoke, dust storm, volcanic dust and haze Paints, starch dispersed in water, gold sol Alloys, coloured glasses, gem stones, ruby glass.

Although eight different types of colloidal solutions are possible, but the most common among them have liquid acting as the dispersion medium while solid or gas as the dispersed phase. Colloidal solutions are also known as colloidal sol.

Properties of colloidal solutions

1.

2.

3.

The important properties of colloidal solutions are briefly discussed below: Colloidal solutions are heterogeneous in nature. Colloidal solutions appear to be homogeneous but are actually heterogeneous in nature. This happens because of particle size (1 nm to 100 nm) which is quite close to particles in true solution. We cannot see the particles in a colloidal solution as we do in case of suspension. But these can be seen under a microscope. Colloidal solutions are a two phase system We have discussed above that the colloidal solutions represent a two phase system. These are dispersed phase and dispersion medium. That is why, the colloidal solutions are of heterogeneous nature. Colloidal particles pass through ordinary filter papers In most of the cases, the colloidal solutions pass through ordinary filter papers like true solutions. This is because of the fine size of the dispersed phase or colloidal particles. Special filter papers known as ultra filter papers have to be used to

separate these particles from the dispersion medium. 4.

Colloidal particles carry charge We have learnt that the dispersed phase particles in a colloidal solution remain dispersed or suspended. They do not come close to one another as in case of suspension. This happens due to the presence

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5.

of some charge (positive or negative) on these particles. Please remember that all the particles belonging to a particular colloidal solution carry the same charge. That is why, these similarly charged particles repel each other and remain dispersed or suspended. Particles in a colloidal solution follow zig-zag path It is normally not possible to see the colloidal particles because of their very small size. However, their path can be seen under a microscope. These particles follow a zig-zag path. You can observe this motion while watching a film in a theater. The beam of light which falls on the screen from behind has dust particles present in it. They follow zig-zag path. Such type of movement of the colloidal particles was noticed for the first by Robert Brown, an English scientist in 1828. This is known as Brownian Movement

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Colloidal solution scatters the beam of light passing through it When a beam of light from a certain source is focussed or passed through a colloidal solution kept 1. in the dark, its path becomes visible while passing through the solution. Along with this, the colloidal particles can also be seen following a zig-zag path. But it does not happen when the same beam is passed through a true solution {e.g., sodium chloride solution). Actually, the particles present in a colloidal solution have size big enough to scatter or disperse the light rays present in the beam as 2. they fall on them. As a result, these rays as well as the colloidal particles becomes visible. This scattering of light by colloidal particles is known as Tyndall effect. This effect is not noticed in a true solution because the particles present in it are too small to scatter the light.

3.

7.

the colloidal solutions have wide range of applications. Only a few out of these are being discussed to generate some interest about this field. Bleeding from a cut can be immediately stopped by applying alum or ferric chloride Blood consists of haemoglobins which are negatively charged colloidal particles. On applying alum or ferric chloride, these particles take up positively charged ions (cations) from these substances. They get their charge removed and get precipitated (or coagulated) As a result, the bleeding stops because blood becomes very thick. Delta is formed when river water comes in contact with sea water for a long period River water is mostly muddy. These mud particles are charged colloidal particles. When river comes in contact with sea, the dissolved salts present in sea water provide ions with charge opposite to the charge on mud particles. These particles get uncharged and combine with each other to form bigger particles. Over the years, deltas appear at these places. Sky appears to be blue in colour When we look at the sky, it appears to be blue in colour. It is for your knowledge that there is no blue colour as such in the sky. Actually, fine particles of dust etc. are always present in the atmosphere. When sun light falls on these particles, they scatter light with a blue colour or tinge. That is why sky is blue.

Colloidals solutions in which only liquids participate are known as emulsions ormative orksheet In the table giving the different types of colloidal solutions, it has been mentioned that the solutions in 1. (i) Which type of solution is milk? (ii) Name a solid - solid type solution. which liquid acts as the dispersed phase and other (iii) Which type of solution is smoke? in which liquid as the dispression medium, are known as emulsions. However, these are not miscible with (i) (ii) (iii) each other. If they mix up or become miscible, we (A) Liquid in Atmosphere Gas in get a true solution and not an emulsion. In an solid liquid emulsion, one of the constituents is generally oily (B) Liquid in Alloys Liquid in while other is water soluble. Thus, the emulsions are oil-water in nature. We know that normally oil gas solid and water form separate layers and are not expected (C) L iquid i n Rubber Soli d in to mix up. But certain substances known as l iquid ga s emulsifiers help in forming a stable emulsions of oil and water. These emulsifiers are generally proteins (D) Gas in Rubber L iquid i n in nature. A few common examples of emulsions liquid l iquid are : Milk, cod-liver oil, both cold and vanishing creams, moisturising creams, paints etc. Applications of colloidal solutions The field of colloids is so vast that it may not be possible to describe the same at this level. Similarly,

F

W

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9th Class Chemistry

188 2.

Match the following : Column - I Column - II A) Soda water i) gas in gas B) H2 in platinum ii) liquid in liquid C) Air iii) solid in solid D) fog iv) gas in solid E) milk v) gas in liquid F) cheese vi) solid in liquid G) smoke vii) solid in gas H) ink viii) liquid in solid I) alloy ix) liquid in gas 3. What is the difference between solutions and colloids ? 4. Classify the following into true solutions and colloidal solutions : Ink, Salt solution, Starch solution, Blood, Sugar solution 5. Explain what happens when a beam of light is passed through a colloidal solution. 6. Which of the two will scatter light : soap solution or sugar solution ? Why ? 7. Which of the following will show Tyndall effect ? Why ? (A) Salt solution (B) Starch solution (C) Milk (D) Copper sulphate solution 8. Classify the following into solutions, suspensions and colloids : Soda-water, Milk, Brine, Blood, Ink, Smoke in air, Chalk water mixture, Milk, Shaving cream, Muddy river water. 9. Explain how does soap help in cleaning dirty clothes ? 10. Fog and cloud are both colloidal in nature. How do they differ ?

4. 5. 6. 7.

6. Saturated, Unsaturated and Supersaturated solutions

1.

Conceptive Worksheet 1.

2.

3.

Which of the following statement/s is/are true? (A) A solution is a homogeneous mixture of one or more components. (B) The major component of a solution is known as solvent. The minor component of a solution is known as solute. (C) In syrup, sugar is the solute and water is the solvent. (D) In brine, salt is the solute and water is the solvent. An example for a liquid in gas type of solution is (A) Soda water (B) Milk (C) Fog (D) Smoke An example for a gas in liquid type of solution is (A) Soda water (B) Milk (C) Alloy (D) Air

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What is a suspension ? Give two examples of suspensions. What is a colloid ? Give two examples of colloids (or colloidal solutions). Colloidal solutions show Tyndall effect but true solutions do not. Discuss. What is the difference between colloids and suspensions?

2.

When we dissolve a solute in a solvent, then a solution is formed. We can dissolve different amounts of solute in the same quantity of the solvent. In this way, we can get many solutions having different concentrations of the same solute. A particular solution may contain less amount of the dissolved solute whereas another solution may contain more amount of the solute in it. For example, we can prepare many salt solutions of different concentrations by dissolving different amounts of salt in the same quantity of water. So, depending upon the amount of solute present, the solutions can be classified into two groups : Unsaturated solutions and Saturated solutions. Let us discuss it in detail. Unsaturated Solution: A solution in which more quantity of solute can be dissolved without raising its temperature, is called an unsaturated solution. For example, if in an aqueous solution of salt, more of salt can be dissolved without raising its temperature, then this salt solution will be an unsaturated solution. Actually, an unsaturated solution contains lesser amount of solute than the maximum amount of solute which can be dissolved in it at that temperature. Saturated Solution: A solution in which no more solute can be dissolved at that temperature, is called a saturated solution. For example, if in an aqueous salt solution, no more salt can be dissolved at that temperature, then that salt solution will be a saturated solution. Thus, a saturated solution contains the maximum amount of solute which can be dissolved in it at that temperature. It is obvious that a saturated solution contains greater amount of solute than an unsaturated solution. (i) A maximum of 32 grams of potassium nitrate can be dissolved in 100 grams of water at a temperature of 20°C. So, a saturated solution of potassium nitrate at 20°C contains 32

Solutions

3.

grams of potassium nitrate dissolved in 100 grams of water. (ii) A maximum of 36 grams of sodium chloride (common salt) can be dissolved in 100 grams of water at a temperature of 20°C. So, a saturated solution of sodium chloride at 20°C contains 36 grams of sodium chloride dissolved in 100 grams of water. In order to test whether a given solution is saturated or not, we should add some more solute to this solution and try to dissolve it by stirring (keeping the temperature constant). If more solute does not dissolve in the given solution, then it will be a saturated solution; but if more solute gets dissolved, then it will be an unsaturated solution. Super Saturated Solution: A solution which has excess of solute dissolved in it than that is required for its saturation at a given temperature is known as a super saturated solution.

Effect of Heating and Cooling on a Saturated Solution A solution is saturated at a particular temperature only. So, if a saturated solution at a particular temperature is heated to a higher temperature, then it becomes unsaturated. This is because the solubility of solute increases on heating and more of solute can be dissolved on raising the temperature of the solution. By cooling the saturated solution slowly. In a saturated solution, few of the solute particles remain undissolved on cooling.

189 (b) A maximum of 36’grams of sodium chloride (common salt) can be dissolved in 100 grams of water at 20°C, therefore, the solubility of sodium chloride (or common salt) in water is 36 grams at 20°C. From the above discussion it is obvious that the solubility of a substance (or solute) refers to its saturated solution. So, we can write a yet another definition of solubility as follows : The solubility of a solute in water at a given temperature is the number of grams of that solute which can be dissolved in 100 grams of water to make a saturated solution at that temperature. The solubility of different substances in water is different. Since the solubility depends on temperature, so while expressing the solubility of a substance, we have to specify the temperature also. The solubilities of some of the substances (or solutes) are given on the next page. All these values of solubilities are ‘per 100 grams of water’. Substance (or solute) 1. Copper sulphate

water (at 20°C) 21 g

2. Potassium nitrate

32 g

3. Potassium chloride

34 g

4. Sodium chloride

36 g

5. Ammonium chloride

37 g

6. Sugar

7. Solubility of a Solution The maximum amount of a solute which can be dissolved in 100 grams of a solvent at a specified temperature is known as the solubility of that solute in that solvent (at that temperature). If the solvent is water, then we can define solubility as follows : The maximum amount of a solute which can be dissolved in 100 grams of water at a given temperature, is the solubility of that solute in water (at that temperature). Please note that solubility is always stated as ‘mass of solute per 100 gram of water’ (or any other solvent). Examples: (a) A maximum of 32 grams of potassium nitrate can be dissolved in 100 grams of water at 20°C, therefore, the solubility of potassium nitrate in water is 32 grams at 20°C.

Solubility in

Solubility =

204 g

Wt. of solute  100 Wt. of solvent

8. Factors effecting solubility Nature of solute and solvent Like dissolves like. Polar solutes dissolves in polar solvents and non polar solutes dissolve in non polar solvents. Non polar solutes does not dissolve in polar solvents and polar solutes does not dissolve in non polar solvents.

Size of solute particles Dissolution is a surface phenomenon like the evaporation. So, increase of surface area of the solute increases the rate of dissolution. The surface area of a solid solute can be increased by converting it into powder. The powdered solute dissolves more easily than the large crystals, as in the former case, www.betoppers.com

9th Class Chemistry

190 larger surface area is in contact with the solvent.

Agitation of the mixture Stirring speeds up dissolution. The reason is, on stirring, the saturated solution surrounding the solute particles diffused away and fresh solvent or unsaturated solution takes it place. Thus, more the solute dissolves in it.

exchange. In such cases, temperature has little effect on solubility. Example: NaCl.

Solubility curves The graphs which indicate the effect of temperature on solubility of a solute are called solubility curves.

Effect of Temperature

(i)

Before understanding the effect of temperature on solubility, let us know about endothermic and exothermic solutions. Endothermic solutions: The solutions that are formed by the absorption of heat from the surroundings are known as endothermic solutions. Example: When ammonium chloride or silver nitrate is dissolved in water, the solution gets cooled. NH 4 Cl

s

 H 2 O      NH 4 Cl

 s   heat

For such solutions, solubility increases on increase of temperature. (ii) Exothermic solutions: The solutions that are formed by the evolution of heat to the surroundings are known as exothermic solutions. Example: When calcium oxide or lithium carbonate is placed in water the solution gets heated.

CaO  s   H O   Ca  OH 2  aq   Heat 2 For such solutions, solubility decreases on increase of temperature.

Why do some solutes dissolve with Utility of solubility curves evolution of heat and some dissolve (a) The solubility of a given salt at any temperature can be read directly from its solubility curve. with absorption of heat in a solvent? The amount of heat change during the formation of a solution depends mainly on two factors: Lattice energy: It is the amount of heat required to separate a mole of the ionic substances into its component positive and negative ions. This is an endothermic process. Heat of hydration: These ions get hydrated. The ions hold the water molecules by ion dipole bonds. Heat is liberated during the process of hydration. Thus it is an exothermic process. Note: (a) If lattice energy > hydration energy, the system cools down. Examples: NaNO3, KNO3, KCl, etc,. (b) If lattice energy < hydration energy, the system heats up. Examples: NaOH, Na2CO3, Na 2SO4, all gases, etc,. (c) If lattice energy = hydration energy, no heat www.betoppers.com

(b) The nature of curve gives an idea about the effect of temperature on the change of solubility. A steep curve indicates a rapid change, while a flat one indicates rather a slow change in solubility with rise of temperature. (c) Substances whose solubility decreases with rise in temperature are : NaOH, Na2CO3, Na2SO4, etc,. (d) Substances whose solubility increases with rise in temperature are: NaNO3, KNO3, KCl, NH4Cl, etc,. (e) Substances whose solubility fairly increases or no change with rise in temperature are: NaCl, Li2SO4. Note: Pressure has no effect on solubility for solidliquid solutions.

Effect of pressure and temperature on solubility of a gas in liquid: (a) Pressure: An increase in pressure on surface of water causes increase in solubility of gas in water. Solubility of gases in water at a fixed temperature can be correlated by Henry’s law.

Solutions Statement: At any temperature the mass of gas dissolved by a fixed volume of liquid is directly proportional to the pressure on the surface of the liquid. Consider the Evolution of gas is seen when a bottle of soda is opened. Soda contains carbon dioxide dissolved in water under pressure. Solubility of carbon dioxide under normal atmospheric pressure is very low, but when subjected to high pressures, it dissolves to a great extent as in the case of soda water in which carbon dioxide is dissolved in water under pressure and the bottle corked. On opening the bottle the gas rapidly bubbles out since the pressure on the surface of the water suddenly decreases and so does the solubility of carbon dioxide gas in water. (b) Temperature: An increase in temperature of water, causes decrease in solubility of gas in water. It is to be known that on boiling, water loses its taste. Let’s understand how this happens. Water contains soluble gases which contributes to the taste of water. On boiling, the temperature of the water increases, thereby the solubility of the dissolved gases [air] decreases and the dissolved gases are hence expelled out. If the above water is once again shaken with air, the water gets back its taste, due to dissolution of the air in water. It is also a fact that Gases are more soluble in cold water than in water at high temperature . Soda contains carbon dioxide dissolved in water under pressure. When a soda bottle kept at ordinary temperature is opened the evolution of carbon dioxide is less compared to similar chilled soda bottle which on opening produces more effervescence. Thus at low temperatures the solubility of the gas is more, compared to higher or ordinary temperatures.

Formative Worksheet 11. Solution-A: A solution which can dissolve some more solute at a given temperature is known as an unsaturated solution. Solution-B: A solution which cannot dissolve any more solute at a given temperature is known as a saturated solution. Solution-C: A solution which has excess of solute dissolved in it than can be present in a saturated solution is known as a super saturated solution.

191 Solution –A

Solution - B

Solution - C

(A) Saturated

Unsaturated

Super saturated

(B)

Unsaturated

Saturated

Super saturated

(C)

S uper s at urate d

Uns a turat ed

S at urate d

Super saturated

U ns atura ted

(D) Saturated

12. There are three types of solutions Solution - A :- The density of it increases on addition of some solute. Solution - B :- The density of it decreases on addition of some solute. Solution - C :- The density remains unchanged on addition of some solute. Identify which of the above three are saturated, unsaturated and supersaturated. 13. In which type of solution, a dynamic equilibrium exists between the solid solute and dissolved solute? (A) Saturated (B) Unsaturated (C) Super saturated (D) None 14. 5g of a salt dissolved in 20g of water to form a saturated solution at 600C. Calculate the solubility of the salt at this temperature. (A) 100 (B) 75 (C) 50 (D) 50 15. When 100g of a saturated solution is evaporated at 500C, 50g of solid is left over. Find the solubility of the substance at 500C. (A) 100 (B) 200 (C) 300 (D) 400 16. Solubility of a substance in a solvent is ‘x’. What is the minimum weight of solution containing ‘y’ g of solute?  x(100  y)  (A)   gm y  

 y(100  x)  (B)   gm x 

 y(100  x)   x(100  y)  (C)  (D)   gm  gm y x    17. Solubility of a substance in a solvent is K. What is the minimum weight of the solvent which can dissolve ‘L’ g of solute, to give a saturated solution?

100  K  g (A)   L 

100  L  g (B)   K 

 K  g (C)  100  L 

 L  g (D)   K  100 

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9th Class Chemistry

192 18. (i)

19.

20.

21.

22.

23.

24.

25.

Cooling favours the solubility of exothermic solutions. True or false? We know that solubility of exothermic solutions decreases on increase of temperature (heating). Therefore the given statement is true. (ii) Heating favours the solubility of endothermic solutions. True or false? We know that solubility increases with increase of temperature (heating). Therefore the given statement is true. (i) (ii) (A) T F (B) F T (C) T T (D) F F Solubility of the salt at 400C is 40. The least weight of water which dissolves 10 grams of substance is __________. (A) 25g (B) 36g (C) 47g (D) 58g Write the effect of temperature on the solubility of following solutions : Solute – A :- During its dissolution, it liberates heat to the surroundings. Solute – B :- During its dissolution, it absorbs heat from the surroundings. Solute – C :- During its dissolution, no heat exchange takes place with the surroundings. The graphs which indicate the effect of temperature on solubility of a solute are called solubility curves. (A) Isothermic curves (B) Isobaric curves (C) solubility curve (D) Isochoric curves The solutions that are formed by the absorption of heat from the surroundings are known as (A) Neutral solution (B) Thermal solutions (C) Exothermic solutions (D) Endothermic solutions Which of the following forms endothermic solutions? (A) NaNO3 (B) NaOH (C) Na2CO3 (D) Na2SO4 The solutions that are formed by the evolution of heat to the surroundings are known as (A) Neutral solution (B) Thermal solutions (C) Exothermic solutions (D) Endothermic solutions Which of the following form exothermic solutions? (A) NaNO3 (B) NaOH (C) Na2CO3 (D) Na2SO4

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Conceptive Worksheet 8.

9.

10.

11.

12.

13.

14.

15.

16. 17. 18. 19.

If a saturated solution is heated, it (A) Changes to unsaturated (B) Changes super saturated (C) Remains same (D) Till 50°C, changes to un saturated and then super saturated How can you obtain a saturated solution from an unsaturated solution? (A) By heating (B) By cooling (C) By increasing the pressure (D) By decreasing the pressure How can you obtain a saturated solution from a super saturated solution? (A) By heating (B) By cooling (C) By disturbing (D) By adding a crystal of the solute What happens when some amount of solute is added to a saturated solution? (A) Changes to unsaturated (B) Changes super saturated (C) Remains same (D) Till 50°C, changes to un saturated and then super saturated A hot solution contains 5g of a substance in 15g of water, at 35 0 C. What is the solubility of the substance at this temperature? (A) 33.32 (B) 66.66 (C) 33.34 (D) 16.67 The solubility of salt at 300C is 50. Calculate the weight of water required to prepare a saturated solution containing 90g of salt. (A) 360g (B) 180g (C) 90g (D) 45g The solubility of a substance in water is 40. What is the maximum amount of solute required to prepare a saturated solution with 50g of water? (A) 65g (B) 50g (C) 35g (D) 20g 30 grams of saturated solution on evaporation gives a residue of 5 grams of solute. Solubility of solute is ____________. (A) 40 (B) 30 (C) 20 (D) 10 What are solubility curves and what is their utility ? What is the effect of temperature on the solubility of a gas in a liquid at constant pressure ? Solubility of a solid in a liquid decreases with increase of pressure. True / False. Why ethyl alcohol, Glucose being non-polar dissolve in water ?

Solutions 20. Name some solutes whose solubility decreases with increase in temperature. (A) NaOH (B) Na2 CO3 (C) Na2SO4 (D) All 21. Name some solutes whose solubility fairy change or does not change with increase in temperature. (A) NaOH (B) Na2 CO3 (C) NaCl (D) Li2SO4 23. Solubility of a gas in liquid decreases with increase of temperature or decrease of pressure. This is called (A) Boyle’s law (B) Charles’ (C) Henry’s law (D) Dalton’s law of solutions 24. (i) What is the nature of solution, if lattice energy is greater than hydration energy? (ii) What is the nature of solution, if lattice energy is less than hydration energy? (i) (ii) (A) Endothermic Endothermic (B) Exothermic Exothermic (C) Endothermic Exothermic (D) Exothermic Endothermic 25. Name some solutes whose solubility increases with increase in temperature. (A) NaNO3 (B) KNO3 (C) KCl (D) NH4Cl.

9. Concentration of a Solution A solution may have a small amount of solute dissolved in it while may have a large amount of solute dissolved in it. The solution having small amount of solute is said to have low concentration. It is known as a dilute solution. The solution having a large amount of solute is said to be of high concentration. It is known as a concentrated solution.

193 We can now define the concentration of a solution as follows : The concentration of a solution is the amount of solute present in a given quantity of the solution. The concentration of a solution can be expressed in a number of different ways. The most common way of expressing the concentration of a solution is the ‘percentage method’. The percentage method of expressing the concentration of a solution refers to the ‘percentage of solute’ present in the solution. The percentage of solute can be ‘by mass’ or ‘by volume’.

Percentage by weight If the solution is of a ‘solid solute’ dissolved in a liquid, then we consider the ‘weight percentage of solute’ in calculating the concentration of solution. So, in the case of a solid solute dissolved in a liquid solvent : The weight percentage of a solution is defined as the weight of solute in grams present in 100 grams of the solution. For example, a 20 per cent solution of sodium chloride means that 20 grams of sodium chloride are present in 100 grams of the solution. Please note that the 100 grams of solution also include 20 grams of the sodium chloride . This means that the 100 grams of sodium chloride solution contain 100 - 20 = 80 grams of water in it. Thus, we can prepare a 20 per cent solution of common salt by dissolving 20 grams of common salt in 80 grams of water (so that the total mass of the solution becomes 20+ 80= 100 grams). Note : The weight percentage of a solution refers to the weight of solute in 100 grams of the solution and not in 100 grams of the solvent. We can calculate the concentration of a solution in terms of weight percentage of solute by using the following formula : Weight percentage 

Weight of solute  100 Weight of solution

The mass of solution is equal to the mass of solute plus the mass of solvent. That is Weight of solution = Weight of solute + Weight of solvent So, we can obtain the weight of solution by adding the weight of solute and the weight of solvent. In the above given example: Weight of solute (sodium chloride) = 20 g Weight of solvent (water) = 80g

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9th Class Chemistry

194 So, Weight of solution

=

Weight of solute + Weight of solvent = 20+ 80 = 100 g Now, putting these values of weight of solute and weight of solvent in the above formula, we get

20  100 100 = 20percent (by weight) Thus, the concentration of this sodium chloride solution is 20 per cent (or 20 %) by weight. Please note that if the concentration is in terms of weight, then the words ‘by weight’ are usually not written with it. For example, in the above case we just say that it is a 20per cent solution of sodium chloride ’. Since sodium chloride is a solid, so it is understood that the percentage is by weight. This is because we do not consider the volume of solids in making solutions. It is to be noted that The weight percentage of a solution remain unchanged with change in temperature. Note: We can also calculate the weight volume percentage of a solution. It is defined as the amount of solute in grams present in 100 ml of solution. Weight percentage 

Wt.Volume % =

Wt. of solute ×100 Vol. of solution

Percentage by Volume If the solution is of a ‘liquid solute’ dissolved in a liquid solvent, then we usually consider the ‘volume percentage of solute’ in determining the concentration of solution. So, in the case of a liquid solute dissolved in a liquid solvent : The volume percentage (concentration) of a solution is defined as the volume of solute in millilitres present in 100 millilitres of the solution. For example, a 10 per cent solution of alcohol means that 10 millilitres of alcohol are present in 100 millilitres of solution. Please note that the 100 millilitres volume of solution also includes 10 millilitres volume of alcohol. This means that the 100 millilitres of alcohol solution contain 100 - 10 = 90 millilitres of water in it. Thus, we can prepare a 10 per cent solution of alcohol by mixing 10 mL of alcohol in 90 mL of water (so that the total volume of the solution becomes 10+ 90 = 100 mL). Note: The concentration of solution refers to the volume of liquid solute in 100 mL of solution and www.betoppers.com

not in 100 mL of solvent. In general, we can calculate the concentration of a solution in terms of volume percentage of solute by using the formula :

Volume of solute  100 Volume of solution In the above example: Volume of solute (alcohol) = 10 mL And, Volume of solvent (water) = 90 mL So, Volume of solution = Volume of solute + Volume of solvent = 10 + 90 = 100 mL Now, putting these values of ‘volume of solute’ and ‘volume of solution’ in the above formula, we get: Volume percent=

10  100 100 = 10 per cent (by volume) Thus the concentration this alcohol solution is 10 per cent or that it is a 10 % alcohol solution (by volume).

Volume percent



Formative Worksheet 26. 10 grams of NaOH is dissolved in 100 grams of water at 25°C to form a saturated solution. Find the solubility and weight percentage of the solution. Solubility Weight (A) 10 9.09 (B) 20 18.18 (C) 9.09 10 (D) 18.18 20 27. A solution with weight percentage of sodium hydroxide of 6% is prepared by adding the following mass of water (in grams) to 200 grams of that solution with a weight percent of NaOH of 30% (A) 300 g (B) 500 g (C) 800 g (D) 1000 g 28. A research student has only 300 g of valuable solvent and wishes to make solution of 10% by weight from solute A. How many grams of ‘A’ is weighed out? (A) 30g (B) 33.33g (C) 70g (D) 66.66g 28. If solubility of a solute is ‘S’, find its percentage by weight. (A)

100S 100  S

(B)

100  S 100S

Solutions

195 2.

3.

4.

5.

Conceptive Worksheet

29.

30.

6.

(A)

Temperature Temperature

Temperature Temperature

7.

8. 9.

1.

Calculate the masses of glucose and water required to prepare 180 g of 18% (w/w) glucose solution. Mass of water Mass of glucose (A) 147.6 g 32.4 g (B) 295.2 g 64.8 g (C) 442.8 g 97.2 g (D) 590.4 g 129.6 g

11. 12.

13.

Temperature

d) (D)

(C)

10.

Summative Worksheet

(B) Solubility

28.

find the solubility of the solute. (A) 25 (B) 50 (C) 75 (D) 100 If 50 ml of a liquid solute is dissolved in 500 ml of liquid solvent, then find its % by volume (v/v). (A) 6.06 (B) 7.07 (C) 8.08 (D) 9.09 Find the following in 15% (w/v) of solution if the density of solution is 1.06 g/ml. a) Weight of solute b) Weight of solvent Weight of solute Weight of solvent (A) 25 g 75 g (B) 50 g 50 g (C) 15 g 91 g (D) 20 g Temperature Temperature 90 g Temperature Temperature An aqueous solution of glucose is 10% (w/v). The volume in which 1 g mole of it is dissolved will be______. (Formula of glucose is C6H12O6) (A) 450 ml (B) 900 ml (C) 1800 ml (D) 3600 ml How many grams of NaCl have to be dissolved in 54.0g of H2O to give a 10.0% weight solution? (A) 3 g (B) 6 g (C) 9 g (D) 18 g

Solubility

27.

1 its solubility,, 2

Solubility

26. If percentage by weight of a solute is

If 2.2 g of oxalic acid is dissolved in 500 ml of solution (density = 1.1 g/ml), what is the weight percent of oxalic acid in solution? (A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4 Calculate the mass percentage of K 2 CO 3 in a solution that is made by dissolving 15g of K2CO3 in 60g of H2O. (A) 10 (B) 20 (C) 30 (D) 40 If the solubility of a solute in a solution is 50, find the percentage by weight of the solution. (A) 33.33 (B) 66.66 (C) 25 (D) 75 Solubility of a solid at 300C is 30 and at 800C is 50. If a saturated solution is made with 50g of solvent at 800C and cooled to 300C, the weight of solute that comes out of the solution is _________. (A) 30g (B) 20g (C) 10g (D) 5g Solubility of a substance remains unchanged on heating or cooling. Plot its solubility curve.

Solubility

100S2 (100  S)2 (D) 100  S 100S 29. Calculate the volume of solvent required to prepare 20 % by volume (v/v) solution by using 70 ml of H2 SO 4 . (A) 280 ml (B) 560 ml (C) 1120 ml (D) 2240 ml 30. 10 g of a solute is dissolved in 90 g of solvent. If the density of solution is 2g/ml, find the weight-volume percentage. (A) 5 g/ml (B) 10 g/ml (C) 15 g/ml (D) 20 g/ml

(C)

Temperature

Solubility of a solute is 25. Minimum weight of solvent required to dissolve 10g of that solute is _________. (A) 90g (B) 65g (C) 40g (D) 15g A solution contains 30 g of sugar dissolved in 370 g of water. Calculate the concentration of this solution. If 110 g of salt is percent in 550g of solution, calculate the mass percentage of the solution A solution contains 30 g of sugar dissolved in 370 g of water. Calculate the concentration of this solution. If 2 mL of acetone is present in 45 mL of its aqueous solution calculate the concentration of this solution The solubility of a salt in water is 22g/litre at 250C and 60 g/litre at 800C. A student prepares 500 cc of saturated solution at 800C cools to 250C and adds a tiny crystal of that salt. How many grams of salt comes out? The solubility of a salt is 'S', then what is the ratio of amount of solute to the amount of solvent at same www.betoppers.com

9th Class Chemistry

196

What does the following solubility curves mean:

2.

3. 4.

5.

6. 7.

From the list of following list of salts: NaNO3 , NaOH, NaCl, KNO3, Na2CO3, Li2SO4, NH4Cl, Na2SO4,KCl, etc,. State the salts whose solubility (a) increases, (b) decreases, (c) is fairly independent of or slightly increases with rise in temperature of water. The density of a solution P increases, and that of a solution Q decreases on addition of solute. What type of solutions are P and Q? How can you convert a saturated solution to unsaturated and supersaturated solutions? Identify the solvents required for the following: (a) to remove grease stains (b) to remove paint stains (c) coffee or tea stains. How does the following factors affect solubility? (a) Nature of solute and solvent (b) size of solute particles (c) agitation of the mixture Some solutions are formed by the evolution of heat and some with absorption of heat. Explain. Solution - I can be prepared by dissolving a solute P in a solvent Q. Solution - II can be prepared by dissolving a solute R in S. The lattice energy of P is greater than the hydration energy of Q and the lattice energy of R is less than the hydration energy of S. Then, categorize P and Q into exothermic and endothermic solutions.

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temperature

temperature

Graph - I

Graph - II

Solubility

HOTS Worksheet 1.

Solubility

8.

Solubility

temperature? 14. Calculate the percentage weight of K2 CO3 in a solution that is made of dissolving 15 g of K2CO3 in 60 grams of H2O. 15. The solubility of a solute is 30 at certain temperature. Then, the weight percent of the solute in its saturated solution is _________________.

temperature Graph - III

9. 10.

11.

12.

13.

14. 15.

Give some examples of solutes that exhibit the above graph properties. The solubility of a salt in water is 22g/litre at 250C and 60 g/litre at 800C. A student prepares 500 cc of saturated solution at 800C cools to 250C and adds a tiny crystal of that salt. How many grams of salt comes out? The solubility of a salt is 'S', then what is the ratio of amount of solute to the amount of solvent at same temperature? 30 g of a saturated solution on evaporation gave a residue of 5 g. Solubility of the solute is _____ . Solubility of a solute is 15 . To dissolve 3 g of that solute the minimum weight of water required is _______. Decrease of pressure decreases the solubility of a gas in a liquid. Explain. How does an increase in temperature affect (A) The solubility of NaCl (B) The solubility of CaSO4 in water?

Solutions

197

IIT JEE Worksheet 1.

Sugar is soluble in water due to (A) High solvation energy (B) Ionic character of sugar (C) High dipole moment of sugar (D) Hydrogen bond formation with water. 2. Sodium sulphate is soluble in water whereas barium sulphate is sparingly soluble because. (A) The hydration energy of sodium sulphate is more than its lattice energy. (B) The lattice energy of barium sulphate is more than its hydration energy. (C) The lattice energy has no role to play in solubility. (D) The hydration energy of sodium sulphate is less than its lattice energy. 3. When 45 g of solution is cooled from 600C to 300C, the mass of solute deposited is (A) 4.5 g (B) 4 g (C) 5 g (D) 4.75 g 4. For which of the following pairs would you expect solubility to be the greatest? (A) O2 - water (B) sugar - water (C) sugar - benzene (D) O2 – N2 5. Water acts as a good solvent for ionic compounds because of (A) Its dipole moment (B) Its bent shape (C) Its high dielectric constant (D) Its high boiling point 6. Solubility of gases in liquids is governed by _____ law. 7. Deep sea divers use ( O 2 + He) mixture in preference to ( O2 + N2) mixture. This is because (A) Helium is lighter than nitrogen (B) Helium is less soluble in blood than nitrogen (C) Helium is more soluble in blood than nitrogen (D) Helium provides a better inert atmosphere than nitrogen 8. The amount of solute dissolved in 100 grams of solvent at a given temperature is known as ______. 9. The solubility of NaCl ________ with increase of temperature. 10. Soda water is an example for _______ type of solution.



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9th Class Chemistry

B y t h e e n d o f t h i s c h a p t e r, y o u w i l l u n d e rs t a n d      

Introduction Allotropes of Carbon Diamond Graphite Wood Charcoal Coal and Its types

CARBON MONOXIDE  Introduction  Laboratory preparation  Physical properties  Chemical properties  Uses & Tests for Carbon monoxide

1. Introduction

Occurrence of carbon Carbon is one of the very select group elements that can exist free in the nature. In the elementary state, carbon occurs as diamond, graphite and amorphous carbon. In the combined state it occurs as follows :

ii.

Oxides : CO2 (0.03% by volume in air) and CO that occurs in volcanic gases and in furnace gases. Carbonates:Limestone-CaCO3, MagnesiteMgCO 3 ;Dolomite- MgCO 3 ; Calamine -

ZnCO3 , Spathic iron- FeCO3. iii. Hydrocarbons : Methane, acetylene, benzene, petroleum, natural gases etc. each form during preparation.

•

Complex compounds : With hydrogen , nitrogen and sulphur, carbon forms complex compounds in plants and animals. v. In natural water: The carbon dioxide dissolved in water exist as carbonic acid. vi. Carbon is also an important constituent of a wide variety of industrial and commercial products like sugar, paper, leather, marble, petroleum, silk, medicines, insecticides, dyes and perfumes. iv.

The study of chemistry that deals with the study of carbon compounds made by and derived from living organisms is called organic chemistry. The study of all other compounds is grouped under inorganic chemistry. Several forms of carbon are also important sources of energy.

i.

CARBON DIOXIDE  Occurrence  Preparation  Physical properties  Chemical properties

2. Allotropes of Carbon (i) Allotropy : The property, by virtue of which an element exists in more than one form in the same physical state, having the same chemical properties but different physical properties, is called allotropy. (ii) Reasons for allotropy : a) The different methods by which each form is or was prepared. b) Different atomic arrangements in the molecules of each form. c) Different amounts of energy associated with

Allotropic forms of carbon : Carbon

Crystalline

Diamond

Chapter - 9

Compounds of Carbon

Learning Outcomes

Amorphous

Graphite

Coal

Coke

Wood charcoal

Charcoal

Lamp black

Gas carbon

Sugar charcoal

Bone charcoal

9th Class Chemistry

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3. Diamond

It is also used to drill holes in rocks. (iv) Eye surgeons use fine edged diamonds for removing cataract. (v) Diamonds are used for making dies for drawing fine wires. (vi) It is used in watches as a bearing. (vii) Dust of diamond is used for polishing other diamonds. The glasses in telescopes are shaped using diamond dust. (viii) They are used as needles in high quality record players. (ix) It is also used in making laser beams. (x) Black diamonds are used in precision instruments.

Structure of Diamond In diamond, each carbon atom is attached to four other carbon atoms. Each of the four valence electrons of the carbon atoms are bonded strongly with another carbon atom by covalent bonds. Thus, in a regular diamond structure, each carbon atom is held firmly in place by four bonds (Tetrahedral units). These carbon -carbon bonds are the strongest bonds to be found in any substance at ordinary temperatures.

4. Graphite Structure of Graphite In graphite, the carbon atoms are arranged in flat planes of hexagonal rings, stacked one on another.

Chemical Properties of Diamond

(i)

Diamond is chemically inert and is unaffected by organic solvents, acids and alkalis under ordinary conditions. However, the following properties may be noted: Reaction with fluorine : Diamond reacts with fluorine gas at 1023 K temperature and forms carbon tetrafluoride.

Each carbon atom is attached to just three others within the plane. As such, only three out of the four valence electrons are involved in carboncarbon bonding. The remaining electron wanders freely between the planes. This free electron accounts for the electrical conductivity of graphite. The links between the adjacent layers are held by weak Vanderwaal’s forces. Therefore, the planes can slide over each other easily, and make graphite soft, slippery and a good lubricant.

C  2F2  CF4 (ii) Reaction with oxygen or air : When diamond comes in contact with air at 9000C temperature, it burns to form carbon dioxide gas. Ash is not formed. It burns in the presence of oxygen at 8000C. 0

800 C C  2O2   CO2 (iii) Reaction with sodium carbonate : In the absence of air, it reacts with sodium carbonate at high temperatures. Sodium oxide and carbon monoxide are the products formed.

Chemical Properties of Graphite Chemically, graphite is slightly more reactive than diamond. This is because the reactants are able to penetrate between the hexagonal layers of carbon atoms in graphite. It is unaffected by ordinary solvents, dilute acids or fused alkalis. However,

C  Na 2CO3  Na 2 O  2CO

Uses of Diamond (i) It is used in jewellery. (ii) Diamond checks harmful U.V radiations. Hence, used in making of windows for spacecrafts. (iii) Diamond is a very hard substance. Hence, used for cutting and grinding hard materials. www.betoppers.com

1.

chromic oxide ( Cr2O3 ) oxidises it to CO2. Reaction with sodium carbonate : When Graphite fuses with Na2CO3, sodium oxide and carbon monoxide are formed.  C  Na 2 CO3   Na 2 O  CO

Compounds of Carbon

2.

201

Reaction with Oxygen : Graphite burns in the presence of oxygen at 7000C and forms carbon  dioxide and carbon monoxide. A small amount of ash is collected after burning. 0

700 C 3C  2O2   CO 2  2CO  

Uses of Graphite 



In making lead of ‘lead’ pencils : Graphite marks paper black. The weakly held layers of carbon atoms in graphite easily slide over each other and are left behind on the paper. Graphite is mixed with clay in various proportions to obtain ‘lead’ of different hardness. Therefore, no real lead is present in it. In the manufacture of electrodes : The presence of free, middle electrons in the crystal of graphite,

makes graphite a good conductor of electricity. In the manufacture of refractory crucibles : Graphite is a good conductor of heat, it is resistant to chemicals and has a high melting point of about 35000C.



Used as a lubricant for heated machine parts : Graphite is nonvolatile and slippery. The slippery nature is due to the sliding of the weakly hold carbon layers.



In nuclear reactors as moderators : Graphite slows down and absorbs the fast moving neutrons. This keeps the reactors under control.



At high pressure, temperature and in the presence of catalyst, graphite changes into diamond.

Comparision of physical properties of Diamond and Graphite Property 1. Nature 2. Physical state 3.

Structure

4. 5. 6.

H ardness Specific gravity Electrical conductivity Thermal conductivity Reaction with oxygen

7. 8.

Diamond Brittle solid Transparent with extraordinary brilliance Compact, three dimensional structure in which the atoms are held firmly by strong covalent bonds. Hardest known substance 3.52 Bad conductor of electricity Bad conductor of heat

Soft and slippery 2.3 Good conductor of electricity Good conductor of heat

Catches fire at 800 0 C

Catches fire at 700 0 C

5. Wood Charcoal Wood charcoal is obtained as the residue on iv) destructive distillation of wood. The process of heating wood out of contact with air and distilling the products is destructive distillation of wood.

Physical Properties of Wood Charcoal i) ii) iii)

Graphite Flaky solid Black opaque solid with metallic luster. Layers of arranged sheets held by weak physical forces

Wood charcoal is a soft, black, porous material. It is brittle and tasteless. Its density is 1.5 g/cm3. It floats on water due to the air trapped in its pores, which lowers its density to about 0.2 g/cm3. If air is removed from the pores of wood charcoal by boiling

or by means of an exhaust pump, the charcoal will gradually settle down. Charcoal is a good adsorbent for gases, particularly for poisonous and foul smelling gases. The adsorption capacity of wood charcoal is increased by heating above 900°C by steam because, this treatment opens up the pores of the charcoal. The resulting charcoal is called activated charcoal.

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9th Class Chemistry

202

Chemical Properties of Wood Charcoal Reactant 1. Air [O2]

Conditions

Equation

Burns in

C + O2  CO2 + 

i) Free supply of air and catches fire at 400°C

2C + O2  2CO

ii) Limited supply of air forms carbon monoxide and incomplete combustion. 2. Hydrogen

Reacts with hydrogen when sparked electrically to form methane

C + 2H2 CH4

3. Sulphur

Reacts with Sulphur vapour at high temperature to form carbon disulphide.

 C + 2S   CS 2

4. Calcium

It reacts at very high temperature to form calcium carbide.

 Ca + 2C   CaC2

Reducing Nature of Woodcharcoal. Wood charcoal re duces

Substance to

Equation

be re duce d i) Metallic oxide to metal when heated at high temperatures.

Zinc oxide

ZnO + C  Zn + CO

Fe rric oxide Le ad oxide

Fe 2O3 + 3C  2Fe + 3CO

Coppe r oxide

CuO + C  Cu + CO

PbO + C  Pb + CO CaO + 3C  CaC2 + CO SiO2 + 3C  SiC + 2CO

ii) Calcium and silicon oxides to their corresponding carbides when heated electrically.

Calcium oxide

CaO + 3C  CaC2 + CO

Silicon dioxide

SiO2 + 3C  SiC + 2CO

iii) Hot conc. Sulphuric acid to Sulphur dioxide [conc.]

Sulphuric acid

2H 2SO 4 + C  CO2 + 2H 2O + 2SO2 [conc.]

iv) Nitric acid to nitrogen dioxide on boiling

Nitric acid

4HNO3 + CCO2 + 2H2O +4NO2

v) Superheated steam to hydrogen

Supe rhe ate d ste am

1000 C C + H 2O   CO + H2

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Compounds of Carbon

203

Uses of Wood Charcoal Application 1. Fuel Excellent household fuel, used for starting fir es. 2. Gunpowder [Mixture of charcoal, nitre and sulphur] 3. Decolourising agent For sugar and organic liquid 4. Deodorant and adsorbent used in a) Gas masks

Form used Logs of charcoal

Reason for use Burns without any smoke. Low ignition temperature.

Finely divided charcoal

Its low ignition temperature helps in easy ignition of the gunpowder mixture.

Powdered charcoal

Good adsorbing agent.

Powdered charcoal

Adsorbs poisonous gases, hence used for military and industrial applications. Removes foul odours from sewers Removes gases produced in the stomach by indigestion.

b) Sewers

Powdered charcoal

c) Indigestion and related gastric disorders

Activated charcoal

d) Dental powders

Activated charcoal

Removes foul gases from gums.

6. Coal and Its types Coal is a hard, black solid formed over millions of 1. years from the fossils of plants. Coal is mainly carbon but also contains oxygen, hydrogen, nitrogen and sulphur as well. Peat: It is the first stage product of the decomposition of vegetable matter and may contain not more than 60% carbon. Lignite: It is harder than peat. It contains 60% to 70% carbon. It has poor calorific value and disintegrates when exposed to air. Bituminous coal: It is the most common variety of coal. Bituminous coal is of high, medium and low 2. varieties in decreasing order of carbon content. High grade bituminous coal contain about 90% carbon. On heating, it gives off both volatile and non-volatile material. It is black and hard. Anthracite coal: It contains 90-95% carbon. It is the best form of coal. It is hard, dense and black. It is found at only a few places in the world.

3.

Match the following: Column - I 1) Carbon that occurs in combined state 2) Carbon that occurs in combined state in 3) Carbon that occurs in combined state 4) Carbon that occurs in free state

Formative Worksheet Statement A : Allotropy is the phenomenon in which an element exhibits different physical forms with same chemical properties. Statement B : Various physical forms of an element with different physical properties and same chemical properties are called allotropic forms. (A) Statement ‘A’ is correct but ‘B’ is incorrect. (B) Statements ‘A’ and ‘B’ are correct. (C) Statement ‘A’ is incorrect but ‘B’ is correct. (D) Both statements are incorrect. Assertion : Black diamonds are generally used for cutting and drilling purposes. Reason : In diamond, stable and rigid crystal lattice is present. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

Column - II p) Vegetable oil in gaseous state q) Spathic iron liquid state r) Graphite in solid state s) Carbonic acid www.betoppers.com

9th Class Chemistry

204 (A) (B) (C) (D) 4.

5.

(1) p s s q

(2) q p r s

(3) r q q p

(4) s r p r

Which of the following statements is/are correct? (A) Diamond contains no free electrons in the crystal. (B) Diamond is insoluble in all known solvents. (C) Distance between ‘C–C’ bonds in diamond is less. (D) Pure diamond is colourless in nature. Identify A, B in the following reactions: 3500ºC i) A + 3C (s)   SiC(s) + 2CO (g)

Conceptive Worksheet 1.

ii)

6.

7.

8.

9.

3500ºC SiC(s)   B + graphite (A) (B) (A) SiO4 Si (B) Si SiC (C) SiO2 Si (D) SiO2 C In Acheson’s process ‘A’ is heated with silica in an electrical furnace at 3700K to obtain an intermediate ‘B’ and finally it gives ‘C’ and ‘D’. ‘D’ goes out as vapour. Then, (A) ‘A’ is an element and ‘B’ is its oxide. (B) ‘B’ is an oxide and ‘C’ is an element. (C) ‘A’ and ‘C’ are allotropes. (D) ‘B’ and ‘C’ are allotropes. Statement A : Graphite cannot withstand very high temperatures. Statement B : Graphite is used as a moderator by absorbing excess neutrons in reactors. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. Which of the following elements is added to turn artificial diamond into semi conductors? (A) Beryllium (B) Boron (C) Magnesium (D) Calcium Among diamond and graphite which is: i) More inert ii) More stable? (i) (ii) (A) Graphite Diamond (B) Diamond Graphite (C) Graphite Graphite (D) Diamond Diamond

10. Assertion: Wood charcoal is an excellent house hold fuel. Reason : Wood charcoal has low ignition point of 4000C and it burns without any smoke. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

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2.

3.

4.

5.

6.

What happens when diamond is heated to more than 18000C in vaccum? (A) It forms CO2 (B) It changes into graphite (C) It gives CO and CO2 (D) It changes into fullerene Which of the following statement(s) is/are correct? (A) Black lead is also called as graphite. (B) Graphite is thermodynamically unstable. (C) Lead pencils contain graphite but not lead. (D) All the three statements are correct. Assertion : Graphite is used in making acid resistant electrodes for electrolytic cells.] Reason : Graphite is chemically inactive with acids. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. The important constituent of gun powder is: (A) Charcoal (B) Nitre (C) Coal (D) Sulphur What are the products formed in the reaction of graphite and sodium carbonate? (A) N2O2 and NO2 (B) Na2O and 2CO (C) Na2O and CO2 (D) None of these 0

800 C Assertion : C  s  + O2 )g   CO2  g 

Reason : Diamond is a pure form of carbon. Hence, no residue is formed on heating. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

Compounds of Carbon 7.

i) ii) a) b) c)

What is the solubility of fullerence in CS2? The solid form of ‘C60’ molecule is: (i) (ii) Insoluble C – 60 Soluble Fullerite Soluble C – 60

CARBON MONOXIDE 7. Introduction

8. (i)

205 HC OOH

 formic acid 

 H 2 SO 4  aq  conc

0 100  C CO  g   H SO . H O  aq  2 4 2

9. Physical properties (i) (ii) (iii) (iv)

Carbon monoxide is a colourless, odourless and tasteless gas. Vapour density of carbon monoxide is 14. It is slightly lighter than air, It is almost insoluble in water. Under high pressure and low temperature, it gets liquefied. Liquid carbon monoxide has boiling point of –192 °C. Carbon monoxide is highly poisonous in nature, 1% of this gas in air can kill a person within 10 minutes.

Carbon monoxide was first prepared by Lasson in 1776, by heating zinc oxide with wood charcoal. In (v) I880, Cruik Shank established that carbon monoxide is a compound of carbon and oxygen and not an element hydrogen. Poisonous nature of CO Occurrence Carbon monoxide is 300 times more soluble in blood (i) Carbon monoxide is found in traces in the than oxygen. Thus, when it comes in contact with atmosphere in the regions having active blood, it reacts with haemoglobin (red blood volcanoes. corpuscles). It binds itself with haemoglobin to form (ii) When the organic matter decays in swamps carboxy haemoglobin, a compound which cannot in the absence of air, carbon monoxide gas absorb oxygen. Thus, red blood corpuscles lose the is formed. capacity to carry oxygen to the cells and tissues (iii) Carbon monoxide is formed when wood, Thus, the person dies for want of oxygen. charcoal, kerosene oil or any other fuel Dangerous action of carbon monoxide containing carbon burns in the limited supply - more than chlorine of air. Chlorine is greenish yellow gas having a suffocating (iv) All automobile engines produce traces of and pungent smell, which can be easily detected carbon monoxide, when fuel burns in limited and hence, the person can move away from the supply of air. area polluted with chlorine Laboratory preparation However, carbon monoxide is a colourless and From Charcoal: odourless gas which cannot be detected. Thus, it In laboratory, carbon monoxide is prepared by silently kills a person and is sometimes called silent heating charcoal with carbon dioxide. killer. C  s   CO 2  g    2CO  g   t  Red hot 

(ii) From Oxalic acid: Carbon monoxide is obtained by heating the reaction mixture containing oxalic acid and concentrated sulphuric acid

 COOH 2  s   H 2 SO4   H SO .H O  aq   CO  g   CO  g  2 4 2 2 (iii) From formic acid: Carbon monoxide is obtained by warmingf o r m i c acid with conc. sulphuric acid at 1000C. The dehydration results in the formation of carbon monoxide gas.

10. Chemical properties (i)

Combustibility : Carbon monoxide is a combustible gas, but does not support combustion. It burns silently in air with pale blue flame. However, a mixture of carbon monoxide and air burns spontaneously, producing an explosive reaction. As carbon monoxide burns without any smoke and produces a large amount of heat energy, it serves as an excellent fuel. (ii) Action with Litmus: It is neutral to litmus, i,e., it neither turns blue litmus red nor red litmus blue and hence, has no acidic or basic properties. www.betoppers.com

9th Class Chemistry

206

(iii) Action with Non-metals:   Cu s  CO g CuO  s   CO  g    (a) Action with oxygen or air : It burns in air or oxygen 2  producing carbon dioxide gas and large amount of (b) Action with steam: heat energy. Fe O Cr O 2 3 2 3  CO  g   H  g  CO  g   H O    2CO  t 2 0 C 5000 C 2 2 2CO  O 2  450 2  Steam  (b) Action with hydrogen : Two volumes of hydrogen and one volume of carbon monoxide react to form (vi) Action with compounds: methyl alcohol when passed over catalyst (zinc (a) Action with water: High pressure oxide + finely divided copper) at 4500C. H O     CO  g   HCOOH  aq  2 ZnO Cu CO  g   2 H  g    CH OH    (b) Action with ammonical cuprous chloride: 2 3 4500 C

 Methyl alcohol  CuCl     CO  g   H O   CuCl.CO.2 H O  aq  2 2 (c) Action with chlorine : When equal volumes of (c) Action with soda lime: carbon monoxide and chlorine gas are exposed CaO to sunlight, they react lo form extremely poisonous CO  g   NaOH  HCOONa  s  gas phosgene or carbonyl chloride. Soda lime Sodium formate Sunlight CO  g   Cl  g    2

COCl  g  2  Carbonyl chloride 

11. Uses & Tests for Carbon monoxide

(iv) Action with metals: When finely divided metals like nickel, chromium and iron are heated and the carbon monoxide gas is passed over them, they form their respective carbonyls i.e., addition compounds (i) of metals with carbon monoxide.  Ni  s   4CO  g    Ni  CO 4     Nickel carbonyl   Cr  s   6CO  g   

(ii)

Cr  CO 6    Chromium carbonyl  

 Fe  s   5CO  g    Fe  CO 5     Iron carbonyl 

Note: The above carbonyls are volatile liquids and are highly poisonous in nature. They can decompose to form metals and carbon monoxide on strong heating and hence can be used in the purification of metals. (v) Reducing properties of carbon monoxide: (a) Action with oxides of less active metals: When carbon monoxide is passed over heated oxides of less active metals, it reduces them to form their respective free metals and itself gets oxidized to carbon dioxide.  Fe 2O3  s   3CO  g    Fe  s   3CO 2  g 

  Zn s  CO g ZnO  s   CO  g    2    Pb s  CO g PbO  s   CO  g    2  www.betoppers.com

(iii)

(iv)

(v)

Uses Carbon monoxide is a combustible gas which burns without smoke and produces a large amount of heat. It is an important constituent of coal gas, water gas and producer gas, which are used as fuels. Nickel can be removed from the mixture of other metal by passing carbon monoxide over it, when it forms vapour of nickel carbonyl at 70 °C. These vapours on heating to 150 °C decompose to form nickel metal. Thus it is used in extraction of metals. Carbon monoxide is extensively used in the manufacture of sodium formate, methyl alcohol, formic acid and synthetic petrol, etc. The ores of the metals in the form of their oxides, (Fe2O3; PbO; CuO; SnO) can be reduced to their respective metals, when carbon monoxide is passed over heated metallic oxides. Phosgene (COC12) is extremely poisonous which can be used as war gas.

Tests (i)

Carbon monoxide does not turn limewater milky. However, if carbon monoxide is ignited, it burns with a pale blue flame. The residual gas, on testing with limewater, turns it milky. (ii) When a filter paper soaked in palladium chloride or platinum chloride is taken in the jar of carbon monoxide, first it turns pink, then green and finally black. The colour change is proportional to the concentration of carbon monoxide, e.g., if a gas

Compounds of Carbon

207

A B C (A) CO CO CO 2 (B) CO CO 2 CO Treatments of atients poisoned by (C) CO 2 CO CO 2 Carbon monoxide (D) CO 2 CO 2 CO The unconscious victim of carbon monoxide should 15. Carbon monoxide is collected by: be immediately brought out in the open air. (A) Upward displacement of water. He should be given artificial respiration with (B) Downward displacement of water. carbogen (a mixture of 95% oxygen and 5% carbon (C) Downward displacement of air. dioxide), till normal breathing is restored. (D) None of the above. In the factories, where the percentage of carbon 16. Which one of the following statement(s) is/are true? monoxide is more than safe limits, the workers should A) Carbon monoxide is found in traces in the be asked to wear gas masks. In the gas masks, the atmosphere in the regions having active air is breathed in through hopcalite (a mixture of volcanoes. 50% MnO2, 30% CuO, 15% Co2O3 and 5% Ag2O), B) When the organic matter decays in swamps in when carbon monoxide is oxidised to carbon dioxide. the absence of air, carbon monoxide gas is formed. C) All automobile engines produce traces of carbon ormative orksheet monoxide when fuel burns in limited supply of Initially carbon monoxide was prepared by: air. (A) Reduction of a metallic oxide (A) only ‘A’ is correct (B) only ‘B’ is correct (B) Oxidation of a metallic oxide (C) only ‘C’ is correct (D) All are correct (C) Reduction of a non-metallic oxide 17. The most poisonous gas which causes death by (D) Oxidation of a non-metallic oxide affecting respiratory problems is: Assertion(A) : It is not advisable to sleep in a room (A) CO2 (B) CO (C) H2 (D) O2 where coke or wood is burning and doors and windows are closed. 18. C  s  + CO2  g    2CO  g  Reason(R): CO produced, is a poisonous gas. The above reaction is _________ reaction. (A) Both assertion and reason are correct and (A) Endothermic reason is the correct explanation of assertion. (B) Exothermic (B) Both assertion and reason are correct but (C) Either endothermic or exothermic reason is not the correct explanation of (D) None of the above assertion. 19. During the preparation of carbonmonoxide gas from (C) Assertion is correct and reason is incorrect. formic acid at 100°C with conc. H2SO4 the impurity (D) Assertion is incorrect and reason is correct. formed is_________ and is absorbed by________ The oxide that is used for the preparation of carbon (A) Sulphur dioxide, Calcium hydroxide monoxide for the first time is: (B) Carbon dioxide, Caustic potash (A) ZnO (B) CaO (C) Na2O (D) MgO (C) Sulphur trioxide, aqueous sodium hydroxide sample contains very little amount of carbon monoxide, then colour change will be only pink.

(i) (ii)

(iii)

F 11.

12.

13.

14. (i)

W

HCOOH  aq  + H SO  aq  2 4  conc 

Δ  A g + H SO .H O aq    2 4 2   COOH

(ii)

(s) + H 2SO4  aq 

COOH

  CO g + B g + H SO .H O aq      2 4 2    Cg  , (iii) CO2  g  + Coke  s   Identify A, B, C respectively

(D) None 20. Assertion (A) : Carbon monoxide gas is collected in the fume cupboard. Reason (R) : It is very pleasant in nature. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 21. An inorganic compound (A) made of two most occuring elements in the earth’s crust and used in building construction when made to react with carbon, forms a poisonous gas (B) which is most www.betoppers.com

9th Class Chemistry

208 stable diatomic molecule, compounds (A) and (B) are: (A) SiO2, CO2 (B) SiO2, CO (C) SiO2, N2 (D) CaO, CO2 SiO2 is a compound of oxygen and silicon, the two most abundant elements of earth’s crust and is used in building construction.

SiO 2 + 2 C   Si+ 2 CO (poisonous gas and a stable diatomic molecule).

Conceptive Worksheet 8.

9.

10.

11.

12.

13.

14.

The most poisonous gas present in the exhaust fumes of vehicles is: (A) CO2 (B) CO (C) O2 (D) N2 Match the following (related to CO): Scientist Discovery p) De Lassone i) Established the formula to be CO. q) Cruik shank ii) Prepared CO for the first time. r) Dalton iii) Found that it was a compound of carbon and oxygen only. p q r (A) ii iii i (B) i ii iii (C) iii ii i (D) ii i iii Nature of carbon monoxide is: (A) Acidic oxide (B) Basic oxide (C) Neutral oxide (D) None Statement A: Carbon monoxide is a gas with molecular weight 28. Statement B: Its vapour density is 14. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. In the preparation of carbon monoxide from formic acid and oxalic acid, the action of H2SO4 is: (A) Dehydrating agent (B) Oxidising agent (C) Reducing agent (D) None of the above The action of coke on carbon dioxide is ______ process. (A) Reduction (B) Oxidation (C) Neutralisation (D) None Assertion(A) : Carbon monoxide is almost insoluble in water. Reason(R) : At 0°C, 1 litre (1000 cc) of water can dissolve 35 cc of carbon monoxide.

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(A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 15. Conditions favourable for the liquefication of carbon monoxide gas is: (A) Low pressure and low temperature. (B) High pressure and high temperature. (C) Low pressure and high temperature. (D) High pressure and low temperature. 16. Liquid carbon monoxide has boiling point of: (A) – 135°C (B) – 146°C (C) – 175°C (D) –192°C 17. The catalyst used in the below reaction is:

g   CH OH    2   450°C 3 (A) Nickel (finely divided) (B) Palladium (finely divided) (C) ZnO + Cu (D) Platinum (finely divided) 18. When equal volumes of carbon monoxide and chlorine gas are exposed to sunlight, they react to form extremely poisonous gas called_________. (A) Carbon dioxide (B) Phosgene (C) Marsh gas (D) None of the above 19. Which of the following statement(s) is/are correct about metal carbonyls? A) When finely divided metals like nickel, chromium and iron are heated and the carbon monoxide gas is passed, their respective carbonyls are formed. B) The formed metal carbonyls are volatile liquids. C) They are highly poisonous in nature. (A) ‘A’ and ‘B’ are correct (B) ‘B’ and ‘C’ are correct (C) ‘C’ and ‘A’ are correct (D) All are correct. 20. Assertion (A) : Metal carbonyls can be used in purification of metals. Reason (R) : Metal carbonyls can decompose to form metals and carbon monoxide on strong heating. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. CO  g  + 2H

Compounds of Carbon

209

21. Carbon monoxide combines with _____ of blood 28. When carbon monoxide is passed over heated soda and gives a compound named___. lime (a mixture of equal parts of NaOH and CaO) (A) Platelets, carbon dioxide it reacts to form: (B) Haemoglobin, carboxy haemoglobin (A) Sodium acetate (B) Sodium propionate (C) Plasma, oxygen, haemoglobin (C) Sodium formate (D) None (D) None of the above 22. Which of the following statement(s) is/are true about Carbon monoxide? 1)

It is a combustible gas, but does not support combustion.

CARBON DIOXIDE

12. Occurrence

2) It burns silently in air with pale blue flame.

(i)

Carbon dioxide occurs in free state in air, as it constitutes 0.03% of it by volume. 3) It burns without any smoke and produces a (ii) In combined state, a vast amount of carbon dioxide large amount of heat energy. is locked up in the form of mineral carbonates, on (A) Only ‘1’ is true (B) Only ‘2’ is true the surface of earth, e.g., limestone (CaCO 3 ); magnesite (MgCO3), dolomite (MgCO3. CaCO3), (C) Only ‘3’ is true (D) All are true etc. In addition to it, it is found in the form of calcium 23. The density and melting point of carbon monoxide carbonate in sea shells. are respectively: (A) 1.46 g/litre, – 402°C

13. Preparation General Methods of Preparation

(B) 1.37 g/litre, – 330°C (i)

(C) 1.25 g/litre, – 200°C (D) 1.80 g/litre, – 180°C 24. In the below reaction carbon monoxide acts as a:

Δ  2Fe s + 3CO g Fe O  s  + CO  g    2 3 2  (A) Oxidising agent (B) Reducing agent (C) Catalyst (D) None A CO  g  + H 2O  g    B + H2  g  450°C-500°C 25.  steam 

Identify ‘A’ and ‘B’ A

B

(A) Raney Nickel

Carbon monoxide

(B) Platinum gauze

Carbonic acid

(C) Fe2O3 + Cr2O3

Carbon dioxide

(D) Palladium

Methane

26. Which one of the following does not undergo thermal decomposition at any temperature?

By strongly heating bicarbonates of metals: When bicarbonates of metals are heated strongly, they decompose to form their respective metallic carbonates, carbon dioxide gas and water.   Na CO  H O  CO 2NaHCO3  2 3 2 2

  K CO  H O  CO 2KHCO3  2 3 2 2  Ca HCO3  CaCO3  H 2 O  CO2 2

   Mg  HCO3   MgCO3  H 2O  CO 2 2

(ii) By burning charcoal or coke or compounds containing carbon: When charcoal, coke, methane gas, ethylene gas, acetylene gas, alcohol, etc., are burnt in air or oxygen they form carbon dioxide as one of the products.   CO   C  O 2  2 CH 4  2O2   CO 2  2H 2O  t

(A) Carbon dioxide

(B) Li2CO3

C 2 H 4  3O2   2CO 2  2H 2O  t

(C) CO

(D) None

2C 2 H 2  5O 2   4CO2  2H 2O  t

27. Which one among the following forms addition product with carbon monoxide? (A) CuCl + 2H2O

(B) CuCl

(C) Cu2Cl2

(D) None

(iii) By strongly heating carbonates of metals: With the exception of sodium carbonate and potassium carbonate, all metallic carbonates on heating strongly decompose to form their respective www.betoppers.com

9th Class Chemistry

210 metallic oxides and carbon dioxide.   CaO  CO CaCO3  2

  ZnO  CO ZnCO3  2 (iv) By treating carbonates and bicarbonates of metals with dilute mineral acids When bicarbonates or carbonates of metals are treated with dilute mineral acids, they react without heating to form their respective metallic salts, water and carbon dioxide gas.   Na SO  H O  CO Na2CO3  H 2 SO4  2 4 2 2

  CaCl  H O  CO CaCO3  2HCl  2 2 2

Laboratory Preparation Carbon dioxide gas is prepared in laboratory by treating marble chips with dilute hydrochloric acid, when the reaction takes place at the room temperature.   CaCl  H O  CO CaCO3  2HCl  2 2 2

14. Physical properties

magnesium ribbon is taken in the jar of carbon dioxide, it continues burning in it. Why burning magnesium continues burning in carbon dioxide ? When burning magnesium ribbon is taken in the jar of carbon dioxide, it continues burning, because the heat of burning magnesium decomposes it to carbon and oxygen. The oxygen so liberated sustains the burning of magnesium,

2Mg  CO 2   2MgO  C II. Action with litmus: Carbon dioxide turns moist blue litmus paper scarlet or pink and hence, is acidic in nature. It can react with alkalis and forms salt and water as the products. III. Acidic Properties or Action with Alkalis : (a) Action with water : It dissolves in water under pressure and forms carbonic acid as the only product. CO2 (g) + H2O   H2CO3(Carbonic acid) (b) Action with dilute caustic alkalis (NaOH sol. or KOH sol] : (i) With limited amount of carbon dioxide : When limited amount of carbon dioxide gas is passed through excess of caustic soda solution or caustic potash solution, it reacts to form their respective carbonates and water. The carbonates so formed are soluble in water.

(i) Carbon dioxide is a colourless gas. (ii) Pure carbon dioxide is an odourless gas. However, impure CO2 has a faint acid odour due to the presence of HCl vapours. 2NaOH  excess   CO2  less  (iii) Pure carbon dioxide has a faint sour taste. When a few drops of water are added to a jar of carbon   Na CO  H O 2 3 2 dioxide and then tasted, the water has a sour taste. (iv) Carbon dioxide is fairly soluble in water. One volume 2 KOH  excess   CO  less  2 of water can dissolve one volume of carbon dioxide gas.   K CO  H O 2 3 2 (v) Carbon dioxide gas is heavier than air, its vapour (ii) With excess amount of carbon dioxide : When excess density being 22. whereas that of air is 14.4. amount of carbon dioxide is passed through dilute (vi) When carbon dioxide at room temperature (20 0C) caustic alkali solutions, it reacts to form their insoluble and at a pressure of 70 atmospheres, is allowed to bicarbonates, as the only product. expand through a nozzle, it solidifies to form solid NaOH  CO 2   NaHCO 3 carbon dioxide, commonly called dry ice. Dry ice has nothing in common with ice formed KOH  CO 2   KHCO 3 from water, except that it is snow white. It does not melt but changes directly into gaseous carbon (c) Action with lime water: (i) With limited amount of carbon dioxide : When dioxide. The temperature of solid CO2; is 78 °C. limited amount of carbon dioxide is passed through 15. Chemical properties lime water, it turns milky, on account of the formation of insoluble calcium carbonate, which is white in I. Combustibility : Carbon dioxide gas is neither colour. combustible nor does it support combustion. If a burning candle or burning wooden splint is taken in Ca  OH 2  CO 2   CaCO3  H 2O the jar of carbon dioxide, it goes off and the gas does not catch fire. However, if a burning www.betoppers.com

Compounds of Carbon

211

(ii) With excess amount of carbon dioxide : When excess (ii) Mortar is a mixture of slaked lime [Ca(OH)2], sand amount of carbon dioxide is passed through and water. When applied to join bricks, it slowly limewater, it first turns milky and then the milkiness hardens due to the absorption of carbon dioxide disappears. It is because carbon dioxide reacts with from the atmosphere and changes to calcium calcium carbonate (insoluble) to form calcium carbonate which binds the bricks. bicarbonate, which is soluble in water. Ca  OH 2  CO 2   CaCO3  H 2O Ca OH 2  2CO 2   Ca HCO3 (iii) Health salts are the mixtures of dry citric acid or 2 tartaric acid and sodium hydrogen carbonate. On (d) Action with ammonium hydroxide : It reacts with addition of water to above mixtures, the acids react ammonium hydroxide solution to form ammonium with sodium hydrogen carbonate to liberate carbon carbonate and water. dioxide gas which forms an effervescent drink. On 2NH 4OH  CO 2   NH 4 CO3  H 2O consuming the drink, the gas pressure in stomach is 2 relieved. (e) Action with dry ammonia gas : When the mixture (iv) A mixture of 95% of oxygen and 5% of carbon of 2 volumesof dry ammonia and 1 volume of dry dioxide is marketed as Carbogen. Carbon dioxide carbon dioxide are compressed to 150 atmospheric present in the carbogen stimulates respiratory pressure at a temperature of 2500C, they react to system. Thus, carbogen is used for resuscitation in form urea which is an important artificial fertilizer the cases of (a) gas poisoning (b) shocks (c) 150 atm pneumonia (d) suffocation. 2NH3  CO2   NH 2 CO  s   H 2 O 0 2 250 C (v) When carbon dioxide at a pressure of 20 urea atmospheres is dissolved in water, it forms soda IV. Action with Metals water or plain soda. However, if carbon dioxide is (a) Burning sodium, potassium and calcium continue dissolved in sugar solution containing various burning in carbon dioxide to form their respective flavours or colours, it forms soft drinks commonly metallic carbonate and free carbon. consumed by us. 4Na  3CO 2   2Na 2 CO 3  C (vi) Used in the preservation of food stuffs.

















4K  3CO 2   2K 2CO 3  C

17. Tests

(b) Burning magnesium and aluminium continue burning (i) in carbon dioxide to form their respective oxides and free carbon (ii) 2Mg  CO   2MgO  C

It turns clear limewater milky, when passed through it in limited amount. However, if it is passed in a large amount, the milkiness disappears. If a burning match stick or candle is taken inside 2 the jar of carbon dioxide, it gets extinguished. 4Al  3CO 2   2Al 2O 3  3C (iii) If burning magnesium is taken inside carbon dioxide cylinder, it continues burning, producing white V. Action with Non-metals powder of magnesium oxide, which contains black The only non-metal which reacts with carbon dioxide specks of carbon. is carbon. When carbon dioxide gas is passed through red hot charcoal or coke, it is reduced to ormative orksheet carbon monoxide. 22. Assertion (A) : Carbon monoxide can be  CO 2  C   2CO   distinguished from other oxides of carbon. Reason (R) : Carbon monoxide burns with pale 16. Uses blue flame. (i) During photosynthesis, green plants absorb carbon (A) Both assertion and reason are correct and dioxide gas. In the presence of sunlight and reason is the correct explanation of assertion. chlorophyll, the carbon dioxide reacts with water to (B) Both assertion and reason are correct but reason form glucose and oxygen. The oxygen is released is not the correct explanation of assertion. in the atmosphere. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Sunlight 6CO2  6H 2O  Chlorophyll C6 H12O 6  6O 2

F

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9th Class Chemistry

212 23. When a filter paper soaked in palladium chloride or platinum chloride is taken in the jar of carbon monoxide, first it turns to colour ‘A’ then to colour ‘B’ and finally to colour ‘C’. Identify ‘A’ , ‘B’ and ‘C’ respectively. (A) Pink, Red, and Yellow (B) Pink, Green and Black (C) Green, Orange and Black (D) None of the above 24. When a mixture of air and steam is passed over red hot coke, the outgoing gas contains: (A) Producer gas (B) Water gas (C) Coal gas (D) Mixture of (a) and (b) 25. Match the following (related to carbon dioxide): Scientist Discovery p) Van Helmont i) Named it as acid carbonique. q) Joseph Black ii) Discovered CO2 gas formed when charcoal is burnt in air. r) Lavoisier iii) Prepared CO2 by heating magnesium carbonate. p q r (A) i ii iii (B) ii iii i (C) iii i ii (D) i iii ii 26. Carbogen is a mixture of: (A) 95% carbon dioxide and 5% oxygen (B) 95% oxygen and 5% carbon dioxide (C) 95% oxygen and 5% nitrogen (D) 95% carbon dioxide and 5% hydrogen. 27. Hopcalite is a mixture of: (A) 50% MnO2, 30% CuO, 15 % Co2O3, 5% Ag2 O. (B) 30% MnO2, 50% CuO, 7 % Co2O3, 13% Ag2 O. (C) 13% MnO2, 30% CuO, 50 % Co2O3, 7% Ag2 O. (D) None of the above. 28. Assertion (A) : A person exposed to carbon monoxide dies due to lack of oxygen. Reason (R) : If carbon monoxide is inhaled, the haemoglobin combines with it forming a stable compound, carboxy-haemoglobin which prevents the haemoglobin from taking up oxygen. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. www.betoppers.com

29. Match the following (related to carbon dioxide): Scientist Discovery p) Van Helmont i)Named it as acid carbonique. q) Joseph Black ii) Discovered CO2 gas formed when charcoal is burnt in air. r) Lavoisier iii) Prepared CO2 by heating magnesium carbonate. p q r (A) i ii iii (B) ii iii i (C) iii i ii (D) i iii ii 30. Δ  Na SO aq + H O  + A Na CO  s  + H SO  aq   2 3 2 4 2 4  2  





Δ  B + 2H O  + 2CO g Ca HCO3  s  + 2HNO3  aq   2 2   2 

Identify ‘A’ and ‘B’ in the given equations. A B (A) Carbon monoxide Calcium oxide (B) Carbon dioxide Calcium nitrate (C) Carbon dioxide Calcium oxide (D) None of the above 31. Take an empty glass trough and fix in it two candles of different heights and light them. Allow the carbon dioxide to flow into the trough from the delivery tube. It is observed that the flame of smaller candle goes off first, followed by the bigger candle. Which of the following observations are correct according to the given experiment? A) CO2 is lighter than air. B) It can be poured out of a gas jar like a liquid. C) It does not support combustion. (A) A and B (B) B and C (C) A and C (D) None 32. ‘A’ on heating gives ‘B’, a solid ‘C’ and water. ‘B’ on heating gives again ‘C’, which is a gas. ‘C’ when allow through lime water it gives ‘B’. Identify A, B and C respectively. A B C (A) Calcium bicarbonate Calcium carbonate Carbon dioxide (B) Calcium hydroxide Calcium carbonate Carbon monoxide (C) Calcium bicarbonate Calcium oxide Carbon dioxide (D) Calcium hydroxide Calcium oxide Oxygen

Compounds of Carbon

213

33. Which one of the following about dry ice is true? (A) It is very cool to touch. (B) It melts when it is heated. (C) It is snow white in colour. (D) It is poisonous in nature.

Ferric oxide 37. Carbon dioxide gas is _______ in nature. (A) Acidic (B) Basic (C) Amphoteric (D) None 38. When few drops of water are added to a jar of carbon dioxide and then tasted, the water has a onceptive orksheet _______ taste. (A) Bitter (B) Sour ‘A’ and ‘B’ are two unknown oxides of carbon. (C) Sweet (D) None ‘A’ burns with blue flame to give ‘B’. 39. Invert a cylinder filled with carbon dioxide in a Identify A and B. trough of water and leave it undisturbed for one (A) CO and CO2 (B) CO and H2CO3 hour. It is seen that level of water________. (C) CO2 and CO (D) None of the above (A) Decreases (B) Remains unchanged It is dangerous to: (C) Increases (D) None of these (A) Sleep in a closed room with a coal fire burning. 40. Dry ice is: (B) Smoke or be in the vicinity of smokers. (A) Solid carbon monoxide. (C) Both ‘a’ and ‘b’. (B) Liquid carbon monoxide. (D) None of the above. (C) Solid carbon dioxide. Carbon dioxide occurs freely in the atmosphere to (D) Liquid carbon dioxide. the extent of _______ by volume. 41. Carbon dioxide is: (A) 0.05% (B) 0.04% (A) A supporter of combustion. (C) 0.03% (D) 0.02% (B) Non-supporter of combustion. In the combined state, a vast amount of carbon (C) Combustible. dioxide is locked up in the form of mineral. (D) None of the above. (A) Carbonates (B) Oxides 42. Dry ice provides temperature as low as: (C) Halides (D) None of the above (A) – 62°C (B) – 54°C Carbon dioxide is ______ in nature. (C) – 40°C (D) – 38°C (A) Pleasant (B) Poisonous 43. Assertion (A) : Small percentage of carbon dioxide (C) Non-poisonous (D) None is mixed with oxygen in the production of carbogen. The action of hopcalite is: Reason (R) : During over diffusion of O2 into the (A) To reduce carbon monoxide lungs, victims will feel respiratory discomfort. (B) To reduce carbon dioxide (A) Both assertion and reason are correct and (C) To oxidise carbon monoxide reason is the correct explanation of assertion. (D) To oxidise carbon dioxide (B) Both assertion and reason are correct but reason Carbon dioxide occurs in the atmosphere: is not the correct explanation of assertion. (A) In free state . (C) Assertion is correct and reason is incorrect. (B) In the combined state. (D) Assertion is incorrect and reason is correct. (C) Both ‘a’ and ‘b’ are correct. 44. Which one of the following statement(s) is/are true (D) None of these. about CO2? Identify A, B and C in the given equations: (A) It regenerates oxygen into the atmosphere Δ A) CaCO3  s  + 2HCl  aq    A + H 2 O  l  + CO 2 during photosynthesis. (B) As carbogen it stimulates respiratory system. Δ  MgCO 3  s  + H 2 O  l  + B B) Mg  HCO 3 2  aq   (C) Both ‘A’ and ‘N’. (D) None of the above. Δ  C + 3CO 2  g  C) Fe 2  CO3 3  s   45. Composition of baking powder is: (A) Sodium hydrogen carbonate + calcium A B C dihydrogen phosphate and starch. (A) Carbon monoxide; Carbon monoxide; (B) Sodium carbonate + calcium phosphate and Ferrous oxide starch. (B) Chlorine; Magnesium oxide; Ferric oxide (C) Sodium hydrogen carbonate + calcium (C) Carbon tetrachloride; Carbon monoxide; phosphate and starch. Ferric oxide (D) None. (D) Calcium chloride; Carbon dioxide ;

C 29.

30.

31.

30.

33.

34.

35.

36.

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214

46. Assertion (A) : A foam type fire extinguisher can Reason (R) : Opening the valve, there results in a be used for fighting an oil fire but a soda-acid type fall in pressure which causes the liquid carbon cannot. dioxide to solidify and be liberated in the form of Reason (R) : In foam type extinguishers, the viscous white snow. foam ejected out covers the burning oil fire, cuts (A) Both assertion and reason are correct and off the air supply and the fire gets extinguished. reason is the correct explanation of assertion. (A) Both assertion and reason are correct and (B) Both assertion and reason are correct but reason reason is the correct explanation of assertion. is not the correct explanation of assertion. (B) Both assertion and reason are correct but reason (C) Assertion is correct and reason is incorrect. is not the correct explanation of assertion. (D) Assertion is incorrect and reason is correct. (C) Assertion is correct and reason is incorrect. 48. The commonly liberated gas in both soda-acid type (D) Assertion is incorrect and reason is correct. and foam type fire extinguisher is: 47. Assertion (A) : Liquid carbon dioxide fire (A) Oxygen (B) Carbon monoxide extinguisher may be used for both electrical and oil (C) Carbon dioxide (D) None of the above fires.

Separation of carbon dioxide and carbon monoxide from their mixture Method by passage of mixture through Equation Concentrated potassium hydroxide solution: KOH  CO2  KHCO3  CO2 dissolves forming potassium bicarbonate.  CO remains undissolved and is collected.  Recover y of CO2 from KHCO3 by the addition of acid. 2KHCO3  H2 SO4  K 2 SO4  2H 2O  2CO2 Water under pressure  CO2 dissolves forming carbonic acid. CO2  H 2O  H2CO3  CO remains undissolved and is collected.  Recover y of CO2 from H2CO3 by warming. H 2CO3  H 2O  CO2 

Comparioson between carbon monoxide and carbon dioxide: Property Carbon dioxide 1. Taste Slightly acidic taste 2. Nature Colourless and odourless 3. Density 22 (air 14.4) 4. Solubility in water Sparingly soluble at ordinary pressure 5. Liquefaction Liquefies at ordinary temperature 6. Combustibility Neither burns nor supports combustion 7. Action on

litmus

8. Nature by oxides

9. Action by caustic a lkali 10. Action on lime water 11. Oxidising/Reducing action. 12. Action with chlorine 13. Reaction with nickel 14. Physiological property

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Being weakly acidic, turns blue litmus scarlet red Acidic oxide. It neutralises alkali solutions forming carbonates and bicarbonates Absorbed to form alkali carbonate and hydrogen carbonates Turns it milky due to formation of CaCO3 Oxidising agent: Oxidises red hot carbon to action CO. No action None Non-poisonous

Carbon monoxide Tasteless Colourless and odourless 14 (air 14.4) Insoluble Difficult to liquefy Burns with a blue flame but does not support combustion Neutral, no effect on litmus Neutral. It does not react with alkalis No action No effect Reducing agent; reduces heated metallic oxides to respective metals Combines to form phosgene Forms volatile nickel carbonyls Highly poisonous; causes death

Compounds of Carbon

215

To demonstration that carbon dioxide contains 37. On passing CO and CO 2 through ammonical carbon. cuprous chloride solution. Take a magnesium ribbon and put over Bunsn flame (A) Carbon monoxide dissolves in ammonical till it catches fire. cuprous chloride solution forming addition compound. Put the burning magnesium in jar of carbon doxide. Magnesium contains burning with a dazzling white (B) Carbon dioxide bubbles out and is collected. flame, forming residue of magnesium oxide which is (C) Both ‘a’ and ‘b’ are true. seen to contain black specks of carbon. (D) None of the above. Collect the residue in a clean beaker and dissolve it 38. On passing two unknown gases ‘C’ and ‘D’ through in dilute HCl. Filter the solution the black powder of water under pressure, ‘D’ dissolves in water to give carbon is left on filter paper. carbonic acid where as C does not. How can ‘D’ is recovered? ormative orksheet C D Recovery 34. Which of the following statement(s) is/are correct? (A) CO2 CO by dissolving in water (i) CO is obtained by the downward displacement (B) CO CO 2 by warming of water. (C) CO CO 2 by treating with HCl (ii) CO2 is heavier than air and can be poured from (D) None a gas jar is required. 39. Air does not normally contain carbon monoxide but (A) (i) is correct. (B) (ii) is correct. it contains carbon dioxide. The reason is: (C) Both (i) and (ii) are correct. (A) Carbon monoxide decomposes to carbon and (D) None of the above. oxygen. 35. ‘A’ and ‘B’ are two oxides of carbon. ‘A’ is a (B) Carbon monoxide gets oxidised to carbon poisonous gas and ‘B’ is a non-poisonous gas. On dioxide. exposing to ‘A’, the victim should be given artificial (C) Carbon dioxide is reduced to carbon monoxide. respiration with carbogen which contains ‘B’ in (D) None. small fraction. Identify A, B and the function of ‘B’ here. 40. Which one of the following on heating with H2SO4 gives both CO and CO2? AB Function of ‘B’ (A) Formic acid (B) Oxalic acid (A) CO CO 2 acts as oxidising agent. (C) Metal carbonates (D) None (B) CO2 CO acts as reducing agent. 41. Lower the burning magnesium ribbon in the jar of (C) CO CO 2 it avoids respiratory carbon dioxide. It is observed that magnesium discomfort during over continues burning with a dazzling white flame, diffusion of O2 takes place. forming residue of magnesium oxide and ________. (D) None of the above. (A) Oxygen is liberated 36. Match the following: (B) Black specks of carbon p) Carbon monoxide(i) is used as a refrigerant (C) Both ‘A’ and ‘N’ (D) None q) Carbon dioxide (ii) forms metal carbonyls 42. Which of the following is true? r) Dry carbon (iii) forms metal carbonates (A) If excess amount of CO2 is passed through dioxide (iv) almost insoluble in water lime water, it first turns milky and then milkiness even at high pressure disappears. (v) fairly soluble in water (B) The CO 2 absorbed by caustic potash is p q r recovered by the addition of an acid. (A) i, ii iii, iv, i (C) CO forms a poisonous gas with chlorine. (B) ii, iv v, iii i (D) CO forms carbonyls with finely divided metals (C) v, vi i, ii iii like Ni, Cr and Fe. (D) iv v ii, i, iii

F

W

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9th Class Chemistry

216

Summative Worksheet

Conceptive Worksheet 49. Statement A : CO2 is non-combustible and nonsupporter of combustion. Statement B : CO is combustible, but not a supporter of combustion. (A) Statement ‘A’ is true but ‘B’ is false. (B) Statement ‘A’ is false but ‘B’ is true. (C) Both statements ‘A’ and ‘B’ are true. (D) Both statements ‘A’ and ‘B’ are false. 50. Match the following: i) Powerful reducing agent p) CO ii) Oxidises active metals q) CO 2 iii) Used as a good fuel iv) Neutral v) Acidic vi) Used in carbogen p q (A) i, ii, iii iv, v, vi (B) i, iii, iv ii, v, vi (C) i, iv, vi ii, iii, v (D) i, ii iii, iv, v, vi 51. Assertion(A): Carbon monoxide is an excellent gaseous fuel whereas CO2 is not. Reason(R) : Combustion of carbon monoxide is an exothermic reaction. That is it proceeds through liberation of heat. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct and reason is not the correct explanation of assertion. (C) Assertion is correct, reason is incorrect. (D) Assertion is incorrect, reason is correct. 52. CO does not turn lime water milky on direct passage. But, on ignition, it turns lime water milky because. (A) CO is converted to H2CO3 (B) CO gas is converted to CO2 (C) CO2 is reduced (D) None of the above 53. By passing CO and CO2 through caustic soda solution both are distinguished because: (A) Both give different colours. (B) Both have different tastes. (C) CO2 dissolves in caustic soda whereas CO does not. (D) None. www.betoppers.com

1.

2.

3.

4. 5.

6. 7. 8. 9. 10 11.

12. 13. 14. 15. 16. 17. 18. 19. 20.

Diamond is the hardest natural substance known whereas graphite is one of the softest substance known. Explain. The basic structural difference between diamond and graphite contributes to their differences in densities. Give reason. The crystal lattice of diamond is built up from a tetrahedral unit while that of graphite from a hexagonal unit. Explain. Diamond is a non conductor of electricity while graphite is a good conductor. Explain. Diamond is chemically inert and does not react with most chemicals while Graphite is comparatively less inert. Give reason. Graphite is preferred to ordinary lubricating oils for heated machine parts. Give reason. Black diamonds are generally used for cutting and drilling purposes. Give reason. Clay is mixed with graphite for use in lead pencils. Give reason. Air does not normally contain carbon monoxide. Give reason. Exhaust fumes of cars release poisonous fumes. Explain. A pale blue flame is seen on top of a charcoal oven, but no such flame is seen in the middle of the oven. Explain Why a conc. solution of KOH is preferred to other alkalis for absorption of CO2? Why carbon monoxide is not collected by downward displacement of air? In what way oxy- haemoglobin differs from carboxyl- haemoglobin? Why workers in certain factories are enforced to wear gas masks containing ‘ hopcalite mixture’?. Why an electrical fire cannot be extinguished using a soda- acid or foam type of fire extinguisher?. Carbon dioxide gas and not chlorine finds application in extinguishing fires. Give reason. How dry ice is advantageous compared to ordinary ice? The term “ lead pencil” is misnomer. Explain. The density of charcoal is 1.5 g/cm3 . Yet it floats on water. Explain.

Compounds of Carbon 21. The following is a list of certain properties of diamond and graphite. (A) Graphite is a good conductor of electricity and inactive with acids. (B) Diamond has high refractive index and dispersion power. (C) Graphite absorbs excess neutrons. (D) Graphite is resistant to chemicals and has good thermal conductivity. (E) Hexagonal layers in graphite slide over one another and graphite has a high melting point. (F) C-C bonds in diamonds are very strong and the crystal lattice stable and rigid. (G) Hexagonal layers in graphite are held by weak forces and hence graphite leaves a mark on paper. Choose from the list A to G the most appropriate property on which the following applications of diamond and graphite are based. 1. As a lubricant for heated machine parts. 2. In dry cells. 3. In jewellery 4. As a tip for deep boring drills 5. In nuclear reactors 6. During manufacture of high grade steel 7. In lead pencils. 22. Coal, coke and lamp black are three forms of amorphous carbon. State which of the three forms is: A) Formed by carbonization of vegetable matter. B) Obtained by incomplete combustion of carbonaceous fuels in a limited supply of air. C) Used in the production of water gas and producer gas. E) Prepared by heating bituminous coal to high temperatures in the absence of air. F) Black, light and used in the manufacture of typewriter ribbons. 23. Starting from wood caracal (carbon) how would you obtain the following. (write balanced equations) A) An acidic gas [ using a neutral gas] B) A hydrocarbon [using hydrogen] C) An inflammable liquid [ using sulphur] D) Water gas [using steam] E) Sulphur dioxide [using an acid]

217 24. Complete the table by selecting the correct word in each case: Property of carbon monoxide 1. Comparative density 2. Solubility in water 3. Liquefaction 4. Combustibility

5. Nature

(a) 2.2 times heavier than air (b) almost as heavy as air (a) fairly (b) highly (c) slightly (a) easy to liquefy (b) difficult (a) Noncombustible but supporter of combustion. (b) Combustible but non supporter of combustion (a) Acidic (b) Basic (c) Neutral

25. Thermal decomposition or action of dil.acids on metal carbonates and bicarbonates ( write balanced equations) A) A carbonate which on thermal decomposition leaves a black residue. B) A bicarbonate which on thermal decomposition gives a soluble carbonate. C) A carbonate which reacts with an acid to give sodium sulphate as the regulator salt. D) Thermal decomposition of a bicarbonate of a divalent metal. E) A bicarbonate which reacts with an acid to give soluble Ca(NO3 )2 as the residual salt.

HOTS Worksheet I. 1. 2. 3. 4.

5. 6. 7. 8. 9.

Give reasons for the following: Allotropes are chemically identical. Diamond can scratch iron but the reverse is not true. Diamond is hard, whereas graphite is soft. Though density of wood charcoal is greater than water, it still floats on water. However, if wood charcoal is boiled, it sinks in water. Charcoal is used in gunpowder preparation. Inhaling of exhaust fumes of vehicles is hazardous to health. Activated charcoal is preferred to wood charcoal in the preparation of dental powders. Diamond is more inert than graphite. Diamond is a bad conductor of electricity while graphite is a good conductor of electricity. www.betoppers.com

9th Class Chemistry

218 10. Graphite as well as diamond both are the allotropes of carbon, but graphite is used as a lubricant, where as diamond is used as an abrasive. 11. A pale blue flame is seen on top of a charcoal oven, but no such flame is seen in the middle of the oven. 12. A concentrated solution of KOH is preferred to other alkalies for absorption of CO2. 13. Carbon monoxide is not collected by the downward displacement of air. 14. Sleeping in a room with all forms of ventilation closed and a coal fire smouldering in one corner could be fatal. 15. Limewater cannot be used to distinguish between SO2 and CO2 gas. 16. CaCO 3 is preferred to lead carbonate for the laboratory preparation of CO2 using dilute acid. 17. Formation of calcium chloride as the residual salt formed on reaction of CaCO3 with dil. acid is preferred to the formation of CaSO4 as the residual salt on reaction with dil.H2SO4. 18. Carbon dioxide does not support combustion but a burning magnesium ribbon continues to burn in it. 19. Dilute HCl is preferred to dil.H 2 SO 4 for the preparation of CO2 from marble. 20. On passage of CO2 through KOH solution no milkiness is seen but when passed through lime water the latter turns milky. II. Write the balanced chemical equations for the following changes: 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

31. Complete the following : conc.H SO 2 4  (A) + (B) + (C) H C O  2 2 4 Δ Gas Gas Liquid Gas (A) burns with blue flame and gas (B) turns lime water milky. Gas(A) on oxidation gives gas (B). Liquid (C) produces blue colour when comes in contact with anhydrous copper sulphate. (A) + Cl2  (D)

NH 3  (E) 

(B)

NH3 

Identify the compounds from (A) to (E) ? 32. A mixture of two gases is formed when an organic acid is heated with conc. H2SO4. When the gaseous mixture is passed through KOH solution, one gas is absorbed. The unabsorbed gas combines with chlorine and forms a poisonous gas. The organic acid and the two gases evolved with conc. H2SO4. are ? 33.

 CaCO3  (A) (S) + (B)(g)

+ carbon heat

(C)(S) + (D)(g)

(C)(S) + H2O  (E) (g)

Identify the compounds from A to E. 34. A white compound (A) on red heating decomposes Action of wood charcoal on ferric oxide and calcium to give a residue (B) and two gases (C) and (D). oxide. The gas (C) burns with a blue flame. The gas (D) Action of wood charcoal on sulphuric acid and nitric turns lime water milky. The residue (B) on dissolving acid. in water gives heat and solution is alkaline to litmus.The compound (A) imparts brick-red colour Fusion of graphite and sodium carbonate. Action of wood charcoal on calcium and sulphur. to flame. Identify (A) to (D). Superheated steam is passed over wood charcoal. Nickel is heated and carbon monoxide is passed through it. orksheet Laboratory preparation of CO2 from marble chips. 1. Carborundum is obtained when silica is heated at high temperature with Preparation of carbon monoxide from formic acid. (A) Carbon (B) Carbon monoxide Combustion of ethene. (C) Carbondioxide (D) Calcium Carbonate Action of NaOH with (a) limited amount of CO2 and (b) excess amount of CO2 2. Moderate electrical conductivity is shown by (A) Silica (B) Graphite (C) Diamond (D) Carborundum

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IIT JEE W

Compounds of Carbon 3.

Lead pencil contains (A) Pb (B) FeS (C) Graphite (D) Phosphorus 4. Which of the following is a semiconductor (A) C (B) Pb (C) Ge (D) Sn 5. The inert form of carbon is (A) Diamond (B) Graphite (C) Coal (D) Charcoal 6. Synthesis gas is a mixture of (A) Steam and CO (B) CO and N2 (C) H2 and CO (D) H2 and CH4 7. Poisonous gas present in the exhaust fumes of car is (A) CH4 (B) C2H2 (C) CO (D) CO2 8. Galena is an ore of (A) Gallium (B) Lead (C) Tin (D) Germanium 9. Identify the correct statement with respect to carbon monoxide? (A) It combines with water to form carbonic acid. (B) It reacts with haemoglobin in red blood cells (C) It is a powerful oxidising agent (D) It is used to prepare aerated drinks. 10. Incomplete combustion of petrol or diesel oil in automobile engines can be best detected by testing the fuel gases for the presence of (A) Carbon monoxide and water vapour (B) CO (C) NO2 (D) SO2 11. Which of the following types of forces bind together the carbon atoms in diamond? (A) Ionic (B) Covalent (C) Dipolar (D) Vanderwaals

219 12. Which one of the following oxides is neutral? (A) CO (B) SnO2 (C) ZnO (D) SiO2 13. Which of the following is true for diamond? (A) It is a good conductor of electricity (B) It is soft (C) It is a bad conductor of heat (D) It is made up of C,H and O 14. Carbon forms large number of compounds because it has (A) Fixed valency (B) Non-metallic nature (C) High I.P (D) Property of catenation 15. The gas which burns with a blue flame is? a) CO b) O2 c) N2 d) CO2 16. Freon -12 is used as a (A) Refrigerant (B) Insecticide (C) Fungicide (D) Herbicide. 17. Carbogen is (A) Pure form of carbon (B) Lignite (C) Bituminous (D) Mixture of O2 & CO2 18. Higher percentage of carbon is found in (A) Anthracite (B) Lignite (C) Bituminous (D) Peat 19. The green house effect is caused by (A) NO2 (B) NO (C) CO (D) CO2 20. Drikold is (A) Solid CO2 (B) Ether and dry ice (C) Dry ice and alcohol (D) Dry ice and acetone

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B y t h e e n d o f t h i s c h a p t e r, y o u w i l l u n d e rs t a n d      

Occurrence Bio data Preparation Physical properties Chemical properties Nitrogen cycle

 Uses  Tests AMMONIA  Preparation  Physical properties  Chemical properties

 Uses  Tests NITRIC ACID  Occurrence  Laboratory Preparation  Manufacture of Nitric acid

NITROGEN 1. Introduction In 1772, Daniel Rutherford, a Scottish physician, discovered a gas that did not support life. He found that animals died in it and he called it 'Metaphitic Air'. The properties of this gas were also studied by the French chemist Antoine Lavoisier, and in 1775, he named it 'Azote' meaning lifeless. It was Antoine Chaptal, another French scientist, who in 1823 named this gas 'Nitrogen', a name by which it is known today. The name was based on 'nitre' (KNO3) of which it is an essential constituent. Nitrogen is a gas with no colour, taste or smell. It forms part of the proteins in every living cell. Like oxygen, nitrogen in the air is made up of a molecule consisting of two atoms. Although nitrogen is itself unreactive, it's significance lies in the fact that it is an essential constituent of proteins which is required by all living organisms for their growth. By processes natural as well as artificial, nitrogen is made available to plants.

Position in the Periodic Table 

 

   

Physical properties Chemical properties Tests Uses

Chapter - 10

Compounds of Nitrogen

Learning Outcomes

2. Occurrence 







In the free state, nitrogen constitutes nearly 78% of the air. Three - fourths by weight and four-fifths by volume of air is nitrogen. In the combined state, Nitrogen is an important constituent of proteins. Proteins form the bulk of plant and animal tissues. Nitrogen occurs as a constituent in several minerals such as nitre (KNO3) and chile salt petre (NaNO3). Nitrogen forms many compounds with oxygen. These include : * Nitric oxide which is formed during thunder and lighting and forms part of the nitrogen cycle. * Nitrogen dioxide which come out of car, exhausts and pollute the environment. Dinitrogen oxide (N2O) is a sweet smelling gas used as an anaesthetic. It is called 'laughing gas' because, it makes some patients laugh before they become unconscious. Though this oxide of nitrogen was initially used as a party gag, later scientists realised its medical usefulness.

3. Bio data 

Biodata of nitrogen: Symbol Atomic weight Atomic number Electronic configuration

: : : :

N 14 7 K L 2 5 Valency : 3, 5 Position in Periodic table : Second period, Group

Nitrogen ranks first in Group 15 of the periodic table. Other members of this group are phosphorus, arsenic, antimony and bismuth. The density and boiling point increase with the rise in atomic number. Non - metallic character decreases while metallic character increases. Valence electrons 15 Nitrogen and Phosphorus : Non - metals Some Important points to note Arsenic and Antimony : Metalloids They all form number of oxides, Nitrogen forms Bismuth : Typical metal  the maximum number of oxides. For example: N2O, NO, N2O3, N2O4, N2O5.

9th Class Chemistry

222  





They all form hydrides, halides and oxyacids. The hydride of nitrogen (ammonia) is the most stable and is basic in nature. Stability and basic nature decreases down the group Hydrides : NH3, PH3, AsH3, SbH3, BiH3 Halide of nitrogen is NCl3 (nitrogen tr ichlor ide). The halides are all covalent. Halides : NCl3, PCl3, AsCl3, SbCl3, BiCl3 NCl3, PCl3, AsCl3 : Liquid SbCl3, BiCl3 : Solid The oxyacid of nitrogen is HNO3 (nitric acid) and HNO2 Acidic property of the oxyacids decreases from nitrogen to bismuth. For example, Oxyacids : HNO3 , H3PO4 , H3 AsO4 , H3SbO4, H3BiO4 . Nitric acid (HNO3) : Strong acid Phosphoric acid (H3PO4) : Weak acid

From Chemical Compounds 1.

By Oxidation of Ammonia : Ammonia is an hydride of nitrogen. When it is heated with oxidising agents like oxygen, chlorine, copper oxide, lead monoxide or bleaching powder, it yields free nitrogen. (a) Ammonia + Oxygen (Ammonia is burnt in an atmosphere of oxygen)

4NH3  3O 2  2N 2  6H 2O (b) Ammonia + chlorine (Chlorine gas is bubbled through a concentrated solution of ammonia). 8NH3  3Cl 2  6NH 4Cl  N 2 (Excess)

(c) Ammonia + copper (II) oxide 3CuO

(Basic oxide)

2NH3 

4. Preparation Laboratory Method (Nitrogen from Air) 

Air is purified in the following manner: Dust is removed by passing the air through filters. Water vapour is absorbed by passing air through drying agents like fused calcium chloride or concentrated sulphuric acid. Carbon dioxide is removed when the air is bubbled through an alkali (e.g., KOH). The air free from dust, moisture and carbon dioxide is passed over copper turnings. Oxygen combines with the red, metallic copper to form black copper oxide.



Nitrogen thus isolated from air, is collected by the downward displacement of water since it is practically insoluble in water and almost as heavy as air. Obtaining Nitrogen from Air : A Summary 1. Filter : Removes dust 2. Drying agent : Removes moisture 3. Alkali : Removes CO2 4. Separation of N2 : (i) Fractional distillation of liquid air, or, from [N2 + O2] (ii) Removal of O2 by a reducing agent.

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3Cu

3Cu

(Reddish metal)

(Reddish metal)

 3H 2O  N 2

 3H 2O  N 2

(d) Ammonia + lead (II) oxide (Ammonia is passed over the metal oxides placed in a combustion tube). 3PbO

( Amphoteric oxide )

Pb + 2NH3  ( Silvery3white + 3H2O + N2 metal )

(e) Ammonia + bleaching powder Concentrated ammonia solution is heated with bleaching powder (paste). 3CaOCl2 + 2NH3  3CaCl2 + 3H2O + N2

From Thermal decomposition of nitrogenous compounds 2.

2Cu  O2  2CuO 

 2NH3 

Ammonium Dichromate: When orange ammonium dichromate is heated, an exothermic reaction occurs which causes the salt to erupt like a volcano. Nitrogen and steam are liberated and green residue of chromic oxide is formed.

Heat Cr O  4H O  N  NH 4 2 Cr2O7  2 3 2 2 (Orange)

(Green)

3.

Ammonium Nitrite : Ammonium nitrite is quite unstable and readily decomposes giving nitrogen and steam. NH4NO2  N2 + 2H2O



Procedure: A concentrate mixture of aqueous ammonium chloride and sodium nitrite is taken in equimolar ratio and heated in a round bottomed flask.

Laboratory preparation

Compounds of Nitrogen



The flask is provided with a thistle funnel for pouring 5. in the mixture, and a delivery tube for the gases to escape. Collection of nitrogen: Nitrogen is collected by the downward displacement of water since, (i) it is only slightly soluble in water. (ii) being almost as heavy as air, nitrogen cannot be collected over air Heat NH C   NaNO   NH NO  NaC  4 2 4 2 NH NO   N  2H O 4 2 2 2

223 Reaction with Metals: When nitrogen is passed over heated, reactive metals, the corresponding metal nitrides are formed. Calcium +

nitrogen

  Calcium nitride

3Ca

N2

  Ca3N2

+

Magnesium + nitrogen   Magnesium nitride 3Mg

+

N2

  Mg3N2

Aluminium + nitrogen   Aluminium nitride 2Al + N2   2AlN The nitrides are decomposed by boiling water to form ammonia and the corresponding hydroxide. Calcium nitride + water

5. Physical Properties (i) It is a colourless, odourless and tasteless gas. (ii) Vapour density of nitrogen is 14 while that of air is 14.4. Thus it is slightly lighter than air. (iii) It is very slightly soluble in water. (iv) Nitrogen liquefies into a colourless liquid when compressed. On further cooling, this can be frozen 6. into a white solid. (v) Nitrogen is a non- poisonous gas. However, it is due to the lack of oxygen that animals suffocate and die in an atmosphere of nitrogen. (vi) Freezing point of nitrogen is – 209.80C. (vii) Boiling point of nitrogen is – 195.80C.

Ca3N2+6H2O   3Ca(OH)2 + 2NH3 Magnesium nitride+ water Mg3N2+6H2O   3Mg(OH)2+2NH3 Aluminium nitride+ water

6. Chemical Properties

N2 +

1.

Oxygen: Nitrogen combines with oxygen to form nitric oxide (NO) under the influence of an electric arc at a temperature of 30000 C. The reaction between nitrogen and oxygen also occurs in nature during thunder and lightning.

2 3.

Combustibility: Non-combustible, and non-supporter of combustion. Nitrogen extinguishes a burning candle. Action on moist litmus: Neutral to moist litmus paper. Chemical reactivity: At ordinary temperatures, nitrogen is chemically unreactive. Under special conditions of temperature and pressure, nitrogen reacts with certain 7. substances. Model of nitrogen molecule

AlN + 3H2O   Al(OH)3 + NH3 Reaction with Non-metals Hydrogen: Nitrogen and hydrogen combine in the presence of electric sparks, or a catalyst like finely divided iron at 4500 C-5000C and 200 atmospheres to form ammonia.

N2 +

3H2   2NH3 +

O2

 2NO 

–

heat

heat

Reaction with calcium carbide: At temperatures between 8000C-10000C nitrogen reacts with calcium carbide to form calcium cyanamide and carbon. This is commercially known as nitrolim, which is an important fertilizer. CaC2 +

N2

 CaCN2 + 

C

7. Nitrogen Cycle

4.

N N N N

Inactivity of nitrogen at ordinary temperature: Nitrogen is a diatomic molecule. The two atoms are held together by three strong covalent bonds. A large amount of energy is required to break apart the molecule. Hence nitrogen reacts only at high temperatures.



The Nitrogen Cycle: The main steps of the nitrogen cycle are: Fixation of atmospheric nitrogen.  Assimilation of the fixed nitrogen by plants.  Transfer of the fixed nitrogen from the plant  ot the soil. Nitrogen is set free from the nitrogenous  compounds and returned to the atmosphere. www.betoppers.com

224 



Nitrogen Fixation: The process of converting atmospheric nitrogen into useful nitrogenous compounds by natural or artificial means is known as fixation of atmospheric nitrogen. Nitrogen fixation may be by natural processes or artificial processes. Natural methods of nitrogen fixation During rain and thunder storms: During thunder and lightning, the high temperatures produced (30000C-50000C) cause the nitrogen and oxygen of the air to combine and form unstable nitric oxide. N2 + O 2   2NO This further reacts with the oxygen of the air to form nitrogen dioxide which dissolves in rain water to form nitric acid. 2NO

+

O2

 

2NO2

4NO2 + 2H2O + O2   4HNO3 Nitric acid enters the soil and dissolves the minerals to form soluble nitrates. Certain bacteria present in the soil, called nitrifying bacteria, make the nitrogen in the form of soluble nitrates available to the plant. MgCO3 + 2HNO3

 







8. Uses 1.

2.

  Mg(NO3)2 + H2O + CO2  Symbiotic bacteria: Roots of leguminous plants have symbiotic bacteria which help convert atmospheric nitrogen into soluble nitrates and make it available to the plant. Important Points to note: i) Nitrogen enters the soil in the following ways: During lightning discharge when it is converted into soluble nitrates. With the help of nitrogen fixing bacteria in leguminous plants. By the addition of natural (compost) and artificial fertilizers into the soil. ii) Nitrogen from the soil enters plants and animals: The root hair of plants absorb the nitrates from the soil. Enzymes present in the plant body convert them into amino acids which are the building blocks of proteins.The plants are eaten by the animals. Thus, nitrogen is the form of plant proteins reaches the body of animals. iii) Nitrogen from plant and animal bodies is returned to the soil and air: In the animal body, the plant proteins are broken down into urea which is excreted in urine. Hydrolysis of urea produces ammonia.

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9th Class Chemistry Artificial fixation of nitrogen: In addition to the conversion of atmospheric nitrogen into assimilable nitrogenous compounds by natural methods, artificial processes are also used to form nitrogenous compounds which can be converted into fertilizers for use by the plants and for the production of ammonia and nitric acid. Nowadays, all the ammonia is manufactured by the Haber's process. Earlier, it was also made by the Birkeland and Eyde process and the Cyanamide process. Both these processes have now become absolute and are no longer used. The ammonia produced is directly converted into fertilizers like ammonium sulphate, ammonium nitrate and urea. Ammonia is converted to nitric acid by the Ostwald's Process. Nitric acid is then used in the production of fertilizers.

3.

4.

5.

6.

7.

8.

In the manufacture of: Ammonia (by Haber’s process), nitric acid and fertilizers such as ammonium sulphate, calcium cyanamide, etc. In high temperature thermometers: The space above mercury is filled with nitrogen to reduce its evaporation. Such thermometers can measure upto 50000C. In canning foodstuffs: Coffee, vegetable ghee etc. retain their flavour and colour better if the ‘air space’ in the can is filled with nitrogen. A layer of inactive nitrogen prevents the oxidation of food and retards bacterial growth. In industries: Being inert, nitrogen is flushed in laboratory and industrial reactions (such as extraction of metals, and nylon) to remove oxygen of the air and to provide an inert atmosphere. Due to its inertness only, it is also used in filling electric bulbs. (Argon is now preferred due to its better heat conductivity). When electric welding is carried out in air, metals tend to get oxidised and hence the welded part can break. To avoid oxidation, an atmosphere of nitrogen is provided. Liquid nitrogen It is used to provide low temperature in the laboratory i.e. as a refrigerant and to store the eye, blood and corneas at low temperature in hospitals. In plants: For protein synthesis.

Compounds of Nitrogen 9.

Diluting agent: It reduces activity of oxygen of the atmosphere and 4. thus controls the rate of combustion.

9. Tests 1.

2.

Nitrogen is chemically inert and there is no positive chemical test for the gas. If a gas extinguishes burning paper or candle or if it is neutral to litmus and does not turn lime water milky, it is most probably nitrogen. 5. When a burning magnesium ribbon is introduced in a jar of nitrogen, it continues burning, forming a very pale yellow powder (Mg3 N2 ). This powder on treating with water gives a very strong smell of ammonia gas. This gas gives dense white fumes when a glass rod dipped in HCl solution is brought near it.

Formative Worksheet 1.

2.

3.

Match the following: (related to nitrogen) Column I Column II 1) Nitrogen occurs in combined p) White of an egg state in the gaseous form in 2) Nitrogen occurs in the free state q) Nitre 3) Nitrogen is found in the combined r) Atmospheric air state in certain animal products. 4) Nitrogen found in certain minerals s) Ammonia like 1 2 3 4 (A) s r q p (B) p q r s (C) s r p q (D) q s p r Statement A : Nitrogen exists as a diatomic molecule. Statement B : Triple bond between nitrogen atoms makes nitrogen inert at room temperature. (A) Statement ‘A’ is correct but ‘B’ is incorrect. (B) Statements ‘A’ and ‘B’ are correct. (C) Statement ‘A’ is incorrect but ‘B’ is correct. (D) Both statements are incorrect. Assertion (A) : The dry nitrogen gas cannot be collected by the displacement of air. Reason (R) : Nitrogen is almost as heavy as the air. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion.

6.

7.

8.

9.

225 (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. When a crystals of compound ‘A’ are heated, they swell up and produce brilliant flashes with the liberation of nitrogen gas. The residue is green in colour due to the formation of chromic oxide. Identify the compound ‘A’. (A) Cr2O3 (B) K2 Cr 2O7 (C) (NH4)2Cr 2O7 (D) NH4NO2 Assertion (A) : Sodium nitrite and ammonium chloride are not heated in dry state. Reason (R) : Decomposition reaction of ammonium nitrite is endothermic and can be controlled. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Nitrogen can be liquefied under ______ pressure and _____ temperature. (A) High and low (B) Low and high (C) High and high (D) Low and low If a burning wooden splinter is taken in the jar of nitrogen, it goes off and the gas does not catch fire. What it indicates ? (A) Nitrogen is combustible, and it supports combustion. (B) Nitrogen is combustible but it does not support combustion. (C) Nitrogen is neither combustible nor supporter of combustion. (D) Nitrogen is not combustible but it is a supporter of combustion. 4500 C/?   N (g) +3H (g)  2NH (g) + Δt 2 2 3 200-900atms  The catalyst and promoter used in the above equation are respectively: (A) Fe and V2O5 (B) Ni and Mo (C) Ni and H2O2 (D) Fe and Mo Which of the following statement(s) is/are correct? (A) N2 reduces the activity of ‘O2’ of the atmosphere and thus, controls the rate of combustion. (B) A layer of inactive nitrogen prevents oxidation of food. (C) Nitrogen is used in filling electric bulbs. (D) Nitrogen is used as a refrigerant in laboratory.

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226

Conceptive Worksheet 1.

2.

3.

4.

5.

6.

7.

Which of the following is chile salt petre? (A) NaNO3 (B) KNO3 (C) Ca(NO3)2 (D) Mg (NO3)2 i) Exhaust of automobiles gives of _______. ii) ________ percentage of volume of nitrogen is present in air. (i) (ii) (A) Ammonia 78% (B) Oxyacids of nitrogen 4% (C) Chloride of nitrogen 21 % (D) Oxides of nitrogen 78 % Which of the following is/are exclusive properties of nitrogen? (A) Nitrogen forms trinegative ion. (B) Nitrogen can form sulphides. (C) Nitrogen is chemically inert at ordinary temperatures. (D) All the three statements are correct. What is the solubility of nitrogen in water? (A) Highly soluble (B) Insoluble (C) Slightly soluble (D) None of these i) Freezing point of nitrogen is____________. ii) Boiling point of nitrogen is ______________. (i) (ii) (A) 1000C 2000C (B) 209.80C 195.80C 0 (C) –100 C –2000 C 0 (D) –209.8 C –195.80C Which of the following is used in preserving blood? (A) Gaseous nitrogen (B) Liquid oxygen (C) Liquid nitrogen (D) Liquid oxygen and liquid nitrogen Assertion (A) : Finely divided iron catalyst does not affect the percentage yield of ammonia in Haber’s process. Reason (R) : Catalyst only accelerates the reaction and does not form a part of the reaction. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

8.

P

i) NH4Cl +

 NH4NO2 + R

(A) (B) (C) (D) 9.

(A) (B) (C) (D)

Q

.

S

ii) CaC2 + N2  + . P Q R S NaNO 2 NaCl CaNCN C NaNO 3 NaCl Ca 2 C 2 C 2NaNO3 2NaCl Ca 2 CN 2 H2 O NaNO 2 2NaCl N 2 2CaCN i) Name a potash fertilizer which is also a constituent of gun powder. ii) A nitrogenous fertilizer also used in manufacture of explosives. (i) (ii) Potassium nitrate Ammonium nitrate Calcium Carbonate Ammonium nitrate Ammonium nitrate Potassium nitrate Calcium cyanamide Calcium Carbonate

AMMONIA 10. Introduction Formula of ammonia : Relative molecular mass : Vapour density :

NH 3 17 8.5

Occurrence Decaying of plants and animals always remind us of unpleasant odour. This unpleasant odour is due to the action of bacteria on nitrogenous organic matter, which releases ammonia gas. Destructive distillation of coal and nitrogenous organic matter too produce ammonia gas as one of the products.

Discovery Joseph Priestley prepared ammonia gas for the first time in 1774 by heating lime with ammonium chloride. He called it ‘alkali air’, as it turned moist red litmus blue. Charles Bethelot analysed the gas and proved it to be a compound of nitrogen and hydrogen.

11. Preparation General methods 1.

Action of alkali on ammonium salts: Any ammonium salt when heated with alkalis or hydroxides of sodium, potassium and calcium, produces ammonia gas.

NH4C  NaOH  NaC  H2O  NH 3 

 NH 4 2 SO4  2 KOH  K2SO4  2H 2O  2 NH 3   NH 4 2 SO4  Ca OH 2  CaSO4  2H 2O  2NH 3  Note : Although all ammonium salts evolve ammonia when heated with alkalis, use of

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Compounds of Nitrogen

227

ammonium nitrate is avoided in the laboratory ii) preparation, because it is explosive in nature. It may decompose to give nitrous oxide and water vapour. NH4NO  3

N O  2H O 2 2 Nitrous oxide  

It cannot be dried by anhydrous calcium chloride, as the gas forms an additive compound with calcium chloride. CaC  4NH  2 3

CaC .4NH 2 3 Additive compound  

iii) Another commonly used drying agent is phosphorus Note: pentoxide, which also reacts with ammonia gas to i) Slaked lime Ca(OH)2 is preferred as the alkali, as it form ammonium phosphate. So, it cannot be used is not deliquescent like NaOH or KOH. for drying ammonia. ii) A higher ratio by weight of the alkali is used to avoid P2O5  6NH3  3H 2O  2 NH 4 PO4 loss of ammonium chloride by sublimation. 3 iii) The reaction is based on the fact that a stronger  ammonium phosphate  alkali like NaOH, KOH or Ca(OH)2 can decompose This is the reason, the gas is dried conveniently over salts of a weaker alkali like NH4OH to, produce calcium oxide, which is a basic oxide. ammonia gas. 2 By catalytic reduction of nitric oxide or Collection of ammonia gas: The gas is collected by the downward displacement nitrogen dioxide: of air in inverted gas jars due to the following reasons Nitric oxide or nitrogen dioxide can be reduced by : hydrogen by passing the mixture over heated i) Ammonia gas (vapour density = 8.5) is lighter than platinum gauze. air (vapour density = 14.4) 2NO  5H2  2NH3  2H2O ii) It is highly soluble in water.



2 NO2  7 H 2  2 NH 3  4 H 2O

Laboratory preparation In the laboratory, ammonia is prepared by heating a mixture of ammonium chloride NH4Cl (also known as salt of ammonia) and dry slaked lime Ca(OH)2 or powdered quick lime CaO. 2 NH 4 C   Ca  OH  2  CaC  2  2 H 2 O  2 NH 3 

2 NH 4 C   CaO  CaC  2  H 2 O  2 NH 3  Procedure: i) A mixture of ammonium chloride (sal ammoniac) and dry slaked lime in the ratio 1:3 by weight is taken in a hard glass flask. The flask is fixed to a retort stand with its mouth downwards in a standing position. This is done to avoid back suction of water into the heated flask which could crack the glass apparatus. The mixture is heated gently. The gas evolved is dried by passing it through lumps of calcium oxide (quick lime) in a drying tower. An inverted gas jar is placed over the drying tower to collect the gas by downward displacement of air. Drying of ammonia gas: i) Ammonia cannot be dried by concentrated sulphuric acid, as the alkaline gas reacts with the acid to form ammonium sulphate. 2NH3  H 2SO 4 

 NH4 2 SO4

 ammonium sulphate 



Preparation from metal nitrides Metallic nitrides such as magnesium nitride (Mg3N2) and aluminium nitride (AlN) are readily decomposed by boiling water at room temperature to give ammonia gas and the hydroxide of the metal, which form a white precipitate.

AN  3H O  Al  OH 3  NH  2 3

Industrial method Haber’s process: Nitrogen unites with hydrogen under controlled conditions to give ammonia.  2NH  92.4kJ N  3H  2 2 3 The reaction is reversible and exothermic. According to the principle of Le Chatelier, the formation of ammonia is favoured by low temperatures and high pressures. At low temperatures, the reactions is slow and so an optimum temperature is maintained. Conditions favourable for the preparation of ammonia by Haber’s process: Temperature = 725 K to 775 K; Pressure = 200 atmospheres. The catalyst used is finely divided iron mixed with a small amount of molybdenum or oxides of potassium and aluminium. NH3 formed is liquefied and removed.

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228

Formative Worksheet 10. Ammonia was called _________ for the first time.

11.

12.

13.

14.

15.

(A) Fixed air (B) Alkali air (C) Nitrogen air (D) None Who analysed the gas and proved ammonia to be a compound of nitrogen and hydrogen? (A) Joseph priestley (B) Lavoisier (C) Charles Bethelot (D) De Lassone Assertion (A) :Ammonium salt when heated with hydroxides of sodium, potassium and calcium produce ammonia gas. Reason (R) : Strong alkali like NaOH, KOH or Ca(OH)2 can not decompose salts of a weaker alkali NH 4 OH. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Assertion (A) :Ammonia gas is heavier than air. Reason (R) : Its vapour density is 8.5. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Assertion (A) :Concentrate H2SO 4 or CaCl2 or P 4O10 should not be used as drying agents for ammonia. Reason (R) : Ammonia gas is dried over quick lime. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. By catalytic reduction of these oxides of Nitrogen, Ammonia gas is produced. (A) NO or NO2 (B) NO2 or N2O5 (C) N2O3 or NO (D) N2O4 or N2O3

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16. Mg3N2 + 6H2O  3Mg (OH)2 + 2A AlN + 3H2O  B + NH3  Identify ‘A’ and ‘B’ in the above equations: A B (A) Nitrogen Aluminium Nitride (B) Ammonia Aluminium hydroxide (C) Ammonia Alumina (D) None 17. The catalyst used in the preparation of ammonia by Haber’s process is: (A) Finely divided iron. (B) Finely divided palladium. (C) Finely divided platinum. (D) None of the above 18. The optimum conditions favourable for the preparation of ammonia are: (A) 725 K – 775 K and 200 – 900 atmospheres. (B) 120K – 200K and 100 atmospheres (C) 900K – 1000K and 1000 atmospheres. (D) None of the above.

Conceptive Worksheet 10. Ammonia gas was prepared for the first time by: (A) Charles Bethelot (B) Joseph Black (C) Joseph priestley (D) Dalton 11. Any ammonium salt when heated with alkalis produces: (A) Nitrogen gas (B) Ammonia gas (C) Hydrogen gas (D) None 12. Ammonia gas is collected: (A) By downward displacement of air. (B) By upward displacement of air. (C) By downward displacement of water. (D) By upward displacement of water. 13. Which one of the following predicts the favourable conditions necessary for getting maximum yield of ammonia by Haber’s process? (A) Henry’s law (B) Le-Chatelier’s principle (C) Law of mass action (D) None 14. The favourable conditions for the formation of ammonia gas in Haber’s process is: (A) Low temperature, Low pressure (B) High temperature, High pressure (C) Low temperature, High pressure (D) High temperature , Low pressure

Compounds of Nitrogen

229

15. In Haber’s process: (A) One volume of Nitrogen reacts with two volumes of Hydrogen. (B) One volume of Nitrogen reacts with three volumes of Hydrogen. (C) Two volumes of Nitrogen react with two volumes of Hydrogen. (D) One volume of Nitrogen react with four volumes of Hydrogen. 16. In Haber’s process molybdenum acts as: (A) Catalyst (B) Promoter (C) Oxidising agent (D) None of the above

17. The formation of ammonia in Haber’s process is an _________ reaction. (A) Endothermic (B) Exothermic (C) Neither endothermic nor exothermic (D) None 18. Haber’s process is a: (A) Reversible reaction (B) Irreversible reaction (C) Redox reaction (D) None of the above

Cyanamide process The calcium cyanamide is formed by the action of nitrogen on calcium carbide at 800 – 10000C. It is then treated with superheated steam at 1800C under a pressure of 3 to 4 atmospheres. Then ammonia is produced.

CaCN2  3H2 O  CaCO3  2NH3  The flow sheet for industrial preparation of ammonia: CaCO3 Limestone

CaO Coke Quick lime

Electric furnace

CaC2

N2 CaCN2 80010000C

Steam 0 180 C

CaCO3++NH3

Nowadays, low temperature (500–6000C) is used under 6-8 atmospheric pressure.

12. Physical properties Besides the gaseous form ammonia is commonly stored in the following forms. Liquid ammonia: It is ammonia gas liquefied at high pressure and ordinary temperature. Liquor ammonia: It is a concentrated solution of ammonia gas in water. A highly saturated solution of ammonia in water is called liquor ammonia fortis. The term fortis means ‘strong’. Aqueous ammonia solution or ammonium hydroxide solution: It is a dilute solution of liquor ammonia in water. (i) Ammonia gas has a slight alkaline taste. (ii) Its pH is 11.77 at 18°C. (iii) Ammonia is a colourless gas with a characteristic pungent odour. It brings tears into the eyes. (iv) It is lighter than air. (v) It is highly soluble in water. One volume of water dissolves 1300 volumes of ammonia at 0°C and 1 atmosphere.

The high solubility is due to the hydrogen bonding. The solubility of ammonia increases with increase of pressure and decreases with increase of temperature. H H

– + H N

– O

H H (iv) It can be easily liquefied at room temperature by the application of pressure. The liquid ammonia is colourless and boils at –33°C. It freezes at –78°C. Liquid ammonia has a large heat of vapourisation (327 cal/g). It is, therefore, used in ice-plants. (v) Ammonia molecules link together to form associated molecules through hydrogen bonding.

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 NCl3 + 3HCl NH3 + 3Cl2  Iodine flakes when rubbed with liquor ammonia form H N H N H N, etc. a dark brown precipitate of ammoniated nitrogen H H H iodide which explodes readily on drying.  NH3.NI3 + 3HI Higher melting point and boiling point in comparison 2NH3 + 3I2  to other hydrides of V group are due to hydrogen Hypochlorites and hypobromites oxidise ammonia bonding. to nitrogen. 13. Chemical properties  N2 + 3NaCl + 3H2O 2NH3 + 3NaClO  (i) Stability : It is highly stable. It decomposes into The oxidation of ammonia with bleaching powder nitrogen and hydrogen at red hot or when electric occurs on warming. sparks are passed through it.  3CaCl2 + N2 + 3H2O 3CaOCl2 + 2NH3   2NH3  N2 + 3H2 Thus, ammonia acts as a reducing agent. The (ii) Combustion: Ordinary, ammonia is neither restricted oxidation of NH3 can be done with air, combustible nor a supporter of combustion. when the mixture is passed over heated platinum However, it burns in the presence of oxygen to form gauze at 700-800°C. nitrogen and water.  4NO + 6H2O 4NH3 + 5O2   2N2 + 6H2O 4NH3 + 3O2  This is known as Ostwald’s process and used for (iii) Basic nature : Ammonia is a Lewis base, accepting the manufacture of HNO3. proton to form ammonium ion as it has tendency to (v) Formation of amides : When dry ammonia is donate an electron pair. passed over heated sodium or potassium, amides are formed with evolution of hydrogen. + H H  2NaNH 2 + H2 2Na + 2NH3  + Sodamide H N +H H N H (vi) Reactions of aqueous ammonia : Many metal H H hydroxides are formed which may be precipitated It forms salts with acids. or remain dissolved in the form of complex compound in excess of NH4OH. NH C 4  NH3 + HCl  dense white fumes  Fe(OH)3 + 3NH4Cl FeCl3 + 3NH4OH  ppt. (Ammonium chloride)  (NH4)2 SO4 2NH3 + H2SO4  AlCl3 + 3NH4OH  A(OH)3 + 3NH4Cl ppt. (Ammonium sulphate) It’s solution is a weak base. The solution is described CrCl3 + 3NH4OH  Cr(OH)3 + 3NH4Cl ppt. as aqueous ammonia. It’s ionisation in water is represented as: CuSO 4 + 2NH 4 OH  Cu(OH) 2 + Blue ppt. + –   NH4OH  NH4 + OH NH3 + H2O  NH4 ) 2 SO 4 The solution turns red litmus to blue and Cu(OH) 2 + (NH 4 ) 2 SO 4 + 2NH 4 OH  phenolphthalein to pink. (iv) Oxidation : It is oxidised to nitrogen when passed Cu(NH3 ) 4 SO 4 + 4H2O over heated CuO or PbO. Tetra min e copper sulphate (Deep blue solution )   3CuO + 2NH3 3Cu + N2 + 3H2O CdSO4 + 4NH4OH  [Cd(NH 3 ) 4 ]SO 4 +  3Pb + N2 + 3H2O 3PbO + 2NH3  H

H

H

Both chlorine and bromine oxidise ammonia.  N2 + 6HCl 2NH3 + 3Cl2   6NH4Cl 6NH3 + 6HCl 

 N2 + 6NH4Cl 8NH3 + 3Cl2  ( Excess)

When chlorine is in excess an explosive substance nitrogen trichloride is formed. www.betoppers.com

Cadmium tetra min e sulphate (Colourless solution )

4H2 O AgNO3 + NH4OH  AgOH + NH4NO3 White ppt.

AgOH + 2NH 4 OH  [Ag(NH 3 ) 2 ]OH + Solub le

2H2 O

Compounds of Nitrogen

231

AgCl also dissolves in NH4OH solution. AgCl + 2NH 4 OH 

 Ag(NH3 )2  C

+

Dia min e silver chloride

2H2 O ZnSO 4 + 2NH 4 OH



Zn(OH) 2

+

ppt.

(NH4 )2 SO 4 Zn(OH)2 + (NH4)2SO4 + 2NH4OH

  Zn(NH 3 ) 4 SO 4 + 4H2O Tetra min e zinc sulphate (So luble) Colourless

Nickel salt first gives a green precipitate which dissolves in excess of NH4OH. NiCl2 + 2NH4OH  Ni(OH)2 + 2NH4Cl Ni(OH)2 + 2NH4Cl + 4NH4OH  [Ni(NH3)6]Cl2 + 6H2O It forms a white precipitate with mercuric chloride. HgCl2 + 2NH4OH  HgNH 2 C + NH4Cl + 2H2O Amido mercuric chloride

It forms a grey precipitate with mercurous chloride. Hg 2 Cl 2 + 2NH 4 OH  2Hg  NH 2 C +   Grey

NH4Cl + 2H2O (vii) Reaction with Nessler’s reagent: A reddish brown ppt. is formed. 2KI + HgCl2  HgI2 + 2KCl 2KI + HgI2  K2HgI4 Alkaline solution of K2 HgI4 is called Nessler’s reagent. This gives brown precipitate with NH3 called iodide of Millon’s base. 2K2HgI4 + NH3 + 3KOH



H NHgOHgI 2 + 7KI + 2H2O Brown ppt.

Formative Worksheet 19. Assertion (A) :Ammonia is a Lewis base. Reason (R) : Lone pair is present on nitrogen and can donate this pair of electrons for bonding. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

20. Statement (A) : Ammonia gas has a slight alkaline taste. Statement (B) : Its pH is 11.77 at 18°C. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. 21. Match the following: (L - II) Column - I Column - II p) Liquid ammonia i) The dilute solution of ammonia in water. q) Liquor ammonia ii) Ammonia gas liquefied at high pressure and ordinary temperature. r) Aqueous ammonia iii) Concentrated solution of ammonia gas in water. Solution or ammonium hydroxide solution p q r (A) ii iii i (B) i ii iii (C) iii ii i (D) iii i ii 22. The boiling point and freezing point of liquid ammonia are respectively: (A) – 42.1°C, – 93.4°C (B) – 33.4°C, – 77.7°C (C) – 28.9°C, – 69.1°C (D) – 13.2°C, – 54.3°C 23. A glass rod dipped in concentrated hydrochloric acid is brought near the mouth of a gas jar containing ammonia. Which of the following statement(s) is/ are correct? A) White dense fumes of NH4Cl is observed. B) It is an example of neutralisation reaction. C) Here NH4Cl acts as a base. (A) ‘A’ and ‘B’ are correct (B) ‘B’ and ‘C’ are correct (C) ‘C’ and ‘A’ are correct (D) All are correct 24. Which one of the following statement(s) is/are true about ammonia? A) It is a colourless gas and has a characteristic pungent odour. B) It is a polar covalent compound. C) Liquid ammonia is neutral to litmus. (A) only ‘A’ is true (B) only ‘B’ is true (C) only ‘C’ is true (D) All are correct

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9th Class Chemistry

232 25. Assertion (A) :Ammonia is able to co-ordinate with many cations and form complexes. Reason (R) : Nitrogen consists no lone pair of electrons. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 26. When ‘A’ reacts with excess ‘B’ and gives ‘C’ which is highly explosive. When excess ‘A’ combines with ‘B’ it gives a compound ‘D’ which have white dense fumes. ‘B’ is a greenish-yellow coloured poisonous gas. Identify ‘A’, ‘B’, ‘C’ and ‘D’. A B C D (A) N 2 HCl NCl3 H2 (B) NH 3 HCl N2 NH4Cl (C) NH 3 Cl2 NCl3 NH4Cl (D) None of the above

Conceptive Worksheet 19. Alkaline solution of ammonium hydroxide turns phenolphthalein ______. (A) Orange (B) Colourless (C) Pink (D) None 20. When dry ammonia gas is passed over heated sodium, it forms ____. (A) Excess ammonia (B) Sodamide (C) Both ‘a’ and ‘b’ (D) None 21. Ammonia acts as a _________ (A) Strong oxidising agent (B) Catalyst (C) Strong reducing agent (D) None 22.

2NH3 + H 2 SO4  A

NH3 + HNO3  B Identify ‘A’ and ‘B’. A B (A) Ammonium hydroxide Nitrogen dioxide (B) Ammonium sulphate Ammonium nitrate (C) Ammonium sulphate Nitrogen pentoxide (D) Sulphurous acid Ammonium nitrate 23. When ammonia is heated with a paste of bleaching powder it yields _________ gas. (A) Hydrogen (B) Chlorine (C) Nitrogen (D) Oxygen

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24. The reactants that are used in cyanamide process is (in presence of steam). (A) Calcium carbide + Nitrogen (B) Calcium oxide + Nitrogen (C) Calcium carbonate + Hydrogen (D) None of the above 25. The mixture of cyanamide and graphite is called as: (A) Ammonal (B) Ignition mixture (C) Thermite mixture (D) Nitrolim

14. Uses (i) (ii)

(iii)

(iv) (v)

(vi) (vii)

Liquid ammonia is used in refrigeration on account of its large heat of evaporation. Ammonia is used in the form of ammonium hydroxide (aq. solution) in the laboratory in qualitative and quantitative analysis. Ammonia is used for the manufacture of nitric acid (Ostwald’s process), sodium bicarbonate (Solvay process) and ammonium compounds. Ammonium sulphate, ammonium calcium phosphate, ammonium calcium nitrate, etc., are used as fertilisers. Ammonium nitrate is used in certain explosives. Ammonia is used for making artificial silk. Liquid hydrogen is not safe to transport in cylinders. Ammonia can be easily liquefied and transported safely in cylinders. Ammonia can be decomposed into hydrogen and nitrogen by passing over heated metallic catalysts. Thus, ammonia is the source for the production of hydrogen at any destination. It is used as a cleansing agent for removing grease. Ammonia is also used in the manufacture of urea which is an excellent fertilizer of nitrogen.

15. Tests (i) It is identified by its characteristic odour. (ii) It turns moist red litmus paper blue and moist turmeric paper brown. (iii) With a drop of concentrate HCl, it forms thick white fumes of ammonium chloride. (iv) With Nessler’s reagent, it forms a reddish brown precipitate or colouration. 2KI + HgCl2  HgI2 + 2KCl 2KI + HgI2  K2HgI4 Alkaline solution of K2 HgI4 is called Nessler’s reagent. This gives brown precipitate with NH3 called iodide of Millon’s base. 2K2HgI4 + NH3 + 3KOH  H 2 NHgOHgI Brown ppt.

+ 7KI + 2H2O (v) Ammonia gas when passed through copper sulphate solution gives a deep blue colour.

Compounds of Nitrogen

233

(vi) It gives a yellow precipitate with chloroplatinic acid. 2NH3 + H2PtCl6 

(NH 4 ) 2 PtC 6 Ammonium chloroplatinate (Yellow )

Formative Worksheet 27. Assertion (A) :Liquid ammonia is largely used in refrigeration. Reason (R) : Liquid ammonia has high latent heat of vaporisation. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 28. ____________ form of ammonia is used in dry cells. (A) Ammonium chloride (B) Ammonium nitrate (C) Ammonium hydroxide (D) None of the above 29. Ammonia gas can be detected by: (A) Benedict’s reagent (B) Tollen’s reagent (C) Fehling’s reagent (D) Nessler’s reagent 30. The formula of Nessler’s reagent is: (A) K3 HgI4 (B) K2HgI2 (C) K2HgI4 (D) KHgI4

Conceptive Worksheet 26. Ammonia is used: (A) In fertilisers. (B) In manufacture of nitric acid. (C) In household detergents. (D) All the above. 27. In laboratory ammonia as ammonium hydroxide is used as (A) Good solvent (B) Oxidising agent (C) Reducing agent (D) None of the above 28. The compound of ammonia used in the revival of unconscious persons is: (A) Ammonium chloride (B) Ammonium hydroxide (C) Ammonium carbonate (D) Ammonium nitrate 29. 2K 2 HgI 4 + NH3 + 3KOH  The  A + 7KI + 2H 2 O formula of ‘A’ and name of the compound in the above reaction are respectively: (A) H2NHgI, Millon’s base (B) H2NHgOHgI, Iodide of millon’s base (C) H2NHgI, Iodide of millon’s base (D) None of the above

NITRIC ACID 16. Introduction Formula of nitric acid : HNO 3 Relative molecular mass : 63 Vapour density : 31.5 Nitric acid is an useful laboratory reagent and an important chemical which is widely used in many industries. The alchemists called it ‘aqua fortis’ (the meaning of the word is ‘strong water’), since in those days, it was the only liquid which could dissolve silver. Nitric acid was first obtained by Glauber in 1648, by distilling nitre (KNO3) with sulphuric acid. Antoine Lavoisier in 1 776, proved that nitric acid contained the element, oxygen. In 1784, Henry Covendish proved that nitric acid also contained the elements hydrogen and nitrogen, in addition to oxygen.

17. Occurrence In the atmosphere, nitric acid is formed in the free state when, just after lightning, rain follows. The series of reactions that lead to the formation of free nitric acid in atmosphere is shown below: lightning N 2  O 2   2NO

 from air 

 nitric oxide 

2NO  O2   2NO 2 4NO2  2H 2 O  O 2   4HNO3 In the combined state, it occurs in the soil as nitrates, the most important of which are : • Nitre or potassium nitrate, KNO3 • Chile salt petre or sodium nitrate, NaNO3.

18. Laboratory preparation Nitric acid is prepared by distilling potassium or sodium nitrate with concentrated sulphuric acid, at 200°C. Procedure : Equal parts by weight, of potassium or sodium nitrate and concentrated sulphuric acid are taken in a stoppered glass retort see in figure below. The neck of the glass retort is introduced into a round-bottomed flask which acts as the receiver. This flask is kept cool by running tap water as shown in the figure. The mixture in the glass retort is heated gently to a temperature of 200°C. Sulphuric acid being a non-volatile acid, produces volatile nitric acid on reacting with potassium or sodium nitrate, as shown below 200C  KHSO  HNO KNO S  H SO  3 2 4 4 3  conc.

 potassium     bisulphate   

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9th Class Chemistry

234

200C  NaHSO  HNO NaNO3  H 2SO4  4 3  conc.

co nc.s ulp h uric a cid + n itre or c hile salt p etre

• •

•

 sodium     bisulphate   

preparation of nitirc acid, because hydrochloric acid is more volatile than nitric acid. Hydrochloric acid, if used to decompose a nitrate, will distilfj. alongwith nitric acid vapour and get collected in the

F

i ecerormative

Worksheet

31. Match the following: (related to Nitric acid) re tort co ld wa te r Scientist Discovery nitric acid p) Glauber i) Proved that nitric acid vap ours contained the element oxygen. q) Antoine Lavoisier ii) Proved that nitric acid n itric a cid also contained element hydrogen and nitrogen. La bor atory p rep aration of n itric ac id s) Henry Cavendish iii) First obtained nitric acid. The nitric acid fomed distills and gets collected as a p q r light yellow liquid in the receiver. Pure nitirc acid is (A) i ii iii a colourless liquid. It has a strength of 98%. The (B) iii i ii yellow colour, in this case, is caused by the nitrogen (C) iii ii i dioxide, a reddish brown gas present in the acid. (D) i iii ii This gas is produced due to thermal decomposition 32. Nitric acid for the first time was prepared by reaction of a portion of nitric acid, which then dissolves in of sulphuric acid with ______ the acid. (A) AgNO3 (B) Ca(NO3)2 4HNO3  2H2O + 4NO2 + O2 (C) KNO3 (D) None of the above When air is bubbled through the yellow nitric acid, 33. The light yellow colour of Nitric acid is due to the the latter turns colourless, because the nitrogen presence of: dioxide present in the acid gets oxidised to nitric (A) Nitric oxide (B) Oxygen acid. This reaction is the reverse of the (C) Nitrogen dioxide(D) None of the above decomposition of nitirc acid. 34. Statement (A) :At high temperature Nitric acid may Note: The following precautionary measures are decompose. taken during the preparation of nitric acid in the Statement (B) :At high temperatures KHSO4 react laboratory. with sodium nitre to liberate Nitric acid vapours. An all-glass retort is used because nitric acid (A) ‘A’ is true, ‘B’ is false. vapours attack rubber and cork. (B) ‘A’ is false, ‘B’ is true. Care is taken to keep the temperature of the reaction (C) Both ‘A’ and ‘B’ are true. mixture at or below 200°C, because nitirc acid (D) Both ‘A’ and ‘B’ are false. formed may decompose. 35. Assertion (A) :Concentrate Hydrochloric acid At higher temperatures (i.e. greater than 200°C), cannot be used in place of sulphuric acid in the potassium or sodium bisulphate react with potassium preparation of Nitric acid. or sodium nitrate to liberate nitric acid vapours. Reason (R) : Concentrate Hydrochloric acid is However, the potassium or sodium sulphate formed, more volatile than Nitric acid. stick to the glass and cannot be removed easily. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. above KNO3 + KHSO4  200°C K SO + HNO (B) Both assertion and reason are correct but 3  4 4 reason is not the correct explanation of (potassium assertion. sulphate) (C) Assertion is correct and reason is incorrect. above (D) Assertion is incorrect and reason is correct. 200°C  Na SO + HNO NaNO3 + NaHSO4  2 4 3 (sodium sulphate) Concentrated hydrochloric acid cannot be used in place of concentrated sulphuric acid in the

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Compounds of Nitrogen 36. Identify ‘A’ and ‘B’ in the following reactions:

KNO3 + KHSO4   A + HNO3 200°C KNO3 + H 2 SO 4   KHSO4 + B A B (A) Potassium bisulphate Nitrous acid (B) Potassium sulphate Nitric acid (C) Potassium nitrite Nitrogen dioxide (D) Potassium sulphate Nitrous acid 37. The nature of Nitric acid produced during its preparation by laboratory process is: (A) Volatile (B) Non-volatile (C) Solid (D) Liquid

Conceptive Worksheet 30. The name ‘aqua fortis’ was given to ________ (A) Sulphuric acid (B) Nitric acid (C) Hydrochloric acid (D) Perchloric acid 31. In the atmosphere Nitric acid is formed: (A) In free state (B) In combined state (C) Both ‘a’ and ‘b’ (D) None 32. In soil Nitric acid occurs in combined state as: (A) Carbonates (B) Nitrates (C) Nitrites (D) Silicates 33. The impurity formed during the preparation of Nitric acid is: (A) Nitrous acid (B) Nitrogen trioxide (C) Nitric oxide (D) Nitrogen dioxide 34. The acid that formed during the rainfall after lightening is: (A) Nitrous acid (B) Nitric acid (C) Hydrochloric acid (D) Perchloric acid 35. In laboratory preparation of Nitric acid, Sodium nitrate and sulphuric acid reacts: (A) Equal parts by weight. (B) More weight of sodium nitrate is taken. (C) More weight of sulphuric acid is taken. (D) None of the above. 36. Assertion (A) :The yellow colour of Nitric acid turns colourless. Reason (R) : When air is bubbled the nitrogen dioxide present in the acid gets oxidised to Nitric acid (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

235

19. Manufacture of Nitric Acid Oswald’s Process Nitric acid is manufactured by Ostwald’s process. Catalytic oxidation of ammonia is the main reaction in this process. Priniciple : Ammonia gas is oxidised by the oxygen present in air, in presence of platinum gauze as catalyst, at 800°C, to form nitric oxide. (Nowadays, a catalyst consisting of 90% platinum and 10% rhodium is being used in the manufacture.) 800  C 4NH 3  5O 2   4NO  6H 2 O pt

Nitric oxide formed is then cooled and reacted with more of oxygen to form nitrogen dioxide. 50 C  2NO2 2NO + O2  Nitrogen dioxide is converted to nitric acid by mixing it with water and air.

4NO2  2H 2 O  O2   4HNO3 Water and air  

Procedure : The different stages in the manufacture of ammonia are described below: Oxidation of ammonia in the catalytic chamber : Air freed from carbon dioxide and dust particles is dried alongwith ammonia gas. They are then mixed in the ratio of 10 parts by volume of air to 1 part by volume of ammonia. The mixture is passed into a chamber containing platinum gauze which acts as the catalyst, at about SOOT, Ammonia is oxidised to form nitric oxide. 4NH3 + 5O2  4NO + 6H2O This reaction is exothermic and the heat that is produced maintains the temperature of the catalyst chamber. Therefore only initial heating is required. Oxidation of nitric oxide in the oxidation chamber : The hot gases are then passed through a heat exchanger where they are

cooled. Then they pass through a cooling tower where the temperature of the gases is further lowered. www.betoppers.com

9th Class Chemistry

236 Air is blown into the mixture of gases which is again cooled in a second cooling tower known as the oxidising tower. The temperature of the mixture of gases falls below 50°C. In the oxidising tower, nitric oxide gets oxidised to nitrogen dioxide.

acid formed here is collected at the base of this tower. 4NO2 + O2 + 2H2O  4HNO3 Concentration of the acid : The nitric acid so obtained is distilled to get commercial nitric acid which is 68% HNC3. The mixture is further distilled with concentrated H 2 SO 4 to yield 98% HNO 3 , under reduced pressure. [Concentrated H 2 SO 4 absorbs water.]

2NO + O2  2NO2 Absorption of nitrogen dioxide : The excess of air and nitrogen dioxide pass into an absorption tower in which water trickles down from the top. Nitric

NH 3  air  



1:10 ratio by volume

Catalytic chamber Pt



NO   O2

Cooling tower

NO  

800C

O2

Oxidation chamber

NO2    O2

Absorption tower ( with oxygen 

  HNO3

water )

Birkeland - Eyde process In this process air is blown into an electric arc when the nitrogen and oxygen of the air chemically react forming nitric oxide. This gas is cooled when it reacts with the oxygen present producing nitrogen dioxide. It is then absorbed in water in presence of excess of air when nitric acid is formed.   N 2  O 2   2NO ; H  90.4 kj/ mol Electric arc

2NO  O 2   2NO 2 4NO2  2H 2 O  O 2   4HNO3

Nitric acid is a very strong acid and a powerful oxidising agent. Salts of nitric acid are called nitrates.

Formative Worksheet 38. The catalyst used during the preparation of Nitric acid by ostwald’s process is: (A) Raney Nickel (B) Platinum guaze + Rhodium (C) Palladium (D) None of the above 39. Identify ‘A’, ‘B’ and ‘C’ in the given equations.

800°C  A + 6H O 4NH3 + 5O 2  2 Catalyst

50ºC B 2NO + O  2 NO + H O + O  C 2 2 2 www.betoppers.com

A B (A) Nitrogen dioxideNitric acid (B) Nitric acid Nitric oxide

C Nitric oxide Nitrogen dioxide (C) Nitrous acid Nitrogen dioxide Nitric oxide (D) Nitric oxide Nitrogen dioxide Nitric acid 40. Statement (A) : Pure Nitric acid is a colourless liquid with choking odour. Statement (B) : Nitric acid is hygroscopic in nature. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. 41. Assertion (A) : In Ostwald’s process, in the first step, yield increases by decreasing temperature. Reason (R) : It is an exothermic reaction. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 42. The yellow colour often shown by Nitric acid can be removed by: (A) Bubbling air through the warm acid. (B) Boiling the acid. (C) Passing ammonia through acid. (D) Adding a little Mg powder.

Compounds of Nitrogen 43. Industrial preparation of Nitric acid by Ostwald’s process involves: (A) Oxidation of NH3 (B) Reduction of NH3 (C) Hydrogenation of NH3 (D) Hydrolysis of NH3

Conceptive Worksheet 37. The starting material used for the preparation of Nitric acid in ostwald’s process is: (A) Nitrous oxide (B) Ammonia gas (C) Nitrogen dioxide (D) None 38. The oxidation of ammonia in the catalytic chamber is a ________ process. (A) Endothermic (B) Neutralisation (C) Reduction (D) Exothermic 39. In Ostwald’s process ammonia and air reacts in _______ ratio by volume. (A) 1 : 5 (B) 1 : 7 (C) 1 : 9 (D) 1 : 10 40. The starting materials used in the preparation of Nitric acid in Birkeland - Eyde process is: (A) Nitrogen and Hydrogen (B) Nitrogen and Oxygen (C) Oxygen and Hydrogen (D) None 41. The specific gravity of pure Nitric acid is: (A) 1.54 (B) 1.36 (C) 1.28 (D) 1.13 42. Assertion (A) : The action of Nitric acid on skin turns skin to stained yellow. Reason (R) : Xanthoproteic acid is formed. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. 43. Commercial nitric acid is brown in colour due to dissolved ________ (A) Nitric oxide (B) Nitrous oxide (C) Nitrogen trioxide (D) Nitrogen dioxide 44. A solution of 68% HNO3 forms a constant boiling mixture at _______ (A) 98°C (B) 100°C (C) 110°C (D) 121°C

237 45. The reaction between nitrogen and oxygen in Birkeland - Eyde process is (A) Reversible, endothermic (B) Reversible, exothermic (C) Reversible, reduction (D) None 46. The boiling point and freezing point of nitric acid are respectively: (A) 84.1°C, – 42°C (B) 93.2°C, – 38°C (C) 76.4°C, – 61°C (D) 86.9°C, – 36°C

20. Physical properties 1.

2.

3. 4.

5.

6.

Pure nitric acid is a colourless liquid with choking odour. It has a sour taste. It is hygroscopic and fumes in moist air to form a mist of small droplets of aqueous nitric acid. Commercial nitric acid is brown in colour, due to dissolved nitrogen dioxide (NO2). Since commercial nitric acid has lesser content of nitric acid, its specific gravity is 1.42, while that of pure nitric acid is 1.54. It is soluble in water. A solution of 68% HNOs (commercial HNO 3 ) forms a constant boiling mixture at 121°C. Therefore further concentration is npt possible by distillation alone. Concentrated nitric acid has corrosive action on the skin. Due to the action of nitric acid on the protein of the skin, the skin gets stained yellow. The yellow stain is due to formation of xanthoproteic acid. Pure nitric acid boils at 86°C and freezes to a white solid at –42°C.

21. Chemical properties 1.

Acidic nature : Nitric acid is a very strong acid. It ionises almost completely in a dilute solution, to produce hydrogen and nitrate ions. HNO3 (aq)

a.

  H+ + NO3

It is a monobasic acid. The acidic properties of nitric acid are shown in the following reactions: Reactions with bases : It neutralises basic oxides to form salt and water. CuO + 2HNO3  Cu(NO3)2 + H2O copper (II) copper (II) oxide (block) nitrate (blue green solution) ZnO + 2HNO3  Zn(NO3)2 + H2O (zinc oxide) www.betoppers.com

9th Class Chemistry

238 b.

c.

Reactions with metal hydroxides or alkalis : It also neutralises metal hydroxides to form salt and water.

[2HNO3  2NO2 + H2O + O] × 2

NaOH + HNO3  NaNO3 + H2O

C + 2[O]  CO2

Mg(OH)2 + 2HNO3  Mg(NO3)2 + 2H2O Reactions with carbonates and bicarbonates : With carbonates and bicarbonates of metals, it liberates carbon dioxide and forms the corresponding metal nitrate and water.

C  4HNO3   2H 2 O  4NO 2  CO 2

ii.

CaCO 3 + 2HNO 3  Ca(NO 3 ) 2 + H 2 O + CO2 

[2HNO3  2NO2 + H2O + O] × 3

(calciumcarbonate) Ca(HCO3)2 + 2HNO3 2H2O + 2CO2  2.

3.

S + 3[O] + H2O  H2SO4 

Ca(NO 3 ) 2 +

(calcium bicarbonate) Stability : Pure nitric acid decomposes in presence of sunlight even at ordinary temperature, into water, nitrogen dioxide and oxygen. 4HNO3  2H2O + 4NO2 + O2 This reaction proves that nitric acid contains oxygen. Oxidising nature : Nitric acid is a strong oxidising agent. On thermal decemposition, the concentrated acid yields nascent oxygen along with nitrogen dioxide and water. 2HNO3  H2O + 2NO2 + [O] (concentrated) Dilute nitric acid produces nascent oxygen, along with nitric oxide and water.

a. i.

With sulphur : Powdered sulphur reacts with hot and concentrated nitric acid to produce reddishbrown fumes of nitrogen dioxide. Sulphur is oxidised to sulphur trioxide by the gain of oxygen which reacts with water to produce sulphuric acid.

2HNO3  H2O + 2NO + 3[O] (dilute) (Note : Fuming nitric acid is an even more powerful oxidising agent). It is because of this reason that nitric acid oxidises compounds like, non-metals, metals, inorganic and organic compounds as shown in the following reactions : Action on non-metals With carbon : Take some powdered charcoal in a test tube and add concentrated nitric acid. When heated, reddish-brown fumes of nitrogen dioxide are produced. Carbon is oxidised to carbon dioxide by gain of oxygen and the nitric acid is reduced to nitrogen dioxide by loss of oxygen.

S  6HNO3   H 2SO4  2H 2 O  6NO 2

iii. With phosphorus : Phosphorus is oxidised to phosphoric acid (HsPOi), and nitric acid is reduced to nitrogen dioxide by loss of oxygen. [2HNO3  2NO2 + H2O + 0] × 10 P4 + 10[O] + 6H2O  4H3PO4 P4  20HNO3   4H 3 PO 4  20NO 2  10H 2 O

b.

Action on metals : By the action of nitric acid on metals, concentrated acid is reduced to nitrogen dioxide, and dilute nitric acid is reduced to nitric oxide. The metal is oxidised to metal nitrate. The reactions of nitric acid with metals depend upon the concentration of the acid, temperature and the nature of the metal. Nitric acid actually plays a double role when it reacts with a metal. It reacts as an acid as well as an oxidising agent. Reaction of copper with dilute nitric acid to produce nitric oxide, can be explained by partial equations, shown below : 2HNO3  2NO + H2O [Cu + O  CuO] × 3 [CuO + 2HNO3  Cu(NO3)2 + H2O] × 3

3Cu  8HNO3   3Cu  NO3 2  2NO  4H 2 O Similarly reaction of copper with concentrated nitirc acid to produce nitrogen dioxide can also be explained by partial equations, shown below : 2HNO3  2NO2 + H2O + O

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Compounds of Nitrogen

239

Cu + O  CuO 2HNO3  Cu(NO3)2 + H2O

v.

Cu + 4HNO3   Cu(NO3 ) 2 + 2H 2 O + 2NO 2

i.

The reactions of other metals with nitric acid also depends on the concentration of the acid. Very dilute acid (1% HNO3) : Very dilute nitric acid (about 1%) reacts with magnesium and manganese, at room temperature to form hydrogen. Mg + 2HNO3  magnesium Mn + 2HNO3  manganese

ii.

3HC  HNO3   NOC  2 H 2 O  2 C   introsyl chloride 

aurric  chloride 

Pt  4 C   PtC 4

 platinum iv    chloride   

Mn(NO3)2 + H2

c. manganese nitrate The above reactions also prove that nitric acid contains hydrogen. i. Dilute nitric acid (50%) : A dilute solution of nitric acid reacts with metals to form nitric oxide, which is a colourless gas. When this gas comes in contact with air, it forms reddish-brown fumes of nitrogen dioxide (NO2). The nitrates of the metals and water are also formed.

ii.

Zn + 4HNO3  Zn(NO3)2 + 2H2O + 2NO2

iv.

Fe + 4HNO3  Fe(NO3)2 + 2H2O + 2NO2 With highly concentrated nitric acid (80%) or fuming nitric acid, iron is rendered passive as a film of an insoluble oxide Fe3O4, (triferric tetra oxide) is formed on its surface. The oxide film can be removed by rubbing the surface with sand paper. i. Vapours of nitric acid : Vapours of nitric acid react with heated copper to produce black copper (II) oxide, gaseous nitrogen and water. 5Cu + 2HNO3  5CuO + N2 + H2O

2HNO3  2NO2 + H20 + [O]

H 2S  2HNO3   2NO2  2H 2 O  S

Cu + 4HNO3  Cu(NO3)2 + 2H2O + 2NO2

iv.

Action on inorganic compounds : A number of inorganic compounds are oxidised by concentrated as well as dilute nitric acid. Concentrated nitric acid oxidises : Hydrogen sulphide to sulphur : When hydrogen sulphide is bubbled through concentrated nitric acid, brown fumes of nitrogen dioxide and a yellow solution, due to formation of sulphur, are obtained.

H2S + [O]  H2O + S

3Cu + 8HNO3  3Cu(NO3)2 + 4H2O + 2NO

3Fe + 8HNO3  3Fe(NO3)2 + 4H2O + 2NO iii. Concentrated nitric acid (hot) : Hot and concentrated nitric acid reacts with metals to form nitrates and reddish brown fumes of nitrogen dioxide.

nascent chlorine

Au  3 C  AuC 3

Mg(NO3)2 + H2 magnesium nitrate

3Zn + 8HNO3  3Zn(NO3)2 + 4H2O + 2NO

This reaction also proves that nitric acid contains nitrogen. As aqua regia : A mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio 3:1 by volume is known as aqua regia. It can dissolve noble metals like gold, platinum etc.

Potassium iodide to iodine. 2HNO3  2NO2 + H2O + O 2Kl + H2O + [O]  2KOH + I2 [KOH + HNO3  KNO3 + H2O] × 2 2K + 4HNO3   2KNO3 + 2NO2 + 2H 2 O +  2

A filter paper dipped in potassium iodide solution turns brown when it is brought close to the mouth of a test tube containing hot nitric acid. The brown colour is due to the deposition of iodine. Dilute nitirc acid too oxidises, inorganic compound, as shown below : Hydrogen sulphide to sulphur. 2HNO3  2NO + H2O + 3[O] [H2S + [O]  H2O + S] × 3 3H 2S  2HNO3   2NO  4H 2 O  3S

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9th Class Chemistry

240 ii.

Potassium iodide to iodine.

22. Tests

2HNO3  2NO + H2O + 3[O]

1.

[2KI + H2O + [O]  2KOH + I2] × 3 [KOH + HNO3  KNO3 + H2O] × 6 6K  8HNO3   6KNO3  2NO  3 2  3H 2 O

2.

iii. Sulphur dioxide to sulphuric acid. 2HNO3  2NO + H2O +3[O] [SO2 + H2O + [O]  H2SO4] × 3

3.

3SO 2 + 2HNO3 + 2H 2 O   2NO + 3H 2SO 7

iv.

Acidified ferrous sulphate to ferric sulphate : In the presence of concentrated sulphuric acid, ferrous sulphate is oxidised to ferric sulphate by dilute nitric acid. Nitric acid is reduced to nitric oxide, which forms a brown ring with ferrous sulphate, when taken in excess. This reaction is used in the ring test to test for nitrates (discussed later). 2HNO3  2NO + H2O + 3[O] [2FeSO4+H2 SO4+[O]  Fe2(SO4 )3+H2O] × 3 Action on organic compounds : Concentrated nitric acid oxidises organic compounds to form carbon dioxide and water.

•

When hot concentrated nitric acid is poured on saw dust, saw dust catches fire. This is due to oxidation of the compounds in saw dust to form carbon dioxide and water. Nitrogen dioxide is also produced due to reduction of nitric acid.

•

Turpentine oil starts burning when poured into fuming nitric acid.

•

Nitric acid attacks proteins to form a yellow substance called xanthoprotein. This is the reason it leaves a yellow stain on the skin and makes wood yellow.

•

Cane sugar, C12H22O11 gets oxidised to oxalic acid. [(COOH)2], by concentrated nitric acid.

4.

1.

C12H22O11 + 18O  6(COOH)2 + 5H2O

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Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O Ring Test : To dilute nitric acid taken in a test tube, freshly prepared ferrous sulphate solution is added. Concentrated H 2 SO 4 is then poured carefully down the sides of the test tube. A dark brown ring is formed at the junction of two layers. It is due to the formation of an additive compound by ferrous sulphate, nitric oxide and water molecules. The compound is called hydrated nitrosoferrous sulphate. The brown ring at the junction of two layers is a sure test for nitric acid or nitrates. 6FeSO4 + 3H2SO4 + 2HNO3

 [Fe(NO) (H2O)5]SO4 hydrated nitrosoferrous sulphate.(brown ring) A yellow stain is formed when a wooden splinter is dipped in concentrated nitric acid.

23. Uses

[2HNO3  H2O + 4NO2 + 90)] × 18

C12 H 22O11 + 36HNO 3   6(COOH) 2 + 36NO 2 + 23H 2 O

4HNO3  2H2O + 4NO2 + O2 (brown fumes) When concentrated nitric acid is heated with copper turnings, brown fumes of nitrogen dioxide evolve profusely.

 3Fe2(SO4)3 + 2NO + 4H2O FeSO4 + NO + 5H2O

6FeSO 4 +3H 2SO 4 + 2HNO3   3Fe 2 (SO4 )3 + 2NO + 4H 2O

d.

When nitric acid is heated, brown fumes of nitrogen dioxide are evolved.

2.

Nitric acid is a chemical which is used extensively in many industries It is used in the large scale preparation of : a. Fertilisers such as aprtnonium nitrate, potassium nitrate, calcium ammonium nitrate (CAN). b. Explosives such as TNT (tri-nitro toluene), nitro-glycerine, and picric acid. c. Nitrobenzene which is an important raw material for the preparation of dyes, perfumes, drugs etc. d. Nitrates of cellulose, which are used in the manufacture of explosives, paints, lacquers and in medicine. As a laboratory reagent, it used as dilute nitric, concentrated nitric acid and aqua regia.

Compounds of Nitrogen 3.

It is used : a. In the purification of gold and silver. b. For etching or writing on copper, brass, stainless steel, bronze ware etc. c. As an oxidiser in rocket fuels.

Formative Worksheet

241 48. Identify ‘A’, ‘B’ and ‘C’ in the following reactions:

3Hg 2 O + 6HNO3   A + 3H 2 O Hg + 4HNO3   B + 2NO2 + 2H 2 O 4Zn + 10HNO3   4C + NH 4 NO3 + 3H 2 O  V.dilute 

A B C A) Mercuric nitrate; mercurous nitrate; zinc nitrate Reason (R) : On thermal decomposition, the B) Mercurous nitrate; mercuric nitrate; zinc concentrated acid yields nascent oxygen along with nitrate nitrogen dioxide and water. C) Mercurous oxide; mercurous nitrate; zinc (A) Both assertion and reason are correct and nitrate reason is the correct explanation of assertion. D) None of the above (B) Both assertion and reason are correct but 49. Ammonal, which is a powerful explosive made up reason is not the correct explanation of with Nitric acid is a mixture of: assertion. (A) Ammonium nitrate + Aluminium powder (C) Assertion is correct and reason is incorrect. (B) Alumina + Ammonium hydroxide (C) Ammonium chloride + Alumina (D) Assertion is incorrect and reason is correct. (D) None 45. Concentrated Nitric acid reacts with Iodine to give: 50. Assertion (A) :HNO 3 is a stronger acid than (A) HI (B) HIO2 (C)HIO3 (D) HIO4 HNO 2 . 46. Which of the following statement is correct about Reason (R) : In HNO3 there are two nitrogen to Nitric acid? oxygen bonds whereas in HNO2 there is only one. A) Noble metals like gold and platinum are attacked by Nitric acid. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. B) Nitric acid plays a double role in the action (B) Both assertion and reason are correct but of metals i.e., it acts as an acid as well as an reason is not the correct explanation of oxidising agent. assertion. (A) Only ‘A’ is correct (C) Assertion is correct and reason is incorrect. (B) Only ‘B’ is correct (D) Assertion is incorrect and reason is correct. (C) Both ‘A’ and ‘B’ are correct 44. Assertion (A) :Nitric acid acts as a strong oxidising agent.

(D) None of the above 47. Assertion (A) :Noble metals like gold, platinum, iridium etc are dissolved in aqua regia. Reason (R) : Aqua-regia forms nascent chlorine. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

Conceptive Worksheet

47. Nitric acid is a: (A) Strong acid (B) Weak acid (C) Dibasic acid (D) None 48. Statement (A) : Nitric acid ionises almost completely in a dilute solution. Statement (B) : Nitric acid produces hydrogen and Nitrate ions in its dilute solution (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. www.betoppers.com

9th Class Chemistry

242 49. The ratio of Nitric acid to Hydrochloric acid present in aqua regia is: (A) 1 : 3 (B) 3 : 1 (C) 1 : 4 (D) 4 : 1 50. Which of the following statement(s) is/are correct for Nitric acid? A) It is used as a explosive. B) It is used as a laboratory reagent. C) It is not used in the manufacture of fertilisers. (A) ‘A’ and ‘B’ are correct (B) ‘B’ and ‘C’ are correct (C) ‘C’ and ‘A’ are correct (D) None of the above 51. The metal which becomes passive in contact with concentrate HNO3 is: (A) Copper (B) Aluminium (C) Zinc (D) Silver 52. When Nitric acid is heated, the gas that evolved is: (A) Nitric oxide (B) Nitrogen dioxide (C) Nitrous oxide (D) None

11. 12. 13. 14. 15. 16.

17.

18. 19.

Summative Worksheet 1.

2. 3. 4.

5. 6. 7. 8. 9. 10.

When nitric acid is prepared by the action of conc. H2SO4 on KNO3, what is the special feature of the apparatus used? Write the equation for the laboratory preparation of HNO3 from KNO3 and conc. H2SO4. Which gas is produced when KNO3 is heated? Write the equation for the reaction. Write the equation for (i) The preparation of ammonia from ammonium chloride and calcium hydroxide. (ii) The reaction of hydrogen chloride with ammonia. (iii) What are the products formed when ammonia is oxidised with copper oxide. Why are anhydrous CaCl2, P2O5 or conc. H2SO4 not used for drying NH3? Write the equation for the formation of ammonia by the action of water on magnesium nitride. How is ammonia collected? Why is ammonia not collected over water? Which compound is normally used as a drying agent for ammonia? A gas is obtained by boiling a solution of NaNO3 with Zn in an alkaline solution. How does this gas

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20. 21. 22. 23. 24. 25. 26. 27.

28. 29. 30.

affect moist litmus paper. What happens when NH3 burns in air? Name the catalyst used in the Haber synthesis of NH 3 ? Name two common fertilisers manufactured from NH 3 ? What would you see upon adding an NH3 solution to one of CuSO4? How does NH3 react with Cl2? Give equations. In the catalytic oxidation of NH3, the reaction takes place between 7500 C and 9000C. After initially heating the catalyst, why is heating stopped? Write equations for the reactions between (i) Copper and conc. HNO3 (ii) CuO and dil. HNO3 What are the compounds required for the laboratory preparation of nitric acid? Name the gas produced in each of the following reactions. (i) The action of conc. HNO3 on copper. (ii) The heating of ammonium nitrate (name only the nitrogen containing compound) (iii) The warming of ammonium sulphate with NaOH solution. What is fuming HNO3? How is it prepared? What is aqua regia? Which metals give H2 with HNO3? Name two metal nitrates which do not give NO2 on heating. Name the nitrogen compounds formed in the redox reactions of HNO3. Acids generally give H2 with metals, but HNO3 does not. Why? Name the nitrogen compound obtained on heating NH 4 NO 3 . During a thunder storm, the rain water contains HNO3. The HNO3 is formed as a result of three chemical reactions. Write the balanced chemical equations. Give equations for the reactions of HNO3 with CaCO3, MgO and NaHCO3. Give equations for the reactions of conc. HNO3 with carbon, sulphur and copper. Give the formula of the brown compound formed in the brown ring test for nitrate.

Compounds of Nitrogen

HOTS Worksheet 1.

2. 3.

4.

In the lab preparation of NH3, a higher ratio by weight of the alkali is used in proportion to sal ammoniac (NH4Cl). Give reason. In the lab preparation of NH3, quick lime is used as the drying agent for ammonia gas. Give reason. All ammonium salts heated with alkalis evolve ammonia, even then ammonium nitrate is not used in the laboratory preparation of NH3. Explain. Complete and balance the following equations for the general methods of obtaining ammonia by substituting the correct symbols for X and Y. (a) (XY4)2SO4 + Ca(OY)2

  CaSO4 + Y2O + XH3

5. 6. 7. 8. 9. 10. 11.

12.

243 13.

(a)

 XH3 + Y2 Catalytic chamber  Pt   XY + H2Y 8000 C

(b)

 XY + Y2 Oxidising chamber    XY2.

(c)

 XY2 + H2Y + Absorption tower   HXY3. Y2 

14. 15.

(b)

(XY4)2SO4 + NaOY 16.

(c)

  Na2SO4 + Y2O + XH3  Al(OY)3 + XH3 AlX + Y2O 

(d)

 Mg(OY)2 + XH3 Mg3X2 + Y2O 

17.

In Haber's process why a higher ratio of hydrogen to nitrogen is used? In Haber's process why the reactants should be pure and free from impurities. An acidic solid drying agent not used for drying ammonia gas. A metallic nitride of a trivalent metal which on reaction with warm water evolves ammonia gas. A drying agent which forms an addition compound with ammonia. The catalyst other than iron which may be used in Haber's process. Give balanced equations for conversion of ammonia to the following ammonium salts. (a) Ammonium chloride (b) Ammonium sulphate (c) Ammonium nitrate Starting from ammonium hydroxide how would you obtain the following: (a) A white amphoteric hydroxide. (b) A green metallic hydroxide of a divalent metal. (c) A reddish brown metallic hydroxide of a trivalent metal.

The equations given below represent the different reactions in the various chambers in ostwald's process. Complete and balance the equations by substituting the correct symbols for X and Y.

18. 19. 20.

21. 22.

In ostwald's process, the absorption tower is a steel tower lined with flint or quartz. Explain. Nitric acid cannot be concentrated beyond 68% by boiling or distillation. Give reason. Some farmers feel that lighting helps produce a better crop. What is the scientific basis for this belief? What happens when NH 3 is passed into suspension of bleaching powder? Apart from N2 O, which other colourless gas supports combustion? Write an equation for the reaction of burning Mg with NO2. Automobiles are now provided with catalytic converters. What is their role in reducing air pollution? NH3 gas is dried by CaO and not by P2O5 and H2SO4. Explain. Pt NH3 + O2   (A)

(A) + O2   (B) brown fumes (B) + H2O   (C) + (D) (both oxyacid) (C) + I–   (E) (violet vapours) Identify (A) to (E). 23.

 Colourless salt (A) + NaOH   gas (B) giving white fumes with HCl + alkaline solution (C).

(C) + Zn   gas (B)

gas (D)  liquid (E)

   (A)    both triatomic

Identify (A) to (E).

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9th Class Chemistry

244 24.

25.

A few drops of concentrated ammonia solution 5. added to a calcium bicarbonate solution causes a white precipitate to form. Write a balanced equation. Identify salts (A) and (B) based on following 6. reactions:

(A) or (B) 

(A)

26.

NaOH, 

(C), gas

NH3 A

H2O

(D)

7.

 (B)   (E) (gas). (E) Is used as an anaesthetic. Vessels made of metals like Al, Fe, Cr, Ni etc., can be used to store conc. HNO3. Explain.

27.

Make distinction between NO 3 and NO 2 .

28.

What is the effect of heat on NH 4 NO 3 and NH 4 NO 2 ?

9.

 3

 2

29.

NO interferes in the 'ring test' of NO . How

30.

is it removed from the mixture? Write balanced equations for the following: (a) HNO3 turns brown when exposed to air. (b)

8.

10.

HNO3 turns yellow-brown on standing.

IIT JEE Worksheet 1.

2.

3.

4.

NCl5 is not formed because (A) It is unstable (B) Nitrogen is inert (C) Nitrogen does not have d-orbitals (D) Nitrogen has small atomic radius

11.

12.

Nitrogen forms nitride,  NO3  ion more readily.. This is due to 13. (A) Its small size (B) Its high electronegativity (C) both (A) and (B) (D) none 14. Ammonia is dried over (A) Slaked lime (B) Quick lime (C) CaCl2 (D) P 2O5 The reddish brown gas formed when nitric oxide 15. is oxidised by air is (A) N2O5 (B) N2O4 (C) NO2 (D) N2O3

On heating a mixture of NH4Cl and KNO2, we get (A) NH4NO3 (B) KHNO4(NO3 )2 (C) N2 (D) NO Which of the following does not give NO2 on heating? (A) KNO3 (B) Pb(NO3)2 (C) Cu(NO3)2 (D) AgNO3 Laughing gas is (A) Nitrous oxide (B) Nitric oxide (C) Nitrogen trioxide (D) Nitrogen pentoxide FeSO4 forms brown ring with (A) NO2 (B) N2O3 (C) NO (D) N2O5 When conc. H2SO4 is added to dry KNO3, brown fumes evolve. These fumes are due to (A) SO2 (B) SO3 (C) NO2 (D) NO When lightning flash is produced through air, the gas formed is (A) HNO2 (B) HNO3 (C) N2O (D) NO Nitrogen is essential constituent of all (A) Proteins (B) fats (C) proteins and fats (D) none When AgNO3 is heated strongly the products formed are (A) NO and NO2 (B) NO2 and O2 (C) NO2 and N2O (D) NO and O2 Which one of the following nitrogen oxides is the anhydride of nitrous acid. (A) N2 O (B) N2O3 (C) N2O4 (D) NO Which acts both as oxidising as well as reducing agent. (A) HNO3 (B) HNO2 (C) H2SO4 (D) HCl When NH 3 is passed over heated CuO, it is oxidised to (A) N2 (B) NO2 (C) N2O (D) HNO2

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B y t h e e n d o f t h i s c h a p t e r, y o u w i l l u n d e rs t a n d      

Introduction Occurrence of Sulphur dioxide Preparation of SO2 Laboratory methods of preparation Physical properties of Sulphur dioxide Chemical properties of Sulphur dioxide

 Tests for Sulphur dioxide  Chemical properties of  Uses of Sulphur dioxide Sulphuric acid SULPHURIC ACID, H 2SO4 OR OIL OF VITROIL  Uses of  Occurrence Sulphuric acid  Preparation  Tests  Physical properties of Sulphuric acid

SULPHUR DIOXIDE 1.

Introduction • •

Molecular Formula : SO2 Molecular Weight : 32 + 2  16 = 64; vapour density =

• • •

3.

2.

Occurrence of Sulphur dioxide •

Mol.Wt 64   32 2 2

Sulphur dioxide has been known as a fumigating agent from very early times and was used as an insecticide and fungicide. In the year 1774, Priestley first prepared the gas by the action of mercury on concentrated sulphuric acid and called it nutreobic acid air. Lavoisier, in the year 1777, studied the compound in detail and established it as oxide of sulphur.

•

Sulphur dioxide is formed in small amounts wherever coal, coke or even oil is burnt and is thus, released into the atmosphere from coal and oil - based power plants; copper smelting plants and oil refining. Sulphur dioxide is therefore, present in traces in the exhaust of certain industrial units. It also occurs in volcanic gases coming out of volcanoes. It has also been found to occur in certain spring waters and water - falls.

Preparation of SO2 (i)

By burning of sulphur in air or oxygen.

(ii)

burning S(s) + O2 (g)   SO2(g) By the action of dilute acids (hydrochloric or sulphuric acid) on sulphites or bisulphites.

Sulphite/Bisulphites + Dilute Acid   Salt + Water + Sulphur dioxide Na 2 SO 3

(iii)

+

2HCl(aq)

  2NaCl(aq) +

H2 O(l)

Ca(HSO3 )2 (aq) + H2 SO 4 2H2O(l)   CaSO4(aq) + This reaction is not preferred because CaSO4 forms insoluble crust. By the action of hot concentrated sulphuric acid on certain metals.

+

SO2(g)

+

2SO2(g)

Metal Sulphur + (Hot) Concentrated Sulphuric acid   Salt + Water + Sulphur dioxide (iv)

Cu(s) + 2H2SO4(aq) CuSO4(aq) +   By the action of hot concentrated sulphuric acid on sulphur. H2 SO 4

  H2O

+

2H2O(l)

SO2 +

+

SO2(g)

[O]  2

S + 2[O]   SO2 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– S(s)

+

2H2 SO4

Chapter - 11

Compounds of Sulphur

Learning Outcomes

  2H2O(vap) +

3SO2(g)

9th Class Chemistry

246

The above mentioned methods. (i), (ii) are used for the industrial preparation/manufacture of sulphur dioxide. Storage of sulphur dioxide: Since, sulphur dioxide is highly dangerous gas because of its toxic nature, it should never be stored in gaseous form in large volumes. Its leakage may cause tremendous damage to human, plant and animal life. Sulphur dioxide is liquefied, compressed and stored in steel cylinders as it is easier to liquefy.

4.

Laboratory preparation

methods

of

(ii)

By the action of the dilute sulphuric acid or dil. hydrochloric acid on sodium sulphite or sodium bisulphite (sodium hydrogen sulphite) In the laboratory, sulphur dioxide gas may also be prepared by treating sodium sulphite or sodium bisulphite with dilute sulphuric acid or dilute hydrochloric acid. Reactions :

Reaction: Cu(s) + 2H2SO4 (aq)   CuSO4(aq) + 2H2O(l) + SO2(g) (i) By the action of hot concentrated sulphuric acid on copper. Procedure : • In the laboratory, sulphur dioxide is prepared by heating copper turning with concentrated sulphuric acid. • Copper turnings are taken in a round bottomed flask fitted with a thistle funnel and a delivery tube. Concentrated sulphuric acid is added through the thistle funnel and the flask is heated. • The mixture becomes slightly black and the sulphur dioxide formed is passed through concentrated sulphuric acid to dry the gas collected. • Why does mixture become black? The residue left behind is copper sulphate which should be normally blue. It becomes greyish white due to dehydration in the presence of excess of concentrated sulphuric acid. This becomes black as a result of partial reduction of copper (I) sulphate to copper sulphide (CuS) which is black in colour. • Drying of the gas: Sulphuric dioxide gas, as prepared in the laboratory, is accompanied with moisture. the gas can be dried by bubbling it through concentrated sulphuric acid. The gas cannot be dried by passing it over calcium oxide (quick - lime) since the gas reacts with it. •

CaO(s) + SO2(g)   CaSO3(s) Collection of the gas: SO2 is collected by the upward displacement of air because. (i) It is 2.2 times heavier than air. (ii) It is highly soluble in water, hence it cannot be collected over water.

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Na2SO3(aq or s) + Na 2 SO4 (aq) + SO2(g)

H2 SO4 (aq)   H2O(l) +

Na2SO3(aq or s) + 2NaCl(aq) SO2(g)

2HCl(aq)   + H2O(l) +

2NaHSO3 + Na 2 SO4 (aq) + 2SO2(g)

H2 SO4 (aq)   2H2O(l) +

NaHSO3 + HCl(aq)   NaCl(aq) + H2O(l) + SO2(g) Procedure : Sodium sulphite or bisulphite is taken in a round bottomed flask. Dilute hydrochloric acid or dilute sulphuric acid is added through the thistle funnel. Sulphur dioxide is dried by passing through concentrated sulphuric acid and collected by the upward displacement of air.

Formative Worksheet 1.

2.

Assertion (A): Sulphur dioxide is collected by the upward displacement of air Reason (R): Sulphur dioxide is 2.2 times heavier than air. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Assertion (A):SO2 should never be stored in gaseous form in large volumes.

Compounds of Sulphur

3.

4.

5.

6.

Reason (R): Sulphur dioxide is highly dangerous gas because of its toxic nature. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Statement A :The only impurity present in SO2 is water vapour. Statement B :The water vapour is removed by bubbling the gas through concentrated sulphuric acid. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. Which of the following statements are correct? a) Sulphur dioxide occurs in free state in the fumes of volcanic vents. b) Sulphur dioxide occurs in industrial belts where coal and petroleum products are used as industrial fuels. c) In smelting of sulphide ores in metallurgy, sulphur dioxide is released. (A) Only a (B) Only b and c (C) Only a and c (D) a, b and c Assertion(A): Dilute hydrochloric acid is seldom used in the preparation of sulphur dioxide. Reason (R) :Dilute HCl is a volatile acid, vapourises. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct. Which of the following statements are correct in the collection of SO2 gas. i) SO 2 is collected by the upward displacement of air. ii) Sulphur dioxide is 2.2 times heavier than air. iii) Sulphur dioxide is highly soluble in water, hence it cannot be collected over water. (A) Only i and ii (B) Only ii and iii (C) Only i and iii (D) All are correct

247

Conceptive Worksheet 1. 2. 3.

4.

5.

6.

7. 8.

9.

10.

5.

The molecular formula of sulphur dioxide is : (A) S2O (B) SO (C) SO2 (D) S2O3 The vapour density of sulphur dioxide is : (A) 64 (B) 32 (C) 16 (D) 48 Priestley first prepared SO2 by the action of (A) Mercury on con.H2SO4 (B) Sulphur and oxygen (C) Calcium on con.H2SO4 (D) Sulphur and excess of oxygen Priestly called SO2 as : (A) Fixed air (B) Mephytic air (C) Nutreobic air (D) Acid former Sulphur dioxide occurs in (A) Volcanic gases (B) Spring water (C) Water falls (D) All Sulphur dioxide is : (A) Toxic nature (B) Acidic nature (C) Highly soluble in water(D) All the above CaO + SO 2  A

(A) CaSO4 (B) CaSO3 (C) SO3 (D) CaS Sulphur dioxide is collected by (A) Downward displacement of air (B) Upward displacement of air (C) Downward displacement of air (D) Upward displacement of water The chemical formula of sodium bisulphite is : (A) Na 2SO4 (B) Na2SO3 (C) NaHSO3 (D) NaHSO4 Sulphur dioxide is a fumigating agent and used as: (A) An insecticide (B) Fungicide (C) Both 1 and 2 (D) Match industry

Physical properties of Sulphur dioxide (i) (ii)

(iii)

(iv)

Colour : Sulphur dioxide is a colourless gas. Odour and taste : It has characteristic pungent, suffocating odour of burning sulphur. It has sour taste. Vapour density : These gas is 2.2 times heavier than air (vapour density of air = 14.4, vapour density of sulphur dioxide = 32) Solubility : It is highly soluble in water . One volume of water dissolves 80 volumes of the gas. The aqueous solution of the gas is acidic in nature and is known as sulphurous acid.

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9th Class Chemistry

248 2NaOH(aq) Na 2 SO 3 +

The gas is therefore, known as sulphurous anhydride. Anhydride is an oxide of non - metal which when dissolved in water forms an acid. SO2(g) + H2O(l)   H2SO3(aq) (v) Liquefaction : It is liquefied at – 100C at a pressure of 2 atmospheres. On further cooling, it solidifies to snow like solid at – 760C. (vi) Melting point : Solid sulphur dioxide melts at – 760C. (vii) Boiling point : Liquid sulphur dioxide boils at 100C. (viii) Physiological nature : It is highly poisonous. It causes inflammation of lungs, thereby, affecting the respiratory system. It creates asthmatic conditions. It proves fatal when inhaled in large amounts.

6.

Chemical properties of Sulphur dioxide

I.

Acidic nature :

H2 SO 3 H3O+ (aq)

+

H2O(l)   + HSO3– (aq)

–

HSO3 (aq) +

H2O(l)

H3O+ (aq)

  2 3

SO (aq)

+ +

Presence of H3O imparts acidic properties to it. Sulphurous acid is unstable and it exists only in solution. It decomposes to water and sulphur dioxide. H2SO3 (aq) SO2(g)

 

H2O(l)

+

Action with caustic alkalis.

 

Action with lime water, Ca(OH)2. When the gas is passed through lime water, the matter turns milky due to the formation of calcium sulphite. Ca(OH)2 (aq) + SO2(q)   CaSO3 + H2O(l) When excess of gas is passed through lime water, milkiness disappears because of the formation of soluble calcium bisulphite. CaSO3 + H2O(l) + SO2

II.

Dry or liquid sulphur dioxide has no effect litmus. (i) However, it turns moist litmus red; methyl orange pink and phenolphthalein colourless. Sulphur dioxide turns moist litmus first red and then colourless. Sulphur dioxide dissolves in water to (ii) form sulphurous acid. SO2(g) + H2O(l)   H2 SO3 (aq) Sulphurous acid ionises in water to give hydronium ion (H3O+), bisulphite ions, sulphite ions.

+ SO2 H2O(l)

 

Ca(HSO3 )2(aq)

Reducing action : In the presence of moisture, sulphur dioxide behaves as a very powerful reducing agent because in moisture it liberates nascent hydrogen. SO2(g) + 2H2O(l)   H2 SO4 (aq) + 2[H] Reducing action of sulphurous acid is of two types : Reaction with water : Sulphurous acid produces very reactive nascent hydrogen which adds, onto other substances, thus, reducing them. Sulphurous acid takes up oxygen from other substance, thus, reducing them. (a) It reduces chlorine water to hydrochloric acid. 2H2O(l) + SO2(g)   H2SO4(aq) +2[H] 2[H] + Cl2   2HCl –––––––––––––––––––––––––––––––––––– 2H2O(l) + Cl2(g)+ SO2(g)   2HCl(aq) + H2SO4(aq) Colour of chlorine water changes from yellow to colourless. (b) It reduces ferric chloride (yellow) to ferrous chloride (green) and, ferric sulphate (red brown) to ferrous sulphate (green). SO2(g) + 2H2O(l)   H2SO4(aq) +2[H](g) FeCl3(aq) + [H](g)   FeCl2(aq) + HCl × 2 ––––––––––––––––––––––––––––––––––––

It reacts with alkalis to form two series of salts – bisulphites and sulphites.

2FeCl3(aq) + SO2(g) +2H2O   2FeCl2(aq) + H2SO4(aq)+2HCl

NaOH(aq) NaHSO3 (aq)

Fe2(SO4)3 + SO2(g) + 2H2O   2FeSO4 + 2H2SO4

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+

SO2(g)

 

Compounds of Sulphur (c)

249

It reduces the pink coloured potassium permanganate solution to colour solution of manganous sulphate thus, rendering pink solution of potassium permanganate colourless.

7.

(i)

Odour: It has characteristic suffocating pungent odour of burning sulphur. (ii) Action on indicator: It turns moist blue litmus red. It turns methyl orange pink. It turns pink phenolphthalein solution colourless. (iii) It turns orange coloured acidified potassium dichromate solution green, when passed through it.

2KMnO4 + 3H2SO4   K2SO4 + 2MnSO4 + 3H2O + 5[O] SO2 + [O] + H 2 O   [H2SO4] × 5 –––––––––––––––––––––––––––––––––––– 2KMnO4 + 5SO2 + 2H2O   K2SO4+ 2MnSO4 + 2H2SO4 (d) It reduces orange solution of acidified potassium dichromate to green solution of chromic salt, thus, rendering orange solution of acidified potassium dichromate green.

K2Cr2O7(aq) + 4H2SO4   K2SO4 + Cr2(SO4)3 + 4H2O + 3[O] H2O + SO2 + Cr   H2SO4] × 3 ––––––––––––––––––––––––––––––––––––

K2Cr2O7 + 4H2SO4   K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]

K2Cr2O7 + H2SO4 + 3SO2   K2SO4+ Cr2(SO4)3 + H2O (iv) It decolourises pink coloured acidified potassium permanganate solution, when passed through it.

SO2 + [O] + H2O   H2SO4] × 3 –––––––––––––––––––––––––––––––––––– K2Cr2O7 + 3SO2 + H2SO4   K2SO4 + Cr2(SO4)3 + H2O

2KMnO4 + 3H2SO4   K2SO4 + 2MnSO4 + 3H2O + 5(O)

III. Addition reactions. Sulphur has valency 6 (hexavalent) but in sulphur dioxide, only four valencies of sulphur are satisfied by two oxygen atoms. The remaining two valencies are unsaturated and hence, sulphur dioxide is an unsaturated compound. It gives addition products with oxygen, chlorine and lead dioxide. (a) It combines with oxygen at 500°C in the presence of platinized asbestos or vanadium pentoxide to form sulphur trioxide.

H2O + SO2 + (O2)   H2SO4] × 5 –––––––––––––––––––––––––––––––––––– 2KMnO4 + 2H2O(l) + SO2(g)   2MnSO4 + K2SO4 + 2H2SO4 (v) It forms a white precipitate when passed through a solution of barium chloride containing chlorine water. BaCl2(s) + SO2(g) + HCl(ag) + HOCl(aq) +

V2 O5

– H2O(l)   BaSO4( ) + 4HCl(aq) (vi) It bleaches the colour of a moist flower. The bleaching action of sulphur dioxide is due to reduction. Hence, if an oxidising agent e.g., dilute nitric acid is added to the bleached flower, original colour of the flower is restored.

Vanadium pentoxide 2SO2(g) + O2(g)   2SO3(g) 0 450  500 C

(b) It combines with chlorine in the presence of sunlight to form sulphuryl chloride, SO2Cl2. sunlight SO2(g) + Cl2(g)   SO2Cl2(g)

IV.

Bleaching property: Sulphur dioxide is a bleaching agent and bleaches vegetable colouring matter in the presence of moisture. The bleaching action of sulphur dioxide is due to reduction in contrast to that by chlorine which bleaches due to oxidation. In the presence of moisture, sulphur dioxide produces nascent hydrogen which reduces many coloured substances to colourless compounds. The colouring matter combines with nascent hydrogen to form colourless reduction product. SO2 + 2H2O   H2SO4 + 2[H]

Tests for Sulphur dioxide

8.

Uses of Sulphur dioxide (i)

It is used in the manufacture of sulphuric acid (H2SO4) by contact process, and calcium sulphite used in paper industry. (ii) It is used as a bleaching agent. Delicate articles like silk, wool, hair, etc., are bleached by this gas, as chlorine cannot be employed for bleaching these articles. The action of chlorine is drastic while that of sulphur dioxide is milder. www.betoppers.com

9th Class Chemistry

250 (iii) It is used as a disinfectant and household fumigant, since it is a powerful germicide and insecticide. (iv) It is used in refining sugar in sugar industry as it decolourises cane-sugar. (v) Sulphur dioxide is used as an antichlor to remove its excessive chlorine on a bleached article as it reduces halogens to hydrogen halides. SO2(g) + 2H2O(l) + Cl2(g)   2HCl(aq) + H2SO4(aq) vi) Liquid sulphur dioxide is used as a refrigerant (these days, it has been replaced by nonpoisonous freon as it undergoes easy liquefaction and vapourisation). (vii) Liquid sulphur dioxide is used as an organic solvent. (viii) It is used as preservative for dry fruits, meat and wine. (ix) It is used in refining kerosene and other petroleum products.

13.

14.

15.

Formative Worksheet 7.

Sulphur dioxide dissolve in water and form (A) H2SO4 (B) H2 S2O7 (C) H2SO3 (D) H2 S2O8

8

Pt 2SO 2 + O 2  2SO3 3000-4000°C

9.

10.

11.

12.

The above reaction is : (A) Addition reaction (B) Substitution reaction (C) Decomposition reaction (D) Reduction reaction When SO2 is passed through cupric chloride solution. (A) A white precipitate is obtained (B) The solution becomes colourless (C) The solution becomes colourless and a white precipitate of is obtained. (D) No visible change takes place

SO 2 + 2H 2O + Cl2   H 2SO 4 + 2HCl In the above reaction which is reducing agent (A) HCl (B) H2SO4 (C) H2O (D) SO2 SO2 acts as bleaching agent in aqueous solution due to release of : (A) Nacent oxygen (B) Nacent sulphur (C) Nacent hydrogen (D) Nacent SO2 Bleached pr oduct + atmospheric oxygen  represents.

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original colour



16.

17.

18.

19.

(A) Reduction (B) Boiling (C) Catalyst (D) Oxidation Which of the following are correct about SO2? a) SO2 is known as fumigating agent from very early times. b) SO2 is used as insecticide and fungicide. c) SO2 is also called vitriolic acid air. (A) All are correct (B) Only a, b are correct (C) b, c are correct (D) Only a is correct Strongly burning potassium or magnesium continues to burn in SO2 and forms (A) and (B). (A) A B : Metal oxide Metal hydroxide (B) A B : Metal sulphide Acid (C) A B : Metal hydroxide Salt (D) A B : Metal sulphide Metal oxide When SO2 is passed through cupric chloride solution (A) White precipitate is obtained. (B) The solution becomes colourless. (C) The solution becomes colourless and a white precipitate of Cu2Cl2 is obtained. (D) No visible change takes place. Assertion (A): In the presence of moisture sulphur dioxide behaves as a very powerful reducing agent Reason (R): SO2 is a pale yellow coloured gas. (A) Both A & R are correct & R correctly explains A. (B) Both A & R are correct but R is not correctly explains A. (C) A is correct & R is wrong. (D) A is wrong & R is Correct. FeCl3 solution on reaction with SO2 changes to (A) FeCl2 (B) Fe2(SO4)2 (C) Fe2(SO3)3 (D) FeSO4 Which of the following is oxidised by SO2? (A) Mg (B) K 2Cr2O7 (C) H2SO3 (D) H2S2O8 SO2 oxidises (A) Mg (B) K2Cr 2O7 (C) KMnO4 (D) All of these

Compounds of Sulphur 20.

21.

"Sulphur dioxide is a powerful reducing agent. But is also acts as an oxidising agent". State in words or give balanced equations of two such reactions in which it acts as an oxidising agent. B) Sulphuric acid is said to be a dibasic acid. What is meant by the term "dibasic" C) Calculate the vapour density of sulphur dioxide gas. (S = 32, O = 16) Give two reactions to illustrate : A) The reducing property of sulphur dioxide. B) The oxidising property of sulphur dioxide. C) The acidic nature of sulphur dioxide.

251

A)

21.

22.

23.

24.

Conceptive Worksheet 11.

12.

13. 14.

15.

16.

17.

18.

19.

20.

Fountain experiment conducted with sulphur dioxide proves which of the following physical properties? (A) High solubility in water (B) Combustion (C) Heavier than air (D) All the above Sulphur dioxide gas is collected upward displacement of air it is: (A) Lighter than air (B) Neither lighter nor heavier than air (C) Heavier than air (D) As heavy as air A gas which cannot be collected over water is : (A) N2 (B) O2 (C) SO2 (D) PH4 When a colourless gas is passed through bromine water only decolourisation takes place. The gas is: (A) SO2 (B) HBr (C) HCl (D) H2S Which of the following is not linear ? (A) BeCl2 (B) SO2 (C) CO2 (D) CH = CH Sulphur dioxide is Stored in the form of (A) Crystals (B) Gas (C) 1& 2 (D) Liquid Sulphur dioxide is how many times heavier thatn air (A) 4.4 times (B) 2.2 times (C) 5 times (D) 8.8 times Sulphur dioxide dissolve in water and form (A) H2SO4 (B) H2 S2O7 (C) H2SO3 (D) H2 S2O8 The nature of SO2 is : (A) Basic (B) Acidic (C) Neutral (D) None of these The nature of SO2 is :

25.

26.

27.

28.

29.

(A) Basic (B) Acidic (C) Neutral (D) None of these The oxide which acts as a reducing, oxidizing, bleaching agent and a Lewis base is: (A) SO2 (B) SO3 (C) CO2 (D) NO SO2 reduces (A) Mg (B) H2S (C) KMnO4 (D) All of these SO2 reacts with Cl2 to yields : (A) Thionyl chloride (B) Carbonyl chloride (C) Sulphuryl chloride (D) Sulphur monoxide When silent electric discharge is passed through a mixture of S and SO2 under reduced pressure. (A) S2O3 is formed (B) S2O is formed (C) S2O2 is formed (D) S6O is formed Which of the following is used to absorb sulphurdioxide? (A) Con.H2SO4 (B) KOH solution (C) Water (D) Anhydrous When SO2 is passed through acidified K2Cr2O7 solution? (A) The solution turns blue (B) The solution is de colourised (C) SO2 is reduced (D) Green solution of chromic salt is formed When H2S is passed through a solution of SO2 in water, S is precipitated. Hence SO2 acts as : (A) An acid (B) A catalyst (C) A reducing agent (D)An oxidizing agent Reducing action of sulphur dioxide is due to the (A) Nascent hydrogen (B) Nascent Oxygen (C) Double Bond (D) None of these Hydrogen sulphide with lead acctate solution gives (A) Red Precipitate (B) Black Precipitate (C) White Precipitate (D) Green Precipitate

SULPHURIC ACID, H2SO4 OR OIL OF VITRIOL 9.

Introduction Medieval alchemists obtained sulphuric acid by distilling ferrous sulphate crystals (green vitriol) as an oily viscous liquid and called it vitriol. Heat 2FeSO .7H O 4

2



Fe2 O3 + SO2 (g) + SO3 (g) + 14H2 O(l) The Sulphur trioxide liberated was dissolved in water to get sulphuric add. In 1746, John Roebuck, a British chemist, introduced a relatively inexpensive lead chamber process for the large scale production of sulphuric acid. In www.betoppers.com

9th Class Chemistry

252 1831, an English vinegar maker, Peregrine Phillips, got contact process patented for the industrial preparation of sulphuric acid.

10. Occurrence 



In the free state, sulphuric acid is found in traces in certain hot mineral springs. It also occurs near sulphide rock-beds formed by the action of water on sulphide ores. In the combined state, it occurs as sulphates of metals eg., magnesium sulphate, calcium sulphate, barium sulphate, etc.

11. Preparation It may be obtained by dissolving sulphur trioxide in water. SO3(g) + H2O(l)   H2SO4(aq)

Industrial preparation of sulphuric acid by contact process: It may be noted that these days sulphuric acid is mostly prepared by the contact process. The acid produced by this method is free from arsenic impurities and hence, can be safely used for the preparation of edible products. Chemistry of Manufacture : The manufacture of H2SO4 by the contact process involves the following steps : (i) Production of SO2 by burning sulphur or roasting iron pyrites. S8

+

8O2

  8SO2

4FeS2 + 11O2   2Fe 2 O 3 + 8SO2 (ii) Catalytic oxidation of sulphur dioxide by air to give sulphur trioxide. 2SO2

+

O2

 

2SO 3 ;

–1

DH = –196.6 kJ mol (iii) Absorption of sulphur trioxide in 98% sulphuric acid to form oleum. SO3 + H2SO4   H2S2O7 (oleum) (iv) Dilution of oleum to get sulphuric acid of desired concentration. H2 S 2 O 7 + H2O   2H2SO4 Sulphur trioxide is passed up an absorption tower where it meets a descending stream of concentrated (98%) sulphuric acid. This tower is also packed with quartz. SO3 www.betoppers.com

+

H2SO4   H2S2O7

To this oleum, a calculated quantity of water is added to get the acid of desired concentration. H2 S 2 O 7 + H2O   2H2SO4 It may be noted that sulphur trioxide is not directly absorbed in water to form sulphuric acid because it forms a dense fog of sulphuric acid. This may cause much difficulty and hazard to the workers. Favourable conditions for the maximum yield of sulphur trioxide : As discussed, sulphur trioxide (SO 3 ) is manufactured by the direct oxidation of sulphur dioxide (SO2) with atmospheric oxygen. 2SO2(g) + O2(g)   2SO3(g); DH = –196.6 kJ mol–1 This reaction is reversible, exothermic and proceeds with a decrease in volume. According to Le Chatelier’s principle, the favourable conditions for the formation of SO3 are : (i) High concentration of oxygen : Air or oxygen used for the oxidation of sulphur dioxide to sulphur trioxide must be in excess. (ii) High pressure : Since the forward reaction proceeds with decrease in volume, therefore, high pressure will favour the reaction. In actual practice, a pressure of about 2 atmospheres is used. This is because the gases are acidic and corrosion of the plant occurs at high pressure. (iii) Low temperature: Since the forward reaction is exothermic, low temperature will favour the reaction. However, rate of the reaction decreases with decrease in temperature. Therefore, the reaction is carried out at an optimum temperature of 623 – 723 K. (iv) Use of catalyst : To increase the rate of the reaction at low temperature, a catalyst is to be used. The commonly used catalysts are platinum, or vanadium pentoxide, (V2O5). Since Platinum is quite expensive and is easily poisoned by arsenic impurities usually present in SO2 , therefore, divanadium pentoxide is employed because it is not only cheaper but is also not easily poisoned. (v) Purity of gases : Purity of gases is another essential condition for the maximum yield of SO3. The impurities present in the reacting gases act as catalytic poisons and thus decrease the efficiency of the catalyst.

Compounds of Sulphur

Formative Worksheet 22.

23.

24.

25.

26.

31.

32.

33.

34.

35.

36.

37

In manufacture of H2SO4 the nitrated acid from Gay-Lussacs tower chemically : (A) H2SO4 .NO2 (B) H2 SO4 .NO (C) H2 SO4.2NO (D) HSO4.NO The catalyst used in the manufacture of H2SO4 by contact process is (A) Al2O3 (B) Cr2O3 (C) V2O5 (D)MnO2 Aqueous solution of sulphur dioxide on oxidation gives (A) SO3 (B) H2SO4 (C) H2SO3 (D)H2 S 2O 7

Catalyst used in Contact process (A) Ni (B) V2O5 38. (C) Powdered Fe (D) M In contact process, impurities of arsenic are removed by 39. (A) Fe(OH)3 (B) Al(OH)3 (C) Cr(OH)3 (D) Fe2O3 The most favorable conditions for the reaction 2SO2 + O2 2SO3 + 98 KJ/Mole is: (A) High temperature and high pressure 12. Physical properties of (B) High temperature and low pressure sulphuric acid (C) Low temperature and low pressure (D) Low temperature and high pressure (i) Nature: Which one of the gas dissolves in H2SO4 to give Concentrated sulphuric acid is a colourless, oleum? dense, syrupy, hygroscopic liquid. It has a sour (A) SO2 (B) H2S (C) S2O (D) SO3 taste. Being hygroscopic, it absorbs moisture Oleum or fuming H2SO4 is : from the atmosphere. That is why, (A) A mixture of conc. H2SO4 and oil concentrated sulphuric acid is kept in air-tight (B) Sulphuric acid which gives fumes of bottle. sulphur dioxide (ii) Relative density: (C) Sulphuric acid saturated with sulphur 98% concentrated acid has relative density trioxide i.e., H2S2O7 of 1.84 at 15°C. (D) A mixture of sulphuric acid and nitric acid (iii) Boiling point: It forms constant boiling mixture containing onceptive orksheet 98.5% of the acid boiling at 3.38°C. (iv) Freezing point: Which has maximum number of oxo groups? It freezes at 10.4°C to colourless crystals. (A) H2SO4 (B) H2SO3 (v) Solubility: (C) H3 PO3 (D) H3PO4 It is miscible with water in all proportions. Which one is known as oil of vitriol? The dissolution proceeds with evolution of (A) H2S 2O7 (B) H2SO3 heat (exothermic). (C) H2S2O8 (D) H2SO4 Note : If equal volumes of the acid and water In combined state sulphuric acid occurs as: are mixed at ordinary temperature, the (A) Sulphates (B) Sulphites solution attains a temperature of 120°C. (C) Sulphides (D) Oxides Hence, acid should be diluted by pouring it The formula of oleum is: slowly into water with continuous stirring. In (A) H2S 2O5 (B) H2SO4 this way, the heat evolved is distributed (C) H2S2O7 (D) H2SO3 uniformly. The water must not be added to Oleum on treating with water gives: the concentrated acid as it will be suddenly (A) Sulphur dioxide converted into steam causing dangerous (B) Sulphuric acid splashing of the acid. (C) Sulphur trioxide (vi) Conductivity: (D)Hydrogen sulphide Pure anhydrous acid is a bad conductor of Sulphuric acid is prepared by the: electricity and heat. Its aqueous solution (A) Monds process (B) Frasch process (dilute acid) is a good conductor of heat and (C) Contact process (D) Sicilian process electricity. Solution of SO2 in water is known as (vii) Physiological nature: (A) Sulphuric acid Concentrated sulphuric acid is highly (B) Sulphurous acid corrosive in nature and chars the skin black. (C) Hydrosulphuric acid (D) Thiosulphurous acid www.betoppers.com

C 30.

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9th Class Chemistry

254

13. Chemical properties of Sulphuric acid • •

• (i)

(iv) With bisulphites:  Na2SO4 + 2NaHSO3 + H2SO4  2SO2  + 2H2O (v) With sulphites:

The chemical properties of sulphuric acid depend on whether the acid is dilute or concentrated. Sulphuric acid has a strong affinity for water. Thus, when sulphuric acid dissolves in water, a large amount of heat is produced which may spurt the acid out of the container. Hence care must be taken while preparing dilute sulphuric acid from concentrated sulphuric acid (which is 98.3% and is 18M). Therefore, concentrated H2 SO4 is always diluted by adding the acid slowly into water with constant stirring and not by adding water to the acid. The important chemical properties of H2SO4 are discussed under the following categories: Acidic character (Dibasic character): In aqueous solution, H2SO4 ionizes in two steps :

 Na2SO4 +SO2  Na2SO3 + H2SO4  + 2H2O (vi) With sulphides:  Na2SO4 + H2S  Na2S + H2SO4  (b) Being an acid, it reacts with active metals like Zn and Mg to evolve H2 gas.  ZnSO4+ H2  Zn + H2SO4 (dilute) 

+ – H2SO4(aq)   H (aq) + HSO4 (aq)

2 + HSO4–(aq)   H (aq) + SO 4 (aq) Thus, H2SO4 acts as a strong dibasic acid and forms two series of salts - normal sulphates such as sodium sulphate (Na2SO4) and acid sulphates or hydrogen sulphates or bisulphates such as sodium bisulphate (NaHSO4 ). (a) Being an acid dil. sulphuric acid neutralizes alkalis to form salts and water and decomposes carbonates and bicarbonates with the evolution of carbon dioxide; sulphites and bisulphites with the evolution of sulphur dioxide and sulphides with the evolution of hydrogen sulphide. (i) With hydroxides:  2NaHSO4+H2SO4 NaOH +H2SO4  + H2O  Na2SO4 + 2H2O 2NaOH + H2SO4  (ii) With bicarbonates:  Na 2SO4+ 2NaHCO3 + H2SO4  2CO2  +2H2O (iii) With carbonates:

 Na2SO4 +CO2  Na2CO3 + H2SO4  + H2O

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(ii)

(a) (b)

(c)

 MgSO4 +H2  Mg + H2SO4 (dilute)  Dilute sulphuric acid sho ws acidic properties. (i) It neutralises bases, forming salt and water. (ii) On reacting with active metals, It evolves hydrogen. (iii) It decomposes carbonates and bicarbonates to evolve carbon dioxide. (iv) It decomposes sulphites and bisulphites to evolve sulphur dioxide. (v) It decomposes sulphide to evolve hydrogen sulphide. Dehydrating agent: Concentrated H2 SO4 has a strong affinity for water and hence, it acts as strong dehydrating agent. Some important examples are : Charring of sugar: (Carbohydrates) conc. H 2SO 4 C12 H22 O11    12C + 11H2 O It r emoves water of cr ystallization fr om hydrated salts. conc. H 2SO 4 CuSO4 .5H2 O    CuSO4 +5H2 O Drying of gases. Many gases such as O2 , N2 , Cl2 , etc. which do not react with concentrated H2 SO4 can be dried by passing them through a bubbler containing concentr ated H 2 SO 4 . (Exception ammonia) Concentrated sulphuric acid: (i) behaves as an oxidising agent. Oxidises non-metals, metals and Inorganic compounds. (ii) Because of its great affinity for water it behaves as : (a) dehydrating agent : Removes elements of water from some compounds. (b) drying agent: to remove moisture from gases. (iii) It is a non-volatile acid and displaces more volatile adds from their salts.

Compounds of Sulphur

255

(iii) Oxidising agent: Concentrated sulphuric acid behaves as an oxidising agent since it gives nascent oxygen on decomposition. Heat H2 SO4   H2 O + SO2 + [O] This nascent oxygen brings about oxidation of a number of non-metals and metals. (a) It oxidises carbon to carbon dioxide.

H2 SO4   H2 O + SO2 + [O] × 2 C+ 2[O]

  CO2 ––––––––––––––––––––––––––––––––––– C + 2H2 SO4   2SO2 + CO2 + 2H2 O (b) It oxidises sulphur to sulphur dioxide H2 SO4   H2 O+ SO2 + [O] × 16 S8 +16[O]

  8SO2 S 8 + 16H2 SO4   24SO2 + 16H2 O (c) It oxidises metals such as Cu, Pb, Hg, Ag, etc. to their corresponding sulphates, liberating sulphur dioxide gas. H2 SO4   H2 O + SO2 + [O] Cu + [O]

  CuO CuO +H2 SO4   CuSO4 + H 2 O ––––––––––––––––––––––––––––––––––– Cu + 2H2 SO4   CuSO4 + SO2 + 2H 2 O (iv) Action on salts (non-volatile nature) : Being a strong and less volatile acid, sulphuric acid liberates weaker and more volatile acids from their salts. (a) With dilute acid. Car b onates, bicar bona tes, sulphides, s ulphites, thiosulphates and nitrites are decomposed by dilute H2 SO4 at room temperature. For example : Na 2 CO3 + H2 SO4   [Na 2 SO4 + H2 CO3 ]

  Na 2 SO4 + CO2 + H2 O 2NaHCO3 + 2H2 SO4   [2NaHSO4 + 2H2 CO3 ]   2NaHSO4 + 2CO2 + 2H2 O Na 2 S + H2 SO4   Na 2 SO4 + H2 S Na 2 SO3 + H2 SO4   [Na 2 SO4 + H2 SO3 ]

  Na 2 SO4 + SO2 + H2 O

2NaNO2 + H2 SO4   [Na 2 SO4 + 2HNO2 ]

  Na 2 SO4 + NO + NO2 + H2 O (b)

With hot concentrated acid : Chlorides and nitrates, are decomposed by hot concentr ated H 2 SO 4 liberating their corresponding acids. For example :

NaCl + H2 SO4   NaHSO4 + HCl KNO3 + H2 SO4   KHSO4 + HNO3 Precipitation reactions: When the aqueous solutions of barium, strontium, calcium and lead salts are treated with dil. H2 SO4 , white precipitates of their corresponding metal sulphates are formed.

BaCl2 + H2SO4   BaSO4  + 2HCl (CH3 COO)2 Pb + H2 SO4   PbSO4  + 2CH3 COOH CaCl 2 +H2 SO 4   CaSO 4  + 2HCl Reaction with sulphur trioxide: SO 3 dissolves in concentrated H 2 SO 4 forming oleum. SO3 + H2 SO4   H2 S 2 O7

14. Uses of Sulphuric acid Sulphuric acid has a vast variety of applications both in industry as well as in the laboratory. The industrial strength of a nation largely depends on the amount of sulphuric acid it produces and consumes. Therefore, it is rightly called King of chemicals. Some important commercial uses of sulphuric acid are : (i) In chemical industry: Sulphuric acid is employed in the manufacture of hundreds of other chemicals such as hydrochloric acid, nit r ic acid, phosphor ic acid, sulphates and bisulphates, diethyl ether, etc. (ii) In fertilizer industry: Sulphuric acid is used to manufacture fer tilizer s like ammonium sulphate and super­phosphate of lime. (iii) In dyes, drugs, paints and pigments: It is used directly or indirectly in the manufacture of number of chemicals used as dyes, drugs, paints and pigments.

Na 2 S 2 O3 + H2 SO4   [Na 2 SO 4 +H 2 SO 3 +S]

  Na 2 SO4 + SO2 + H2 O + S www.betoppers.com

9th Class Chemistry

256 (iv)

In petroleum refining: Cr ude petroleum is treated with sulphuric acid to remove unwanted sulphur and other tarry compounds. (v) In the manufacture of explosives: A mixture of sulphuric acid and nitric acid called nitr ating mixtur e is used for nitration of organic compounds. This nitr ation pr ocess is used for the manufact ur e of lar ge number of explosives like dynamite, gun cotton, TNT (2, 4, 6-tri-nitrotoluene), picric acid, etc. (vi) In metallurgy: A number of metals like copper and silver are extracted from their ores using sulphuric acid. (vii) In pickling: Sulphuric acid is used for cleaning the surface of metals (pickling) before enamelling, electroplating and galvanizing. (viii) In lead storage batteries: Sulphuric acid undergoes electrolysis in the aqueous stage. (ix) As a laboratory reagent: It is widely used in labor atory as a drying and dehydrating agent.

Formative Worksheet 27.

28.

29.

30.

31.

The most efficient agent for the absorption of SO3 is : (A) 20% H2S2O7 (B) 98% H2SO4 (C) 50% H2SO4 (D) 80% H2SO4 When water is added to conc. H2SO4 the reaction is exothermic beacuse (A) H2SO4 is viscous (B) Hydrates of H2SO4 are formed (C) H2SO4 is corrosive (D) None of these H2SO4 has very corrosive action on skin because (A) It reacts with proteins . (B) It acts as an oxidizing agent. (C) It acts as dehydrating agent. (D) All In the reaction, 2Ag + H2SO4  Ag2SO4+ SO2+ 2H2O H2SO4 acts as (A) Reducing agent (B) Oxidising agent (C) Catalytic agent (D)Dehydrating agent

X FeS2 + Y O2   Z Fe 2 O3 + W SO 2 Find

15. Tests Tests for dilute sulphuric acid: (i) It gives a white precipitate of barium sulphate on adding it to barium chloride solution. BaCl2(aq) + H2SO4(aq)   BaSO4(aq)  + 2HCl(aq) (ii) It gives a white precipitate of lead sulphate when added to lead nitrate solution. Pb(NO3 )2 (aq)+H2SO4(aq)   PbSO4(aq)  + 2HNO3(aq) Tests for concentrated sulphuric acid: (i) On adding sodium chloride to concentrated sulphuric acid, colourless fumes of hydrochloric acid are given out which give dense white fumes with ammonia. NaCl(s) + H2SO4(aq)   NaHSO4(aq) + HCl(aq) HCl + NH 3   NH4 Cl (ii) On heating concentrated sulphuric acid with copper turnings, colourless sulphur dioxide gas with characteristic smell of burning sulphur is given out. Cu(s) + 2H2SO4(aq)   CuSO4(aq) +2H2O(l)+ SO2(g) www.betoppers.com

(A) 32.

X+Y Z+W

10 15

(B)

3 4

i)

A + 4B   8C

ii)

 2D 2C + B 

(C)

15 10

(D)

16 10

D + E  F iii) A is the octa atomic elementary molecule which belongs to 3rd period and VI A group. B is the second most abundant gas in atmosphere. E is commonly called king of chemicals. Identify A to F. A B C D E F (A) S 8 CO 2 SO3 SO2 H2SO3 H2SO4 (B) S 8 O2 SO2 SO3 H2 SO 4 H2 S 2 O 7 (C) S 8 N2 SO2 SO3 H2 SO 4 H2 S 2 O 7 (D) S 8 O2 SO3 SO2 H2 SO3 H2 SO 4

Compounds of Sulphur 33.

X + Y(g)

SO2

257 +Y

Z

38.

39.

When Na2SO3 reacts with dil.H2SO4 it forms (A) SO2 (B) Na2SO4 (C) H2O (D) All The products formed when NaHCO3 reacts with dil.H2SO4 . (A) Na2 SO4 (B) CO2 (C) H2O (D) Both 1 and 3

40.

Pb + 2H 2SO4   A + B  g  + 2C   

+ A

H2O +

34.

35.

36.

37.

H2S 2O7

Y is the second most abundant gas in atmosphere Find the molecular weight of Z and A. Z A (A) 63 108 (B) 36.5 116 (C) 80 98 (D) 54 22 98% conc. sulphuric acid has relative density of ________ at 15°C. (A) 5.84 (B) 3.24 (C) 2.86 (D) 1.84 The freezing point of H2SO4 is _______ (A) 0°C (B) 5°C (C) 10.4°C (D) 15.6°C Assertion(A): H2 SO4 should be diluted by pouring acids slowly into water. Assertion(B): H2 SO4 should be diluted by pouring water into water. Reason (R) : Adding of water to conc. H2SO4 suddenly converts it to steam causing dangerous splashing of the acid. (A) A1 and A2 are correct, R is the correct explanation of A1. (B) A1 and A2 are correct, R is the correct explanation of A2. (C) A1 is correct, A2 is incorrect, R is the correct explanation. (D) A1 and A2 are incorrect, R is correct. State True or False : A: H2SO4 forms a constant boiling mixture containing 98.5% of the acid boiling at 100°C. B: H2SO4 is immisible with water. C: Pure anhydrous H2SO4 is a good conductor of electricity. D: Conc. H2SO4 is non-corrossive in nature. A B C D (A) T F T F (B) F T F T (C) T T T T (D) F F F F

Identify A, B and C. A B (A) PbS SO2 (B) PbSO3 SO3 (C) PbSO4 SO2 (D) All 41.

42.

43.

44.

45.

C H2 O H2 O H2 O

Na 2 CO3 + H 2SO 4   A+ B+ C  s

g 

 

A B C (A) Na 2 S O2 H2 O (B) Na 2 SO 4 CO H2 O (C) Na 2 S CO H2 O (D) Na 2 SO 4 CO 2 H2 O A and B are the gases liberated by the action of H2SO4 on sodium sulphite and Zinc, respectively. Identify A and B. A B (A) H2 S SO2 (B) H 2 SO2 (C) SO2 H2 (D) SO2 SO2 x1, x2, x3 and x4 are the co-efficients of reactants and products formed when dil.H2SO4 is passed through Na2S. Find (x1 + x(B) : (x3 + x(D) (A) 1 : 1 (B) 1 : 2 (C) 2 : 3 (D) 3 : 2 A is an element that belongs to VIA group and 3rd period. A when heated in the presence of oxygen above 444°C and forms a gas B and traces of another gas C. D is a compound formed when C is dissolved in water. When D is passed through sodium sulphite, a solid E a gas B and a liquid F are formed. Identify A to F? A B C D E F (A) S SO3 SO3 H2SO4 Na2SO4 H2O (B) S H2O SO3 H2SO4 Na2SO4 H2O (C) S SO2 SO3 Na2SO4 H2SO4 H2O (D) S SO2 SO3 H2SO4 Na2SO4 H2O Which of the following gases can be dried by H2 SO 4 ? (A) O2 (B) N2 (C) Cl2 (D) All www.betoppers.com

9th Class Chemistry

258 46.

47.

48.

49.

50.

51.

52.

53.

The solid product formed on charring of carbohydrates is _________. (A) Sulphur (B) Carbon (C) Sodium (D) Magnesium Which of the following gases can’t be dried by 55. H2 SO 4 ? (A) N2 (B) O2 (C) Cl2 (D) NH3 Which of the following are oxidised by conc.H2SO 4 ? (A) Non-metals (B) Metals (C) Inorganic compounds (D) All conc.H SO x1 C12H22O11  2 4  x2C + x3H2O 56. Choose the correct option. x1 x2 x3 (A) 11 2 1 (B) 1 11 12 57. (C) 1 12 11 (D) 1 12 10 conc.H SO A  2 4  CuSO4 + 5H2O. Identify A. (A) Green vitriol (B) Blue vitriol (C) White vitriol (D) All 58. A : Conc. sulphuric acid acts as an oxidising agent. B : Sulphuric acid acts as dehydrating agent but not as drying agent. C : Conc. sulphuric acid is a volatile acid. D : Conc. sulphuric acid displaces more volatile acids from their salts. 59. Identify the correct statements (A) Only A and B are correct (B) Only B and C are correct (C) Only C and D are correct (D) Only A and D are correct How many moles of water of crystallisation are removed from 5 moles of blue vitriol when excess of conc.H2SO4 is passed through it? (A) 5 (B) 10 (C) 20 (D) 25

Na 2 CO3 + H 2SO4   A + B + C Identify A, B 60. and C. A B C (A) Na 4 SO 2 CO 2 H2 O (B) Na 2 SO 4 CO 2 H2 O 40. (C) Na 2 SO 3 CO 2 H2 O (D) NaSO4 CO 2 H2 O

A B C (A) HSO4 HCl H2 S 2 O 7 (B) HSO2 HCl H2 S 2 O 7 (C) H4 SO 4 HCl H2 S 2 O 7 (D) H2 SO 4 HCl H2 S 2 O 7 A is the gas formed by the oxidation of SO2. B is a compound formed, when ‘A’ is passed through H2SO4. Identify A and B. A B (A) SO2 H2 S 2 O 7 (B) SO3 H2 S 2 O 7 (C) SO4 H2 S 2 O 7 (D) SO H2 S 2 O 7 A is the gas formed by the action of H2SO4 on sodium carbonate. Find the number of moles of A is 220 g of it. (A) 2 (B) 3 (C) 4 (D) 5 ‘X’ is the gas formed by the decomposition of sodium sulphite by dil.H2SO4. Find the number of molecules of ‘X’ present in 8.5 g of it. (N = Avogadro’s number) (A) 0.25 N (B) 0.5 N (C) N (D) 2N C is the gas liberated by the action of conc. H2SO4 on copper. Find the total number of double covalent bonds and dative bonds respectively present in 64 g of C. (N = Avogadro’s number) (A) N, 2N (B) N, 3N (C) 3N, 4N (D) N, N A is the solid product formed by the action of di.H2SO4 on CaCl2. B is the liquid product formed by the disolution of SO3 in dil.H2SO4. C is the gaseous product formed by the action of dil.H2SO4 on potassium bicarbonate. If 100g CaCl2 of A, B and C are taken then which one will possess maximum number of moles? (A) A (B) B (C) C (D) All contain sam number of moles The molecular weight of the liquid formed by the action of conc.H2SO4 on mercury: (A) 19 (B) 18 (C) 17 (D) 16

Conceptive Worksheet

54.

CaC 2 + A   CaSO4  +2B SO3 + A  C Identify A, B and C

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41.

The number of dative bonds in sulphuric acid molecule is : (A) 0 (B) 1 (C) 2 (D) 4 Basicity of any oxyacid of sulphur is : (A) 3 (B) 4 (C) 2 (D) 1

Compounds of Sulphur 42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

53.

54.

What happens to the colour of litmus paper when a drop of H2SO4 is added to it (A) It turns red to blue (B) It turns blue to red (C) It gets destroyed (D) It is unaffected 98% H2SO4 is : (A) Pyrosulphuric acid (B) Oleum (C) Azeotropic mixture (D) None of these H2SO4 has high boiling point this is due to the presence of (A) H-bonding (B) Affinity towards water (C) Its acidic nature (D) Its covalent nature The boiling point and freezing point of sulphuric acid are : (A) 590K, 283K (B) 283K, 590K (C) 582K, 285K (D) 590K, 273K Choose the metals with which sulphuric acid does not react. (A) Cu, Zn (B) Pb, Na (C) Au, Pt (D) K, Cu Which of the following metal does not liberate hydrogen from sulphuric acid? (A) Iron (B) Zinc (C) Magnesium (D) Copper Copper turning when heated with conc.H2 SO4 gives (A) O2 (B) SO2 (C) SO3 (D) H2S Which of the following gases are liberated when sodium nitrite reacts with dilute sulphuric acid (A) NO (B) NO2 (C) Both 1 and 2 (D) None of these H2SO4 is : (A) Colourless (B) Syrupy (C) Hygroscopic (D) All H2SO4 has ________ taste. (A) Sour (B) Bitter (C) Sweet (D) Salty Sulphuric acid has ______ affinity for water. (A) Weak (B) Strong (C) Very weak (D) No When H 2 SO 4 dissolves in water, ________ amount of heat is produced. (A) Least (B) Large (C) No (D) Considerable Conc. H 2 SO 4 is always diluted by adding ____A_____ to ____B_____. A B (A) Water Acid (B) Acid Water (C) Conc. acid Conc. acid (D) All the above

259 55.

56.

57.

58. 59.

60.

61.

62.

63.

64.

65.

66.

67. 68. 69. 70.

71. 72.

In aqueous solution, H2SO4 ionises in ________ steps. (A) 4 (B) 3 (C) 2 (D) 1 H2SO4 is _________ in nature. (A) Mono basic (B) Di basic (C) Tri basic (D) Tetra basic The number of replacable hydrogen ions present in 1 molecule of H2SO4 is _______. (A) 4 (B) 3 (C) 2 (D) 1 The molecular weight of H2SO4 is _______ (A) 36.5 (B) 63 (C) 98 (D) 106 The products formed, when H2SO4 reacts with NaOH: (A) Salt (B) Water (C) Acid (D) Both A and B Dil.Sulphuric acid reacts with active metals and liberates. (A) H2S (B) SO2 (C) O2 (D) H2 Dil.sulphuric acid decomposes carbonates to evolve. (A) CO (B) CO2 (C) SO2 (D) SO3 Dil.sulphuric acid decomposes ________ to evolve SO2 . (A) Metals (B) Metal carbonates (C) Metal sulphides (D) Metal sulphites _________ gas is liberated by the action of dil.H2SO4 on sodium sulphide. (A) H2S (B) SO2 (C) O2 (D) H2 The molecular weight of the gas liberated by the action of di.H2SO4 on sodium bi sulphite. (A) 16 (B) 32 (C) 48 (D) 64 The moleculear weight of the gas liberated by the action of dil.H2SO4 on sodium bi carbonate. (A) 28 (B) 44 (C) 64 (D) 4 The number of moles of SO2 formed, when excess of H2SO4 is passed through one mole of silver. (A) 1 (B) 2 (C) 3 (D) 4 Sulphuric acid removes ________ from sugar. (A) O2 (B) H2 (C) H2O (D) SO2 Sulphuric acid removes _______ from blue vitiriol. (A) O2 (B) H2 (C) H2O (D) SO2 Conc.sulphuric acid on decomposition give (A) H2 O (B) SO2 (C) [O] (D) All Conc.H2SO4 acts as oxidising agent because of liberated of (A) H2 O (B) SO2 (C) [O] (D) All Conc. H2SO4 oxidises carbon to _________ (A) CO (B) CO2 (C) H2CO3 (D) All Conc.H2SO4 oxidises sulphur to _________ (A) SO2 (B) SO3 (C) CuSO4 (D) All www.betoppers.com

9th Class Chemistry

260

Summative Worksheet 1.

Na 2SO 2 + H 2SO 4   +H 2 O + SO 2 Na 2SO3 + 2HC   2B + H 2 O + SO 2

2.

3.

4. 5. 6. 7.

8.

9.

10.

11.

2C + H 2SO 4   Na 2SO4 + 2H 2 O + 2D Identify A, B, C and D A B C D (A) Na2 SO4 NaCl Na 2 SO 4 SO2 (B) Na2SO4 NaCl NaHSO3 SO2 (C) Na2SO3 Na 2 SO 4 NaCl SO2 (D) Na2SO4 H 2 O 2 NaCl SO2 Write fully balanced equations when : (i) FeS2 (iron pyrites) is heated in air. (ii) Copper turnings are heated with concentrated sulphuric acid. (iii) Potassium bisulphite is treated with dilute hydrochloric acid. (iv) Sulphur is burnt in air or oxygen. State one similarly and three differences between bleaching action of chlorine and sulphur dioxide gas. Why is bleaching done by sulphur dioxide is temporary in nature ? Calcium oxide is not used for drying SO2. Give reason. Explain why sulphur dioxide is called sulphurous anhydride. When sulphur is burnt in air, the gas so formed strongly fumes in moist air. Explain your answer. Write chemical equations. (a) Why is sulphur trioxide formed in this process not absorbed directly in water ? (b) Why is vanadium pentoxide considered a better catalyst than platinised asbestos ? (c) Why is heating of catalyst discontinued the moment the oxidation of sulphur dioxide takes place ? a) In the manufacture of sulphuric acid by contact process, great care is taken to purify the mixture of acid and sulphur dioxide, especially to free it from arsenic impurities. Explain. b) Write balanced equations of the reactions in the preparation of the following : i) ammonia from ammonium chloride. ii) chlorine from bleaching powder. "Sulphuric acid" behaves as an acid when diluted with water. Why is this so ?

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12.

13.

14.

15.

16.

Give two balanced reactions of each type (for i to iii) to show the following properties of sulphuric acid : i) Acidic Nature ii) Hygroscopic nature Oleum or fuming H2SO4 is : (A) A mixture of conc. H2SO4 and oil (B) Sulphuric acid which gives fumes of sulphur dioxide (C) Sulphuric acid saturated with sulphur trioxide i.e., H2S2O7 (D) A mixture of sulphuric acid and nitric acid Match the following w.r.t occurence of H2SO4. Column - I Column - II A) Hot mineral springs p) Free state B) MgSO4 q) Combined state C) Sulphide rock beds D) BaSO4 A B C D (A) q p q p (B) p p q q (C) q p q p (D) p q p q Name the product formed when SO3 is passed through 98% H 2 SO 4 during contact process_______. (A) Sulphurous acid (B) Marshall’s acid (C) Oleum (D) King of chemicals Statement A : Decrease in temperature, increases the yield of SO3. Statement B : Use of pure gases decreases the yield of SO3. (A) ‘A’ is true, ‘B’ is false. (B) ‘A’ is false, ‘B’ is true. (C) Both ‘A’ and ‘B’ are true. (D) Both ‘A’ and ‘B’ are false. Assertion A : Purity of gases is an essential condition for the maximum yield of SO3 during contact process. Reason R : The impurity present in the gases decreases the efficiency of catalyst. (A) Both assertion and reason are correct and reason is the correct explanation of assertion. (B) Both assertion and reason are correct but reason is not the correct explanation of assertion. (C) Assertion is correct and reason is incorrect. (D) Assertion is incorrect and reason is correct.

Compounds of Sulphur 17.

18.

19.

20.

261

X, Y and Z are the gases liberated by the action of dil.H2SO4 on Baking soda, Sodium bisulphite and Iron respectively. Arrange then in the increasing order of their rates of diffusion? (A) X < Y < Z (B) Z < Y < X (C) Y < X < Z (D) X < Z < Y A, B and C are the gases liberated by the action of dil.H2SO4 on Sodium carbonate, Sodium sulphide and Magnesium respectively. Identify A, B and C? A B C (A) CO 2 SO2 H2 (B) CO SO3 H2 S (C) CO SO2 H2 (D) CO 2 SO4 H2 S Find the weight of H2O formed when one mole of sugar is charred by the excess of conc.H2SO4. (A) 189 g (B) 198 g (C) 90 g (D) 36 g x1, x2, x3 and x4 are the respective number of water molecules formed by the action of excess of conc.H2SO4 on washing soda, Blue vitriol, Gree vitriol and White vitriol. Find

24. 25. 26. 27. 28. 29.

30.

x1 + x 2 x3 + x4

14 (A) 15

15 (B) 14

12 (C) 17

17 (D) 12

21.

2NaHCO3 + 2H2SO4   2A + 2B + 2C Identify A, B and C. A B C (A) NaHSO2 CO 2 H2 O (B) NaHSO3 CO 2 H2 O (C) NaHSO4 CO 2 H2 O (D) Na 2 SO 4 CO 2 H2 O

22.

NaC + H 2SO4   A + B. Identify A and B. A B (A) NaHSO2 HCl (B) NaHSO3 HCl (C) NaHSO4 HCl (D) Na 2 SO 4 HCl

23.

BaC 2 + H 2SO4   A  +2B Identify A and B. A B (A) BaSO2 HCl (B) BaSO4 HCl (C) BaSO3 HCl (D) Ba 2 SO 4 HCl

31.

32.

33.

Explain why sulphur dioxide is called sulphurous anhydride. Give a chemical test to distinguish between dilute sulphuric acid and dilute hydrochloric acid. Give the test for sulphate radical. What is the function of conc. sulphuric acid in the preparation of CO from oxalic acid? "Sulphuric acid" behaves as an acid when diluted with water. Why is this so ? Give two balanced reactions of each type (for i to iii) to show the following properties of sulphuric acid : i) Acidic Nature ii) Oxidising agent iii) Non-volatile nature iv) Hygroscopic nature Choose the property of sulphuric acid (A, B, C or D), which is relevant to each of the following : A : Dil acid (typical acid properties) B : Non-volatile acid C : Oxidising agent D : Dehydrating agent i) Preparation of hydrogen chloride ii) Preparation of ethene from ethanol iii) Preparation of copper sulphate from copper oxide A given oily liquid when added to water showed the following properties : The reaction occurring was exothermic and the solution forms a white precipitate, when barium chloride solution was added. a) Name the oily liquid b) Give an equation for its reaction with barium chloride solution. c) When the resulting solution was electrolysed using platinum electrode, name the product at each electrode. A colourless oily liquid was poured slowly into a large volume of cold water. With constant stirring, the temperature of the water rose considerably. On testing the solution, only hydrogen and sulphate ions could be detected. i) How would you carry out the test for the presence of hydrogen gas ? ii) Name the oily liquid. A colourless liquid had the following properties : i) It slowly turned blue copper sulphate into white powder. ii) When warmed with sulphur, a gas smelling of burning sulphur was given off.

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9th Class Chemistry

262

34.

iii) When a little liquid was poured into water, the water became hot. Suggest a name for the colourless liquid and support your answer by explaining each test in turn. Give reasons for the following : i) Sulphuric acid forms two types of salts with NaOH ii) Red brown vapours are produced when concentrated sulphuric acid is added to potassium bromide. iii) A piece of wood becomes black when concentrated sulphuric acid is poured on it. iv) Brisk effervescence is seen when sulphuric acid is added to sodium carbonate.

8.

HOTS Worksheet 1.

2.

3. 4.

5.

6.

7.

Write fully balanced equations when : (a) FeS2 (iron pyrites) is heated in air. (b) Copper turnings are heated with concentrated sulphuric acid. (c) Potassium bisulphite is treated with dilute hydrochloric acid. (d) Sulphur is burnt in air or oxygen. When sulphur is burnt in air, the gas so formed strongly fumes in moist air. Explain your answer. Write chemical equations. Why is bleaching done by sulphur dioxide is temporary in nature ? (a) Why is sulphur trioxide formed in this process not absorbed directly in water ? (b) Why is vanadium pentoxide considered a better catalyst than platinised asbestos ? (c) Why is heating of catalyst discontinued the moment the oxidation of sulphur dioxide takes place ? Explain the following : (a) Why are the wooden shelves on which concentrated sulphuric acid bottles kept, stained black ? (b) Why is concentrated sulphuric acid always added to water and not the water to concentrated sulphuric acid ? Why does the mixture get hot ? What will you observe when concentrated sulphuric acid is poured on (i) sugar crystals (ii) Copper sulphate crystals ? a) An important industrial reaction is represented

9. 10.

11.

12.

by the equation 2SO2 +O2  2SO3. The 13. formation of sulphur trioxide is exothermic. State the conditions of temperature and state the name of catalyst required for the above reaction to take place. www.betoppers.com

b) Describe the change you would observe : a) When concentrated sulphuric acid is added to sugar. b) When concentrated sulphuric acid is added to formic acid, HCOOH, and warmed. c) When burning magnesium is introduced into a gas jar containing sulphur dioxide. Name the products formed and write an equation for the reaction. d) In what capacity is sulphur dioxide acting ? a) A mixture of iron dust (iron fillings) and sulphur was heated. The black solid residue M reacts with dilute sulphuric acid to give a foul smelling gas N. N when ignited burned with a blue flame and deposited a yellow residue O. On passing N into lead (II) nitrate solution, it gave a black precipitate P and a colourless solution Q. Name the substances M, N, O, P and Q and write an equation for each. b) Concentrated sulphuric acid (i) has a great affinity for water (ii) is an oxidising agent (iii) is least volatile acid (iv) is dibasic acid. Give one reaction each with equation to show these properties of sulphuric acid. How would you distinguish between sulphur dioxide and hydrochloric acid gas ? a) Both ammonia and sulphur trioxide are manufactured by catalysed reaction of gases. Give an equation for each of these reactions and name the catalyst used in each case. b) Give two household uses of ammonia. What property of sulphuric acid is made use o in each of the following cases ? Give an equation for reaction in each case : a) Production of hydrogen chloride gas when it reacts with a chloride (say sodium chloride) b) In the preparation of carbon monoxide gas from formic acid. c) As a source of hydrogen by diluting it and adding a strip magnesium. d) In the preparation of sulphur dioxide by warming a mixture of concentrated sulphuric acid and copper turning. Explain briefly how you will obtain sulphur dioxide gas from sodium sulphite. a) Give the formula of the substance which is responsible for tarnish on silver spoon. b) What is name given to the salts of sulphurous acid ?

Compounds of Sulphur

14.

c) Sulphuric acid can behave : i) as an acid ii) as a less volatile acid iii) as an oxidising agent iv) as a drying agent for certain gases. State in words, an example of each case. a) Write down the word equations for each of the following reactions : i) Conc. sulphuric acid + Copper  ii) Conc. sulphuric acid + Sodium chloride  iii) Conc. nitric acid + Copper  iv) Conc. hydrochloric acid + Manganese dioxide 

15.

16.

17.

18.

v) Sodium hydroxide solution + Zinc  b) What is the formula for anhydride of sulphuric acid ? State one similarly and three difference between bleaching action of chlorine and sulphur dioxide gas. a) Give one large scale use of sulphur. b) Write balanced molecular equations for the laboratory preparation of : i) Chlorine by the oxidation of hydrochloric acid. ii) Nitrogen from ammonium nitrite. c) i) "Sulphur dioxide is a powerful reducing agent. But is also acts as an oxidising agent". State in words or give balanced equations of two such reactions in which it acts as an oxidising agent. ii) Sulphuric acid is said to be a dibasic acid. What is meant by the term "dibasic" d) Calculate the vapour density of sulphur dioxide gas. (S = 32, O = 16) a) In the manufacture of sulphuric acid by contact process, great care is taken to purify the mixture of acid and sulphur dioxide, especially to free it from arsenic impurities. Explain. b) Write balanced equations of the reactions in the preparation of the following : i) ammonia from ammonium chloride. ii) chlorine from bleaching powder. a) Concentrated sulphuric acid should be kept in air-tight bottles. Why ? b) Write a balanced equation for the preparation of sulphur dioxide by the action of hot conc. sulphuric acid on sulphur.

263 19.

20.

21.

State how (i) PbS (ii) Ag2S (iii) ZnS (iv) CuS can be obtained from H2S. Give chemical equations only. State the colour of each sulphide. Give two reactions to illustrate : i) The reducing property of sulphur dioxide. ii) The oxidising property of sulphur dioxide. iii) The acidic nature of sulphur dioxide. Complete and balance the following equations : 1. S + H2SO4  2. C + S  (conc) 3. FeS2 + O2  4. PbS + O2  5. ZnS + O2  6. NaHSO3 + HCl  7. Na 2 SO3 +H2SO4  8. KMnO4 + H2O+SO2  9. K2Cr2O7 + H2SO4+SO2  10. Mg+ SO2  11. Ca(OH)2 + SO2  12. CaO + SO2  13. SO2 + Cl2  14. PbO2+SO2  15. FeCl3+H2 O+SO2  16. Ca(HSO3)2+H2SO4  17. H2S + H2SO4  18. Fe + H2SO4  (Dil.) 19. Zn + H2SO4  20. HNO3+SO2  (Dil.)  21.FeSO4.7H2O   

22. H2C2 O4+H2SO4  (conc.) 23. C12 H22 O11 +H2SO4  24. HI + H2SO4  (conc.) (conc.) 25. Cu

+ H2SO4 

26. CuO + H2SO4  (conc.) (Dil.) www.betoppers.com

9th Class Chemistry

264 27. SO2+H2O+Cl2 

22. 23.

24. 25. 26. 27.

28.

29.

30.

28. S + HNO3  (conc.) How can you prove that sulphur dioxide is an acidic oxide ? i) "Sulphur dioxide is a powerful reducing agent. But is also acts as an oxidising agent". State in words or give balanced equation of two such reactions in which it acts as an oxidising agent. 2. ii) Sulphurous acid is said to be dibasic acid. What is meant by the term "dibasic" ? Calcium oxide is not used for drying SO2. Give reason. Give a reason why chlorine is not used to bleach silk. What is observed when hydrogen sulphide gas is passed through lead acetate solution? From the gases : ammonia, hydrogen chloride, hydrogen sulphide, sulphur dioxide, select the 3. following : a) This gas can be oxidised to sulphur. b) This gas decolourises potassium permanganate solution. c) This gas can be obtained by the reaction between copper and concentrated sulphuric acid. Name the agent responsible for : a) Offensive odour of rotten eggs b) Blackening of a painting. d) Blackening of silver coin by atmospheric air. What do you observe when H2S is passed through 4. a solution of (i) Copper sulphate (ii) Zinc sulphate (iii) Lead nitrate ? How can you prove that H2S contains sulphur ?

IIT JEE Worksheet 1.

a) "Sulphur trioxide and water react to form sulphuric acid." Sulphuric acid may act as each of the following : 1) An acid, forming sulphates which are soluble in water. 2) A compound, forming sulphates which are insoluble in water. 3) A dehydrating agent. 4) A drying agent 5) An oxidising agent. Which ONE of the above properties 1 to 5 are shown by sulphuric acid when : i) concentrated sulphuric acid is added to sugar ?

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ii) dilute sulphuric acid reacts with lead nitrate? iii) dilute sulphuric gas is added to sodium hydroxide solution ? iv) hydrogen sulphide gas is oxidising through concentrated sulphuric acid ? b) "Sulphur dioxide acts as an oxidising agent and a reducing agent". Give one balanced equation for the reaction in which sulphur dioxide acts as : i) an oxidising agent ii) a reducing agent a) In contact process, sulphur dioxide can be oxidised to sulphur trioxide in the pressure of platinised asbestos, which acts as a catalyst. The reaction is exothermic. Explain what is meant by the terms in bold italics. b) Silver cannot be used to prepare hydrogen from hydrochloric acid or sulphuric acid. Why ? a) A compound X is warmed with dil. sulphuric acid. It gives off a colourless gas which turns acidified potassium dichromate paper green. Name the anion present in the compound. b) The following statements are correct only, certain conditions. Rewrite each of the statement, including appropriate condition(s) and underline them. i) Sulphur dioxide is a bleaching agent. ii) Oxalic acid and sulphuric acid react to produce carbon monoxide and carbon dioxide. a) What is the meaning of term "water of crystallisation ?" b) i) Name two acids which can be prepared by using sulphuric acid ? ii) Which property of sulphuric acid is used in preparing acids, named by you. c) Write balanced equations for obtaining the following gases from dilute sulphuric acid. i) hydrogen ii) carbon dioxide iii) sulphur dioxide d) Name the another metal sulphide [excluding Iron pyrites] which on roasting produces sulphur catalyst.

Compounds of Sulphur

265

5. Sulphuric

Sulphur

acid

B

Sulphur

C

dioxide

E

6.

D

Sulphur trioxide

F

Sodium sulphite

i)

Name a catalyst used in converting sulphur dioxide to sulphur trioxide as shown in step C. ii) In the manufacture of sulphuric acid, sulphuric trioxide is not directly absorbed in water. Instead a two-step procedure is followed. Write two equations involved in the step D. 7.

a) Copy and complete the following table. Column R has names of gases to be prepared using substance you enter in column P, along with dilute or concentrated sulphuric acid as indicated by you in column Q. Column P

8.

iii) What type of a chemical substance will liberate sulphur dioxide from sodium sulphite in step E? iv) Write an equation for step F, in which sulphur dioxide is converted into sodium sulphate. a) i) Which concentrated acid will oxidise sulphur directly to sulphuric acid ? Write the equations for the reaction. ii) Complete the following sentence choosing the correct word from brackets. "Concentrated sulphuric acid is used in laboratory preparation of nitric acid and hydrochloric acid, because it is –––––––– (less volatile/stronger) in comparison to these acids. b) Write equations for the laboratory preparation of the following : i) Iron sulphate from iron ii) Copper sulphate from copper iii) Lead sulphate from lead nitrate iv) Sodium sulphate from sodium carbonate.

Column Q

Column R

Substance reacted with acid Dilute or conc. H2SO4

Gas

__________ __________ __________

Hydrogen

__________ __________ __________

Carbon dioxide Chlorine

b) Write equations for the laboratory preparation of : i) Sodium sulphate, using dilute sulphuric acid. ii) Lead sulphate, using dilute sulphuric acid. a) Choose the correct word from the brackets to complete the sentences. i) Sulphur can be converted into sulphuric acid by using ––––––– (concentrated/dilute) nitric acid. ii) Write the equation for the reaction in a) (i). iii) Sodium nitrate on reacting with ––––––– –– (concentrated/dilute) sulphuric acid produces nitric acid. iv) Write the equation for the reaction (a) (iii) b) Write equations for the following reactions : i) Dilute sulphuric acid producing hydrogen. ii) Lead nitrate solution and dilute sulphuric acid. c) Explain how you can distinguish between dilute sulphuric acid and dilute nitric acid, using barium chloride solution. www.betoppers.com

9th Class Chemistry

266 9.

10.

11.

12.

a) State how you can obtain : 14. i) sulphur dioxide from sulphur. ii) hydrogen sulphide from iron (II) sulphide. b) Some bacteria obtain their energy by oxidising and producing sulphuric acid as by product. In the laboratory or industrially, the first step in the conversion of sulphur to sulphuric acid is to produce sulphur dioxide gas. The sulphur dioxide gas is converted to sulphur trioxide which reacts with 15. water producing sulphuric acid. i) Name a catalyst which is used industrially to speed up conversion of sulphur dioxide to sulphur trioxide. ii)Write the equation for the reaction in (b), (i), Why does this reaction supply energy ? iii)Name the compound formed when sulphur trioxide reacts with sulphuric acid. a) Write word/words to make the statement given below as correct as possible, "copper sulphate crystals are dehydrated by sulphuric acid. b) i) State the colour of flame, when sulphur burns in air. Name the main product formed. ii) The above product of burning sulphur dissolves in water to form a new compound. Name the compound. a) i) Write an equation for the laboratory preparation of sulphur dioxide from sodium sulphate. ii) How is sulphur dioxide collected ? iii) What does the method of collection tell you about the density of sulphur dioxide ? iv) What do you see, when sulphur dioxide is bubbled through acidified potassium dichromate solution ? b) Write an equation to show the action of sulphur dioxide as (i) a reducing agent (ii) an oxidising agent (iii) an acid anhydride. c) i) What is similarity in the use of sulphur dioxide and chlorine as bleaching agents? ii) When chlorine is involved in bleaching, what is the type of chemical reaction that changes, coloured compound to colourless compound ? iii) What is the reason for not using chlorine for bleaching wool ? The questions below refer to the flow sheet :

Powdered

C

roll sulphur

E Sodium

dioxide D

13.

Sulphur

F

sulphite

Write one equation in each case about changes C, D, E and F. What happens when i) Sulphur dioxide is passed through acidified potassium permanganate solution. ii) Sulphur dioxide gas is treated with hydrogen sulphide ?

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Sulphur was burnt in a gas 'X' and it produced another gas 'Y'. The gas 'Y' was acidic in nature. Gas 'y' oxidised another gas 'Z' to sulphur in aqueous solution. a) Name the gases 'X', 'Y' and 'Z'. b) Write equation for the oxidation of gas 'X' by gas 'Y'. Complete and balance the equations : i) Na2SO3+H2SO4  ii) CaSO3+HCl  (Dil) (Dil) iii) ZnS + O2  iv) FeS2+O2  v) Mg + SO2  vi) Ca(OH)2+SO2  vii) Na2O+SO2  viii) SO2 +O2  ix) SO2+Cl2  x) H2S + SO2  xi) Cl2 + H2O + SO2  xii) Fe2 (SO4)3 +H2 O+SO2 



IIT FOUNDATION Class IX

CHEMISTRY SOLUTIONS

© USN Edutech Private Limited The moral rights of the author’s have been asserted. This Workbook is for personal and non-commercial use only and must not be sold, lent, hired or given to anyone else.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of USN Edutech Private Limited. Any breach will entail legal action and prosecution without further notice.

Utmost care and attention to the details is taken while editing and printing this book. However, USN Edutech Private Limited and the Publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in.

Published by

:

USN Eductech Private Limited Hyderabad, India.

CONTENTS 1.

Language of Chemistry

..........

267 - 276

2.

Elements, Compounds & Mixtures

..........

277 - 278

3.

Atomic Structure

..........

279 - 288

4.

Periodic Classification

..........

289 - 296

5.

Chemical Bonding

..........

297 - 306

6.

Gaseous State

..........

307 - 322

7.

Mole Concept

..........

323 - 348

8.

Solutions

..........

349 - 356

9.

Compounds of Carbon

..........

357 - 362

10.

Compounds of Nitrogen

..........

363 - 366

11.

Compounds of Sulphur

..........

367 - 377

1. LANGUAGE OF CHEMISTRY SOLUTIONS 13. 3

FORMATIVE WORKSHEET

Total weight Number of atoms  weight of singleatom

1. 2.

reduced by 25%. A = Z + n.

3.

has 10 electrons  X has 10 electrons Neutral atom of 'X' has 13 electrons   Number of protons in it is 13  Z = 13  Aluminium. False. Atomic number is the number of electrons present in neutral atom of an element. This is because only in an neutral atom the number of electrons are equal to the number of protons. False. Hydrogen does not contain a neutron. No.of neutrons in Silicon is 28 – 14 = 14 No.of neutrons in Phosphorus is 31 – 15 = 16

4.

5. 6.

N

3

–3

 Ratio 

14 7  i.e., 7 : 8 16 8

7. 8.

Protons never change during a chemical reaction. No. of electrons and protons in nitrate ion NO3–1 = 7 + 24 + 1 = 32, 7 + 24 = 31 No. of electrons and protons in sulphate ion SO4–2 = 16 + 32 + 2 = 50, 16 + 32 = 48. 9. Weight of one atom of an element = Atomic weight × amu. x amu, x × 1.66 × 10–24g , 1.66x × 10–27kg 10. Atomic weight Weight of one atom of that element 1 weight of th of C 12atoms . 12 Now ratio of weight of atom of an element to its 

atomic weight is equal to 1 amu, (or) mass of

1 12

th of C – 12 isotopic atom 11. 2 40  1.66 1024  40 1.66  1024  mass number = 40 and the element is either calcium or Argon.  Number of protons = 20 (or) 18. 12. 3 Weight of He atom = 4 amu = 4 × 1.66 × 10–24g

Atomic weight 

weight of single atom = (Atomic weight) amu = 39 amu = 39 × 1.66 × 10–24g. Total weight= Gram atomic weight = 39g  Number

of

atoms



39g 39  1.66  10 24 g

1024  6.023  1023 1.66 14. (2, 3) Atomic weight of silicon = 28  weight of one atom of silicon = 28amu  Weight of 100 atoms of silicon = 28 × 100 = 2800 amu. 15. 1 i) Sodium  monoatomic (Na) ii) Helium  monoatomic (He) iii) Oxygen  diatomic (O2) iv) Ozone  triatomic (O3) v) sulphur  polyatomic (S8) 16. 4 Molecular weight has no unit as it is a ratio of the mass of single molecule of a compound to the mass 

1 th of C – 12 isotopic atom. 12

of

17. 4 Atomic weight 

weight of one atom 1 weight of th of C  12atom 12



weight of one atom Its atomic weight

= weight of

1 th of C – 12 atom = x 12

Molecular weight =



Weight of one molecule 1 Weight of th of C-12 atom 12

Weight of one molecule Its molecular weight

9th Class Chemistry

268 = Weight of

27.

1 th of C - 12 atoms = y 12

S.No i)

 x = y..

18. (i)

H2SO4

= 2 + 32 + 4 × 16 = 98

(ii) HCl = 1 + 35.5 = 36.5 (iii) NaCl = 23 + 35.5 = 58.5 (iv) HNO3 = 1 + 14 + (3 × 16) = 63 (v) MgCl2 = 24 + 2 × 35.5 = 95 (vi) Na2CO3 = 2 × 23+12+3×16 = 106 (vii) Na2SO4 = 2 × 23+32 +4×16= 142 (viii) SO2 = 32 + 2 × 16 = 64 (ix) CO2 = 12 + 2 × 16 = 44 (x) O3 = 3 × 16 = 48 19. 3 To find out the molecular weight, we need to find the weight of one molecule in terms of amu.  Weight of one molecule in terms of amu

Ion

Formula

Acetute

CH 3COO –

Valency

–

1

ii)

Cyanide

CN

1

iii)

Nitrate

NO3

1

iv)

Carbonate

CO32

2

HO2

1

v)

Hydrogen peroxide

vi)

Chromate

CrO24 

2

vii)

Manganate

MnO 24

2

viii)

Sulphate

SO 32

2

ix)

Oxalate

C 2O 4

2

2

x)

Zincate

ZnO 22 

2

22

weight in grams 1.66  10 = = 100 amu. 1.66×1024 1.66×1024  molecular weight of the compound = 100. 20. 3 The weight of ammonia molecule = 17 amu. = 17 × 1.66 × 10–24g =

28. Set - I Ions i) Hydrogen sulphide

Formula Valency HS – 1

ii) Dichromate

Cr2 O72

2

iii) Sulphate

SO24

2

iv) Superoxide

O2

1

24. 3

v) Hypophosphite

H2 PO2

1

25. 2

Set - II Ions i) Formate

21. False. This is because simple ions are not radicals. But all radicals are complex ions. 22. 2 23. 1

26.

S.No Ion i) Potassium ii) Cuprous iii) iv)

Aurous Stannous

Formula K+ Cu+

Valency 1 1

Au+ Sn+2

1 2

iii) Fluoride

+2

v) Phosphonium

v) vi)

Plumbous Zinc

Pb Zn+2

2 2

vii)

Chromous

Cr+2

2

+2

2 2 2

viii) ix) x)

Mercuric Ferrous Radium

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Hg Fe+2 Ra +2

ii) Bisulphite iv) Ammonium

Formula HCOO –

Valency 1

HSO3 F–

1

NH

 4

1 1

1 PH4  b = y1 + y2 + y3 + y4 + y5 =1+1+1+1+1=5

a 7  b 5 29. a - iii, d - iii, c - iv, d - iv 30. i-e; ii-d; iii-c; iv-i, v-b; vi-c; vii-b, viii-c 31. a1 = 1; a2 = 3 ; a3 = 2, a4 : 4 : a5 = 2 ; a6 = 3  x0 = 1 + 3 + 2 + 4 + 2 + 3 = 15 b1 = 1 ; b 2 = 1 ; b 3 = 2 ; b 4 = 2 ; b 5 = 2 ; b 6 = 1  y=1+1+2+2+2+1=9 x 15 5    y 9 3 

Language of Chemistry Solutions

269

32. 3 33. 3 34. 3 35. 1 36. Metal ‘M’ is known to exist in two valence states i) M2+ (ous state) ii) M3+ (ic state) Name of compound Chemical formulae M2+(ous state) M3+(ic state) i) Metal nitrate M(NO3) 2 M(NO3) 3 ii) Metal ferricyanide M3[Fe(CN)6] 2 M[Fe(CN)6 ] iii) Metal hypophosphite M(H2 PO 2 ) 2 M(H2 PO 2 ) 3 iv)Metal nitride M3N2 MN v)Metal phosphide M3P2 MP. 37. a) Ca 3(PO4 )2 b) Na2S2O3 c) Mg3N2 d) MnO e) BaS f) Mg(NO3)2 38. 4 39. 1, 4 40. 4 41. 3 42. The given reactions is  CaCO3   CaO+ CO2



1

mole

of

 CaCO3  1mole of CaO+1mole of CO2  100 grams of  CaCO3   56g CaO  44g CO2   50g of CaCO3   28g CaO  22g CO2 43. Rewriting Mg + CO2   MgO + C No. of Mg atoms 1 1 No. of C atoms 1 1 No. of O atoms 2 1 We see the number of Mg atoms and C atoms on both sides are equal. The number of oxygen atoms on product side is 1 and on the reactant side is 2. To make oxygen atoms equal multiply MgO by 2 we get, Mg + CO2   MgO + C Now, to make Mg atoms equal on both sides, multiply Mg of L.H.S by two. Thus, each kind of atoms on both sides of the equation are equal. Thus, the balanced equation is 2Mg + CO2   2MgO + C 44. Writing this equation keeping oxygen in the atomic form, we get KClO3   KCl + O In this equation, we see that only the number of oxygen atoms are not equal on both sides. To balance oxygen atoms, multiply O by 3; we get

KClO3   KCl + 3O (Atomic equation) It is a balanced atomic equation. To make oxygen molecular, multiply the remaining constituents by 2. 2KClO3   2KCl + 3O (Atomic equation) It is a balanced molecular equation. 45. Writing the equation, keeping oxygen in the atomic form, C6H6 + O2   CO2 + H2O Step1: To balance C-atoms, multiply CO2 by 6 ; we get C6H6 + O   6CO2 + H2O Step 2: To balance H atoms, multiply H2O by 3; we get C6H6 + O   6CO2 + 3H2O Now, see that there are 15 oxygen atoms on R.H.S. To balance oxygen, multiply O by 15. Thus, we get a balanced atomic equation. C6H6 + 15O   6CO2 + 3H2O (Atomic equation) To make it molecular, multiply the equation by 2. 2C6H6 + 15O2   12CO2 + 6H2O H   ve It is the balanced molecular equation. 46. Writing the equation in the atomic form.

Fe3O4 + H2   Fe + H2O Here the biggest formula is Fe3O4 To balance Fe on both sides, multiply Fe by 3. Fe3O4 + H   3Fe + H2O To balance oxygen on both sides, multiply H2O by 4. Fe3O4 + H   3Fe + 4H2O Now H is left unbalanced. To balance it, multiply H by 8; we get atomic balanced equation. Fe3O4 + 8H   3Fe + 4H2O To make the equation molecular, write 4H2 in place of 8H. Thus, the balanced equation is: Fe3O4 + 4H2   3Fe + 4H2O 47. (i) Change the equation into its atomic form FeS2 + O   Fe2O3 + SO2 (ii) To equalise Fe atoms, multiply FeS2 by 2 2FeS2 + O   Fe2O3 + SO2 www.betoppers.com

9th Class Chemistry

270 (iii) Multiply SO2 by 4 in order to equalise sulphur. 2FeS2 + O   Fe2O3 + 4SO2 (iv) Multiply O by 11 to equalise O atoms. 2FeS2 + 11O   Fe2O3 + 4SO2 Multiply the whole equation by 2 to make it molecular 4FeS2 + 11O2   2Fe2O3 + 8SO2 48. 1 49. 3

50. Only Aluminium 51. 4 52. 4

53. a-q, b-r, c-t, d-p, e-q-s 54. Writing f-numbers and order of balancing each element. Element Frequency number Order of balancing each element

Al 2

C 2

O 5

H 2

N 2

1st

3rd

5th

4th

2nd

Reasons for choosing order of balancing of each element : Aluminium, carbon, hydrogen and nitrogen have same frequency number. However, aluminium being a metal gets 1st place in balancing. Amongst the non-metals with frequency number 2, nitrogen has the highest atomic number followed by carbon and hydrogen. Thus, order of balancing nitrogen is 2nd, carbon is 3rd and hydrogen is 4th. The last element oxygen, which has frequency number 5 gets 5th place in the order of balancing. i) Balancing aluminium : The atoms of aluminium towards the reactants side are 2 in Al2(CO3)3 , but in the products side it is one in Al(NO3)3. Thus, in order to balance aluminium we multiply Al(NO3)3 by 2 as shown in the equation (a). Al2(CO3)3 + HNO3  2Al(NO3)3 + CO2 + H2O ...... (a) ii) Balancing nitrogen : In equation (a) number of nitrogen atoms in HNO3 towards the reactants side is 1. The number of nitrogen atoms in 2Al(NO3)2 towards the products side are 6. Thus, in order to balance nitrogen we multiply HNO3 by 6 as shown in equation (b) Al2(CO3)3 + 6HNO3  2Al(NO3)3 + CO2 + H2O...... (b)

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iii)Balancing carbon : In equation (b) number of carbon atoms in Al2(CO3)3 towards the reactants side are 3. The number of carbon atoms in CO2 towards the products side are 1. Thus, in order to balance carbon we multiply CO2 by 3 as shown in equation (c) Al2(CO3)3 + 6HNO3  2Al(NO3)3 + 3CO2 + H2O ...... (c) iv)Balancing hydrogen : In equation (c) number of hydrogen atoms in 6HNO3 towards the reactants side are 6. The number of hydrogen atoms in H2O towards the products side are 2. Thus, in order to balance hydrogen, we multiply H2O by 3 as shown in equation (d). Al2(CO3)3 + 6HNO3  2Al(NO3)3 + 3CO2 + 3H2O ...... (d) v) Balancing oxygen : In equation (d) total number of oxygen atoms on the side of reactants are 27. Similarly, total number of atoms on the side of products are 27. Thus, equation is fully balanced.

55. Writing f-number and balancing order of each element Element f - number Order of balancing each element

Cu 2 1st

S 3 3rd

O 4 4th

H 2 2nd

Reasons for choosing order of balancing of each element : Copper and hydrogen have same f-number 2. However, copper being a metal gets 1st place in balancing and the hydrogen 2nd place. The sulphur with f-number 3 gets 3rd place and oxygen with f-number 4, gets 4th place. i) Balancing of copper : The reactants and the products have one atom of copper each. Thus, it is already balanced as shown in equation (a). Cu+ H2SO4  CuSO4 + SO2 + H2O..... (a) ii) Balancing of hydrogen : The reactants and the products have two atoms of hydrogen each. Thus, it is already balanced as shown in equation (b). Cu + H2SO4  CuSO4 + SO2 + H2O ...... (b) iii)Balancing of sulphur : Reactants have one atom of sulphur. However, the products have two atoms of sulphur, one in CuSO4 and one in SO2. Thus, in order to balance sulphur, multiply H2SO4 by 2 as shown in equation (c). Cu + 2H2SO4  CuSO4 + SO2 + H2O ..... (c) In equation (c) the multiplication of H2SO4 with 2, has changed number of hydrogen atoms to 4. Thus, in order to re-balance hydrogen, multiply

Language of Chemistry Solutions H2O with 2 on the products side as illustrated in equation (d). Cu + 2H2SO4  CuSO4 + SO2 + 2H2O ... (d) iv)Balancing of oxygen : In equation (d) number of oxygen atoms in 2H2SO4 towards reactants is 8. Similarly, total number of oxygen atoms in the products is 8. Thus, equation is fully balanced.

56. Writing f-numbers and balancing order of each element. Element Pb N O f - number 2 2 4 Balancing order 1st 2nd 3rd of each element i) Balancing lead : The number of atoms of lead are one on each side, i.e., reactants side and products side. Thus, it needs no balancing ii) Balancing of nitrogen : The reactants side has 2 nitrogen atoms and the products side one nitrogen atom. Thus, NO2 in the products side is multiplied by 2 so as to balance nitrogen as shown in equation (a).

Pb(NO3)2  PbO + 2NO2 + O2 ...... (a) iii)Balancing of oxygen : The number of oxygen atoms towards the reactants side are 6 (an even number). The number of oxygen atoms towards the side of products are 7 (an odd number).

If there is a difference of one atom between reactants and products or number of atoms towards the reactants are even and towards the products odd or vice versa and the following condition is met, then apply the rule stated below: Condition : The element which is being balanced must be present in equation in its elementary form. For example, in the above equation the element oxygen is present in the form of compound in Pb(NO3)2, PbO and NO2. However, it exists in elementary form in O2. Rule : If there is a difference of one atom in reactants and products or the number of atoms in the reactants and the products are even and odd numbers, and that particular element exists in pure elementary state in the equation, then without disturbing the atom in elementary form, multiply the whole equation by 2 (do not multiply element by 2). Thus, multiplying equation (a) with 2, without, multiplying O,, the equation is written as 2Pb(NO3)2  2PbO + 4NO2 + O2 ....

271

(b) In the above equation (b), the number of oxygen atoms towards the reactants side are 12. Similarly, number of atoms towards the products side are 12. Thus , equation (b) is fully balanced equation. 57. Writing f-number and balancing order of each element. Element f - number Balancing order of each element

K 2 2nd

Cr 2 1st

O 6 5th

H 2 3rd

S 3 4th

i) Balancing chromium : There are 2 chromium atoms towards the side of reactants and 2 chromium atoms towards the side of products. Thus, it needs no balancing. K2Cr2O7 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O + O2 ii) Balancing potassium : There are 2 potassium atoms towards the side of reactants and 2 potassium atoms towards the side of products. Thus, it needs no balancing. K2Cr2O7 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O + O2 iii)Balancing hydrogen : There are 2 hydrogen atoms towards the side of reactants and 2 hydrogen atoms towards the side of products. Thus, it needs no balancing. K2Cr2O7 + H2SO4  K2SO4 + Cr2(SO4)3 + H2O + O2 iv)Balancing sulphur : There is one sulphur atom towards the side of reactants and 4 sulphur atoms towards the side of products. Thus, H2SO4 is multiplied by 4, in order to balance sulphur atoms. K2Cr2O7 + 4H2SO4  K2SO4 + Cr2 (SO4)3 + H2O + O2 However, on multiplying sulphuric acid (H2SO4) with 4, the number of hydrogen atoms increases to 8. Thus, H 2 O towards products side is multiplied by 4, in order to re­balance hydrogen atoms. K2Cr2O7 + 4H2SO4  K2SO4 + Cr2(SO4)3 + 4H2O + O2 v) Balancing oxygen : The total number of oxygen atoms towards reactants side are (7 + 16) = 23.

The total number of oxygen atoms towards the products side are (4+ 12 + 4 + 2) = 22. Thus, the atoms on the reactants and products side are www.betoppers.com

9th Class Chemistry

272

odd and even numbers respectively. Furthermore oxygen occurs in equation in pure elementary farm. Thus, we will multiply the whole equation by 2, except the element O2. 2K2Cr2O7 + 8H2SO4  2K2SO4 + 2Cr2 (SO4)3 + 8H2O + O2 Recounting oxygen towards reactants side, there are (14 + 32) = 46 atoms. The number of oxygen atoms in compounds towards products side (excluding O2) are (8 + 24 + 8) = 40. Thus, on products side there are six atoms of oxygen less than reactants. They can be made up by multiplying O2 with 3. 2K2Cr2O7 + 8H2SO4  2K2SO4 + 2Cr2(SO4)3+ 8H2O + 3O2 The above equation is a fully balanced equation. 58. 3Mg + 7HClO3  3Mg(ClO3)2 + 3H2O + HCl 59. 4FeS2 + 11O2  2Fe2O3 + 8SO2

CONCEPTIVE WORKSHEET 1. 2. 3. 4. 5.

7.

The atomic number of an element decides its stability 8. OH– has more electrons than protons and more protons than neutrons. 9. A = Z + n. 10. The ratio of neutrons/protons in the atom is higher than pb. 11. C12 .

1 th of C 12 . 12

1 th of C – 12 isotopic atom 12 Atoms of inert gases exist independently. Molecules and ions can exist independently. The smallest particle of matter that can exist independently is a molecule. Homogeneous molecules are made up of same type of atoms (H2, N2, O2, F2) Heterogeneous molecule are made up of different type of atoms (CH4, NH3, CO 2 ) H2, N2, F2, O2. P4  Atomicity is 4. S8  Atomicity is 8. All inert gases exist in monoatomic form. None of them take part actively in a chemical reaction as they are inert.

13. 1 amu = mass of 14. 15. 16. 17.

18. 19. 20. 21.

22. 1 26. 2

i) Protons - 11 ii) Electrons - 11 iii) Electron - 10 ; Protons - 11 1) 4, 4, 5 2) 11, 11, 12 3) 15, 15, 16 4) 18, 18, 22 a) 10 b) 20 a) 6 b) 6 False, It is 35. 6. A = Z + n.

12.

23. 1 27. 3

2

24. 2 25. 1 28.CrO4–2 and Cr2O7–

29. CN – 30. 4 31. 2 32. 4 33. 2 34. 2 35.zero 36.M(SO4)3 37. M2(NO3)2 38. Na3PO 4 39. 4 40. (i) and (f) ; (ii) and (j) ; (iii) and (a); (iv) and (b); (v) and (e); (vi) and (g); (vii) and (c) ; (viii) and (i); (ix) and (d); (x) and (h).

41. Basic Radicals

Sodium Copper(II) Aluminium Zinc Silver Calcium Cobalt Mercurous Magnesium Tin (IV)

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Acid Radicals

Chloride

Carbonate

Nitrate

Sulphate

Phoshate

Oxalate

NaCl CuCl2 AlCl3 ZnCl2 AgCl CaCl2 CoCl2 Hg2Cl2 MgCl2 SnCl4

Na2CO3 CuCO3 Al2(CO3)3 ZnCO3 Ag2CO3 CaCO3 CoCO3 Hg2CO3 MgCO3 Sn(CO3)2

NaNO3 Cu(NO3)2 Al(NO3)3 Zn(NO3)2 AgNO3 Ca(NO3)2 Co(NO3)2 Hg2(NO3)2 Mg(NO3)2 Sn(NO3)4

Na2SO4 CuSO4 Al2(SO4)3 ZnSO4 Ag2SO4 CaSO4 CoSO4 Hg2SO4 MgSO4 Sn(SO4)2

Na3PO4 Cu3(PO4)2 AlPO4 Zn3(PO4)2 Ag3PO4 Ca3(PO4)2 Co3(PO4)2 Hg3(PO4) Mg3(PO4)2 Sn3(PO4)2

Na2C2O4 CuC2O4 Al2(C2O4)3 ZnC2O4 Ag2C2O4 CaC2O4 CoC2O4 Hg2C2O4 MgC2O4 Sn(C2O4)2

Language of Chemistry Solutions 42. 43. 44. 45. 46. 47. 48.

273

Symbols and formulae. Created nor destroyed. Reactants. Products. To yield or To form. Balanced equation. Potassium nitrate.

49. Skelton equation. 50. 1 51. 2 52. 4

53. An element equation is called frequency true. 54. i) 2K2 Cr 2 O7 + 8 H2 SO4  2K2SO 4 + 2Cr 2 (SO4 )3 + 8H2 + 3O2

ii)3CaOCl2 + 2NH3  3CaCl2 + 3H2O + N2 iii) 6FeSO4 + 3H2SO4 + 2HNO3  3Fe2(SO4)3 + 2NO + 4H2O 55. Mg2 (NO3)2  2Hg + 2NO2 + O2

SUMM ATIVE WORKSHEET 1. 2.

3 1, 2, 3

3.

2 protons, 1 neutron and 1 electron.

4.

Hydrogen ( 1 H1 )

5.

Helium



7



2

He 4  , Carbon



6



C12  , Nitrogen

N14  etc.

a) 12

7. 8. 9.

The mass number is 37. 8 protons, 8 neutrons and 10 electrons. 7 protons + 10 electrons.

b) 12

10. Weight of an element = Atomic weight × 1.66 × 10–24g 11. 2, 3 Smallest unit of mass = amu Mass of one electron = 0.00054 amu

1amu No. of electrons = 0.00054amu = 1852.

xg 1024   6.023  1023 x1.66  1024 g 1.66

Glucose



Chemical formula C6 H12 O6

Number of atoms of A = Number of atom of B.  Similarly, the number of atom of any element in its gram atomic weight = 6.023 × 1023. Now, consider any two compounds, say ‘C’ and ‘D’, whose molecular weight are ‘p’ and ‘q’ respectively.  Gram molecular weight of C = ‘p’ g Gram molecular weight of D = ‘q’ g weight of one molecule of C = ‘p’ amu = p × 1.66 × 10–24g weight of one molecule of D = ‘q’ amu = q × 1.66 × 10–24g Number of molecules of C

'p'g 1024   6.023 1023 24 'p' 1.66 10 g 1.66 Number of molecules of D

4  1.66  1024  4  He. 1.66  1024

13. Common name Molecular weight

yg 1024   6.023 1023 y1.66 1024 g 1.66



12. Atomic weight

(i)



 Number of atoms of B

6.



(ii) Sucrose C12 H22 O11 342 (iii) laughing gas N 2 O 44 (iv) Table salt NaCl 58.5 (v) Marsh gas CH 4 16 (vi) Caustic soda NaOH 40 (vii) Baking soda NaHCO3 84 (viii) Blue vitriol CuSO4.5H2 O 233 14. All the given statement are true. 15. Weight of one molecule = (molecular weight) amu Molecular weight of CO2 = 12 + (2 × 16) = 44.  weight of one molecule of CO2 = 44 amu 16. Let us consider two elements, A and B, whose atomic weights are ‘x’ and ‘y’ respectively.  Gram atomic weight of A = ‘x’ g Gram atomic weight of B = ‘y’ g weight of one atom of A = ‘x’ amu = x × 1.66 × 10–24g weight of one atom of B = ‘y’ amu = y × 1.66 × 10–24g  Number of atoms of A

180

'q 'g 1024   6.023 1023 24 'q ' 1.66  10 g 1.66

17. 4 19. Trivalent

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9th Class Chemistry

274 21. 23. 25. 27. 29. 30.

4 22. 3 –2 –2 SO3 and SO4 24. P3– , PO4–3 4 26. 1 3 28. 1 1 a) FeAsO4 b) Ba2 [Fe(CN)6 ] c) Al4 [Fe(CN)6]3 d) Zn3 (AsO4)2 e) Mg3 (AsO4 )2 f) Ni3(AsO4)2 31. a) Ca 3(PO4 )2 b) Na2S2O3 c) Mg3N2 d) MnO e) BaS f) Mg(NO3)2 32.

Cations

Anions

Bicarbonate

Nitrite

Borate

Iodide

Thiosulphate

Cadmium

Cd(HCO3)2

Cd(NO2)2

Cd3(BO3)2

CdI2

CdS2O3

Chromium

Cr(HCO3)3

Cr(NO2)3

CrBO3

CrI3

Cr2(S2O3)3

Nickel

Ni(HCO3)2

Ni(NO2)2

Ni3(BO3)2

NiI2

NiS2O3

Potassium

KHCO3

KNO2

K3BO3

KI

K2S2O3

Lead (II)

Pb(HCO3)2

Pb(NO2)2

Pb3(BO3)2

PbI2

PbS2O3

Barium

Ba(HCO3)2

Ba(NO2)2

Ba3(BO3)2

BaI2

BaS2O3

Ammonium

NH4HCO3

NH4NO2

(NH4)3BO3

NH4I

(NH4)2S2O3

Manganese

Mn(HCO3)2

Mn(NO2)2

Mn3(BO3)2

MnI2

MnS2O3

Iron (III)

Fe(HCO3)3

Fe(NO2)3

FeBO3

FeI3

Fe2(S2O3)3

33. 2

ix)

Mn(OH)2 + 2Na2O2  Na2MnO4 + 2NaOH

34. 2

x)

Al2(SO4) 3 + 6NaOH  2Al(OH)3 + 3Na2SO4

35. Al, N, C, H, O

xi)

2KI + 3H2SO4  2KHSO4 + 2H2O + SO2 + I2

Balance the following skeleton equations.

i)

3Sn + 6HCl + 2NO  3SnCl2 +

2NH2OH ii)

Ca3(PO4)2 + 3SiO2  P2O5 + 3CaSiO3

iii)

2Al2O3 + 9C  Al4C3 + 6CO

iv)

C2 H 4

+

3O2  2CO2 + 2H2O

v)

2C2H 2

+

5O2  4CO2 + 2H2O

vi)

CH 4

+

2O2  CO 2 + 2H2O

+

2Na  2NaNH2 + H2

vii) 2NH3

viii) 2Cr(OH)3 + 3Na2O2  2Na2CrO4 + 2H2O + 2NaOH www.betoppers.com

xii) 2CuFeS2 + O2  Cu2S + 2FeS + SO2 xiii) 2FeS + 3O2  2FeO + 2SO2 xiv) 2Cu2S + 3O2  2Cu2 O

+

2SO2

xv) Cu2S + CuSO4

 3Cu +

2SO2

xvi) 2Cu2O + Cu2S

 6Cu +

SO2

xvii) 2CuSO4 + 4KI

 2CuI + I2 + 2K2SO4

xviii) 2CuCl2

+ 2H2O + SO2 

2CuCl + 2HCl + H2SO44

xix) 3NaBrO + NH2CONH2  3NaBr + CO2 + 2H2O + N2

xx) 3Fe + 4N2O  4N2 + Fe3O4

Language of Chemistry Solutions

275

IIT JEE WORKSHEET

HOTS WORKSHEET I.

II. 1. 2. 3. 4. 5. 6. 7. 8. 9.

1) ZnCl2 2) CuBr2 3) KCl 4) BiI3 5) Ca3[Fe(CN)6]2 6) Cd3 (PO4)2 7) K4[Fe(CN)6 ] 8) NaAlO2 9) Na3[Co(NO2 )6 ] 10) Na2O2 11) BiOCl 12) Zn2[Fe(CN)6 ] 13) NH4CNS 14) Ca(NO3)2 15) NiBr2 16) Cr2(SO4)3 17) Pb(CH3 COO)2 18) NH4NO2 19) Cu2Cl2 20) Hg2(NO3)2 21) Mn3(PO4)2 22) Mn2(SO4)3 23) Na2Cr2O7 24) BaSiO3 25) SnCl4 26) K2ZnO2 27) NaCN 28) KBO2 29) Zn(ClO4)2 30) SrC2O4 Balance the following skeleton equations.  4CuCl + SO2 + 4HCl

1. 4. 7. 10. 13. 16. 19. 22. 25. 28. 31. 32. 33. 34.

3 2. 4 3. 2 5. 1 6. 2 8. 2 9. 1 11. 3 12. 3 14. 2 15. 4 17. 3 18. 4 20. 3 21. 2 23. True 24. True 26. False 27. False 29. False 30. False Sodium hexameta phosphate Metals Pseudo halide ions.

2 1 3 2 2 1 1 True False True (NaPO3)6.

35. (NH4)3 PO4  3NH3 + H2O + HPO3

4CuCl2



+ 2H2O + S  2NaCN 2NaAg(CN)2 + Zn + Zn(CN)2 + 2Ag FeCl3 + 3NH4 CNS  Fe(CNS)3 + 3NH4 Cl Ca 3(PO 4 ) 2 + 2H2 SO4  Ca(H2 PO 4 ) 2 + 2CaSO4  CaCO3 CaCN2 + 3H2 O + 2NH3 4Mg + 10HNO3  4Mg(NO3 )2 + NH4NO3 + 3H2O S + 6HNO3  H2SO4 + 6NO2 + 2H2 O  2Hg + 2NO2 + O 2 Hg2 (NO3 ) 2

AlN + Al(OH) 3 + 10. 3Mn3 O4 + + 9Mn

3H2 O NH 3 8Al

 

4Al2O 3

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9th Class Chemistry

2. ELEMENTS, COMPOUNDS & MIXTURES SOLUTIONS

FORMATIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

11

a

d

a

*

b

A

MT

a

MT

a

c

12

13

14

15

16

17

18

19

20

21

22

b

a

c

MT

MT

a

d

d

MT

c

*

23

24

25

26

27

28

29

30

31

32

33

*

a

*

b

a, b

MT

a, b, c

a

d

a, b

c

34

35

36

37

38

39

40

41

42

43

44

c

d

b

a

b, d

c

c

b

b, c, d

a

d

45

46

47

48

49

50

51

52

53

54

55

b

c

d

d

c

c

a

a

a

c

c

MT = Matching Type ; * = Subjective questions HINTS/ANSWERS TO THE SELECTED QUESTONS 4. Mercury and Gallium. 7. a - iii, b - v, c - iv, d - i, e - ii 9. i - q, ii - r, iii - p, s 15. l,q - i, m, p - iii, n, o - ii 16. p, t - iii, q, s - i, r - ii 20. a - iii, b- v, c- i, d- iv, e - ii 22. Fractional crystallisation 23. One of the components of mixture is either lighter than the other or is soluble in water 25. Using a suitable solvent 28. p - ii, q-iv, r-i, s- iii

CONCEPTIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

11

a, b

a

a

d

b

d

a

*

*

*

c

12

13

14

15

16

17

18

19

20

21

22

*

d

c

d

a,c,d

b, d

c

b

c

a,b,d

b

23

24

25

26

27

28

29

30

31

32

33

c

a, b

b,c,d

a,b

a,b

a,c

d

b

b

a

a,c

34

35

36

37

38

39

40

41

42

43

44

a

c

a

c

c

b

d

b

c

d

b,c

45

46

47

48

49

50

51

52

53

54

55

a

a

d

a

b

b

c

d

b

a,b

b

* = Subjective questions

8. Oxygen

9. Noble gases

10. 60

12. Mixture

9th Class Chemistry

278

SUMM ATIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

a

a

b

d

d

b

abcd

a

b

10

11

12

13

14

15

16

17

18

c

abcd

c

bd

abd

abcd

c

b

b

19

20

21

22

23

24

25

b

abcd

b

acd

c

b

a

HOTS WORKSHEET 1. a 2. b 3. a 4. a 5. a 6. b 7. (i) oxygen (ii) zinc (iii) iodine (iv) antimony (v) bromine (vi) mercury (vii) graphite 8. i-D, ii-A, iii-C, iv-B, v-E 9. i. Compoundii. Mixture iii. Compound iv. Mixture v. Mixture vi. Mixture vii. Compound viii. Compound 10. Sodium chloride is considered a compound because of the following reasons: i) Sodium chloride cannot be separated into its constituents, sodium and chlorine, by any physical methods (such as filtration, evaporation, distillation, sublimation, using magnet, etc.,) ii) The properties of sodium chloride are entirely different from those of its constituents, sodium and chlorine. iii) The composition of sodium chloride is fixed. It contains sodium and chlorine combined together in a fixed proportion. It has a definite formula, NaCl. 11. i. Heterogeneous ii. Homogeneous iii. Homogeneous iv. Homogeneous v. Heterogeneous 12. i) Pure water from sea water:Sea water is placed in the distillation flask. It is heated, using the tripod stand. This is connected to the Leibig’s condenser. Here, the vapour condenses to form pure water. The liquid formed is collected in the receiver. Salt is left behind in the flask. ii) Kerosene oil from a mixture of kerosene

oil and petrol: The two liquids in the mixture of kerosene and petrol are miscible. Hence, they are separated by using fractional distillation method, in which the liquid having the lower boiling point boils off first. It is then cooled and condensed. iii) Lead sulphate from a mixture of lead sulphate and lead chloride: Lead sulphate and lead chloride can be separated by fractional crystallizaiton. The mixture is added to water and boiled. It is cooled and the crystals of less soluble substances are formed and are separated from the solution by filtering it. The left out solution is again heated and cooled resulting in the crystallisation of a more soluble component. 13. a. Fractional distillation b. Evaporation c. Filtration 14 i) Sulphur from a mixture of sulphur and iron filings. The mixture of sulphur and iron filings is spread on a plain surface or dish. A powerful magnet is moved over the mixture. Iron filings are attached to the magnet. Sulphur is left behind on the plain surface. ii) Separate a precipitate of lead sulphate obtained by adding sulphuric acid to a solution of lead nitrate. The mixture is kept in a flask for some time without disturbing it. The precipitate settles down in the flask. The cleared solution is poured off from the top. Thus, the precipitate is separated from the mixture. iii) Oil from a mixture of oil and water. The mixture of oil and water is passed through a separating funnel. Water, being the heavier component, settles in the lower layer and trickles down the separating funnel. The lighter component, oil is left behind in the funnel. iv) Iodine from a mixture of powdered iodine and ammonium chloride. The mixture is taken into an evaporating dish and is heated. Iodine, being highly sublimable, sublimes leaving behind ammonium chloride. v) Powdered charcoal from a mixture of copper oxide and powdered charcoal. The mixture of powdered charcoal and copper oxide is dissolved in water. Copper oxide is settled at the bottom and charcoal floats over water. The charcoal is removed from the top and is separated from water. 15. i - b, ii - d, iii - c, iv - a, v - c

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3. ATOMIC STRUCTURE SOLUTIONS

FORMATIVE WORKSHEET 1. 2.

Both assertion and reason are correct and reason is not the correct explanation of assertion. We know that, mass of electron (me) = 0.0005486 amu mass of proton (mp) = 1.00727 amu me 0.0005486 1   mp 1.00727 1836

3.

Let the number of electrons be x. We know that, Total number of electrons × mass of each electron = Total mass of electrons  x × mass of each electron = Total mass of electrons.

Total mass of electrons __________ (1) Mass of each electron Given, total mass of electrons = 1 cgs unit of mass = 1g. And we know that mass of one electron = 9.1 × 10–28 g. Substituting the above values in equation (1), x

we get  x 

1 9.1 1028

1 1028 10  1027   1.09  1027 electrons 9.1 9.1 Therefore, approximately 1027 electrons make 1 gram. Mass of each electron (me) = 9.1× 10–28 g No. of electrons (n) = 6.023× 1023 Total mass of electrons = n × me = 6.023× 1023 × 9.1× 10–28 = 54.8× 10–5 g 1 Faraday = 96500 coulombs ______________ (1) Let the number of electrons be x.  1 Faraday = x × charge on each electron ____________ (2) Substituting (1) in (2) , we get  96500 C = x × Charge on each electron  96500 C = x × 1.6 × 10–19C (  Charge on each electron = 1.6 × 10–19 C) x

4.

5.

6.

96500  6.03 1023 electrons 1.6  1019 Thus, the charge of 6.03 × 1023 electrons equal to 1 Faraday. x

Specific charge of electron (S1) = 1.76 × 108 C/g Specific charge of hydrogen (S2) = 9.58 × 104 C/g S1 1.76  108 1837  S  9.58  104  1 2

7.

8.

Therefore, the ratio of specific charge (e/m) of an electron to that of a hydrogen ion is 1837 : 1. The charge on Li+ = one positive charge The charge of one positive charge = charge of proton = + 4.8 × 10–10esu. Specific charge of a proton (S1) = 9.58 × 104 C/g Let us now find the specific charge (S 2 ) of  particle. Charge on  - particle (He+2) = 2 × 1.6 × 10–19 C Mass of  - particle = 4 × 1.66 × 10–24 g

S2  

2  1.6  1019  5  104 C / g 24 4  1.6  10

S1 9.58 104 2   S2 1 5  104

Therefore, the ratio of specific charge of proton to that of an alpha particle is 2 : 1. 9. 3 10. Radius of nucleus (rn) = 10–13cm Radius of atom (ra) = 10–8 cm Fraction of atomic volume occupied by nucleus

Volume of nucleus ? Volume of atom Given that nucleus and atom are spherical. 

4 3 We know that, volume of a sphere  r 3 Fraction of atomic volume occupied by nucleus 3 4  1013  V  f   n  34 3 Va  108  3

f 

Vn 1039 1  24  1015  15 Va 10 10

9th Class Chemistry

280 11. The following are the reasons: i) The emission of energy is not continuous but is discontinuous. ii) Electrons revolve around the nucleus in specified paths called orbits. These orbits have constant energy. As long as electrons revolve in these orbits, they neither lose energy nor gain energy. Hence it does not fall inside the nucleus , and the atom does not collapse.



hc  E1   2 E 2 1  ( h and c are constants)

1 6.625  1034  10 34 Js 2  3.14 17. Momentum (P) = ? Kinetic energy (E) = ? Orbit number (n) = 1

L =

0

Therefore, the ratio of energies of two radiations, one with a wavelength of 400 nm and the other with that of 800 nm is 2 : 1 13. Number of photon (n) = ?  = 4000 pm = 4000×10–12 m Energy (E) = 1J E  nh

C       

 E

nhc 

 n

E h C where h is Planck’s constant with a

1  4000  1012  2  1016 6.625  1034  3  108 Therefore, an approximate of 2 × 1016 photons provides the energy of 1J. 14. Number of photons (n) = 3 × 1018 Energy (E) = 1.5 J n

Wavelength    = ? h = 6.63× 10–34 JS

nhc  18

nhc 3  10  6.63 10  E 1.5

34

 3  10

Radius of the orbit (r) = 0.529 A = 0.529 × 10–10m Calculation of momentum (mv): Terms connected are ‘mv’ , r and n

nh nh 1  mv = P =  2 2 r Substituting the values in the above formula, we get mvr =

P

value of 6.625×10–34 Js and C is velocity of light in air or vacuum (3 × 108 m/s)

 

1 1 6.625  10 34  0.529  1010 2  3.14

 1034 



1010 0.529

1  1034  1010 0.529

 1.9  1024 kg  m / s Therefore the momentum of the electron is 1.9 × 10–24 kg-m/s We know that kinetic energy (E) =

1 mv 2 ............ 1 2 Where ‘m’ is mass of electron = 9.1 × 10– 31 kg How to get velocity (v) in terms of given data? We know that momentum P = mv P ............(2) m Substituting (2) in (1), we get v

8

0

= 39.78 × 10–8m (or) 3978 × 10–10m = 3978 A www.betoppers.com

nh and the value of h = 2

6.625 × 10–34 Js

E1 800 2   E 2 400 1

E

nh is not satisfied. 2 And the electron cannot stay in this orbit  Orbit is not possible. 16. Angular momentum (mvr) = L = ? n=1  mvr 

We know that m v r 

12. 1 = 400 nm ;  2 = 800 nm E1 : E2 = ?

E  h 

15. For an intermediate orbit in between n = 2 and n = 3, the value of ‘n’ is fractional

E

1 P2 P2 m 2  E  ............ 3 2 m 2m

Atomic Structure Solutions

281

Substituting the values of ‘P’ and ‘m’ in (3), we get

21.

2

E

 E

P ............ 3 2m

1.9  1024



2

2  9.1 1031



3.61 10 48 2  9.1  1031



3.61  10 48 1031 2  9.1



3.61  1017 18.2

36.1  1018  1.98  10 18 J 18.2 Therefore the energy of the electron is 1.98 × 10-18 J 18. First orbit Second orbit n1 = 1 n2 = 2 r1 = r1 r2 = r2 v1 : v2 = ? 

We know that mvr =

nh 2

nh 1 n  v 2 mr r ( ‘h’ and ‘m’ are constants)

 v=



v1 n1 r2 v r 1 r    1   2  2 v 2 n 2 r1 v 2 2 r1 2r1

19. 107 20. Let first find the neutrons Same + this new row

Atoms A Z A–Z

A 40 20 20

D 19 9 10

C 7 3 4

L S 16 20 8 10 8 10

i) Ratio of no. of neutrons: A : D : C : L : S = 10 : 5 : 2 : 4 : 5

A  2  A  2Z Z From the table, it is clear that for the atoms D and C , A > 2Z iii) From the table, it is clear that for the atoms A, L and S, A = 2Z. ii)

22. No. of neutron (n) = ? Mass number of zinc (A) = 70 Atomic number of zinc (Z) = 30 We know that A = Z + n  n = A – Z = 70 – 30 = 40. Note: During ion formation, only the number of electrons change but not the number of protons or neutrons. 23. Number of neutrons in silicon (n1) = 28 – 14 = 14 Number of neutrons in phosphorus (n2) 31–15 = 16 Ratio =

n1 14  = or 7 : 8 n 2 16

Therefore, the ratio of number of neutrons in silicon to phosphorus is 7 : 8. 24. a) Nitrate ion is NO3–1 Total number of electrons = 7 + ( 3  8) + 1 = 7 + 24 + 1 = 32 Total number of protons = 7 + ( 3  8) = 7 + 24 = 31 Total number of neutrons = 7 + ( 3  8) = 7 + 24 = 31 b) Sulphate ion is SO4–2 www.betoppers.com

9th Class Chemistry

282 Total number of electrons = 16 + ( 4  8) + 2 = 16 + 32 + 2 = 50 Total number of protons = 16 + ( 4  8) = 16 + 32 = 48 Total number of neutrons = 16 + ( 4  8) = 16 + 32 = 48 25. The contribution of electrons to the mass of an atom is very little and negligible.  mass of an atom = mass of protons + mass of neutrons = 6x + 6x ( mass of neutron  mass of proton)  12 = 6x + 6x = 12x ___________ (1) Given that mass attributed by neutron is halved  the mass contributed by neutrons = 2x. New atomic mass = 6x + 3x = 9x _________ (2) From (1) and (2), it is clear that atomic mass in second case is reduced by 25%. 26. HD + = 1H1 + 1D2 – 1 electron  Total no. of protons = 1 + 1 = 2

Total no. of electrons = 1 + 0 = 1 Total no. of neutrons = 1 + 0 = 1 27. Phosphorus  Atomic number (Z) = 15  No. of electrons in it = 15 At first, electrons enter K shell which can accommodate two electrons. The next eight electrons are adjusted in L shell. The remaining five electrons are adjusted in N shell. Thus, electrons configuration of phosphorus is: 2, 8, 5 28. The atomic number of calcium is 20. Hence, it has twenty electrons, out of which two can be accommodated in K– shell and eight in L– shell. Can you guess how these remaining ten electrons are arranged? 29. a) A tripositively charged ion is formed by the loss of three electrons. Therefore, the number of electrons in its neutral atom = No. of electrons in the ion + 3 = 10 + 3 = 13  The atomic number of the element (Z) = 13  The element is aluminium. Mass number of aluminium (A) = 27 No. of neutrons (n) = A – Z = 27 - 13 = 14 b) To find out the formula, we need to know the valency of aluminium. Electronic configuration = 2, 8 , 3 Valency = No. of electrons in outermost orbit =3 Therefore, it forms XCl3, X2O3 with chlorine and oxygen respectively. www.betoppers.com

30. Electron configuration of M2+ is 2, 8, 14 Mass number (A) = 56 Number of neutrons (n) = ? We know that n = A – Z How to get Z? Z = No. of protons = No. of electrons in the neutral atom = 24 + 2 = 26  n = 56 - 26 = 30. Therefore, the no. of neutrons present in M+2 is 30. 31. Number of electrons in N–3 = 7 + 3 = 10 No. of electrons in X+3 = 10  No. of electrons in neutral atom of X = 10 + 3 = 13  No. of protons X = 13  X is Aluminium 32. Total number of electrons in the orbit = 81 Orbit number (n) = ? Total no. of electrons = 81 = 3n3  n3 = 27 or n = 3 Therefore, the orbit that can accommodate 81 electrons is 3. 33.

34.

35. 36.

37.

Electronic configuration K L M N Potassium 19 2 8 8 1 Nitrogen 7 2 5 Chlorine 17 2 8 7 Element

Z

Therefore, the right match is p - 2, q - 1, r - 3 Species No. of Electronic electrons configuration O2– 10 2, 8 K+ 18 2, 8, 8 Al+3 10 2, 8 – F 10 2, 8 From the above table, it is clear that K+ has same number of electrons in its outermost shell and penultimate shell. M shell  n = 3 No. of electrons = 2n2 =2× 32 = 18 Total no. of electrons = 32 32 = 2n2  n = 4 4th shell  N shell. Total no. of electrons = 2 × 72 = 98

38. The number of electrons in NO3 = 7 + 3 × 8 + 1 = 7 + 24 + 1 = 32 39. Isotopes: These are atoms of same element with same atomic number (Z) and different mass number (A). Isobars: These are atoms of different elements with same mass number (A). Isotones: These are atoms of different elements with same number of neutrons (n).

Atomic Structure Solutions

283

Let us first find the Z, A and n of the given species. N u clid e 1 1A 7 3D 3 1C 12 6L 12 5B 8 4E 14 6S 4 2T

2)

Atom 6C

12

6C

13

6C

14

Atom 18Ar

40

n 0

14 7N

14 – 7 = 7

3

7

4

16 8O

16 – 8 = 8

1

3

2

18 9F

18 – 9 = 9

6

12

6

5

12

7

4

8

4

6

14

8

2

4

2

No. of neutrons

From the above charts it is clear that 18Ar40 20Ca42 and 21Sc43 are isotones. 42. X3– is isoelectronic with argon.  The number of electrons in X3– = 18.  The number of protons in X3– = The number of electrons in neutral atom of x = 18 – 3 = 15. The ratio of electrons to neutrons = 1 : 1  The no. of neutrons (n) in X = 15. Mass number (A) of X = 15 + 15 = 30 C14, 7N15, 8O16 are isotonic as they contain same number of neutrons. i.e., 8. 44. Let us find the number of electrons in each set 1) Species No. of electrons 43.

6

NO+

7 + 8 – 1 = 14

C22

6 + 6 + 2 = 14

O2

8 + 8 + 1 = 17

Not isoelectronic 2) Species No. of electrons

12 – 6 = 6

N2

7 + 7 = 14

13 – 6 = 7

C22

6 + 6 + 2 = 14

CO

6 + 8 = 14

NO

7 + 8 = 15

14 – 6 = 8

No. of neutrons 40 – 18 = 22

Not isoelectronic 3) Species No. of electrons

42

42 – 20 = 22

43 21Sc

43 – 21 = 22

CO

6 + 8 = 14

No. of neutrons

NO+

7 + 8 – 1 = 14

40

40 – 18 = 22

CN–

6 + 8 = 14

40

40 – 20 = 20

C22

6 + 6 + 2 = 14

20Ca

3)

No. of neutrons

A 1

From the above chart, it is clear that i) A, C and L, S are pairs of isotopes. ii) L and B are isobars iii) D, E and C, T are pairs of isotones. 40. The no. of electrons in CO = 6 + 8 = 14 Therefore, we need to select the species containing 14 electrons. Let us check the same O2– = 2 × 8 + 1 = 17 N2+ = 2 × 7 – 1 = 13 CN– = 6 + 7 + 1 = 14 O2+ = 2 × 8 – 1 = 15 Therefore, CN– is isoelectronic with CO. 41. Isotones contain same number of neutrons. Let’s check no. of neutrons in each set. 1)

4) Atom

Z 1

Atom 18Ar

20Ca 21Sc

41

41 – 21 = 20

Isoelectronic Therefore the set of species containing CO, NO+, CN–, C22 are isoelectronic with each other..

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9th Class Chemistry

284

27. 28. 29.

the reason why electrons do not fall into the nucleus in spite of inward nuclear pull. According to the law of electro-dynamics, the electron should, therefore, continuously emit radiation and lose energy. As a result of this, a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus and the atom should collapse. But atom and nucleus are stable. 1, 2 3 3 Nuclear model, Solar model, Orbitary model Rutherford was the first scientist to explain the presence of nucleus inside the atom. Hence, his model of atom is nuclear model. Further, he stated that electrons revolve round the nucleus just like planets round the sun. Thus, his model of atom is also called solar or planetary model. It was to detect the the path of alpha rays. The composition of alpha rays is same as that of doubly positively charged helium atom. Rutherford predicted the presence of nucleus and it was discovered by James Chadwick. Neils’ Bohr K, L, M, N, O, ............. or 1, 2, 3, 4,......... As long as an electron revolves in the orbits, its energy remains constant. Theoretically, infinite number of orbits are possible. 2 Yes, by absorbing or by radiating energy. E  nh

30.

E  E 2  E1  nh

CONCEPTIVE WORKSHEET 1. 2. 3. 4.

5. 6.

All the given statements are true according to Dalton’s atomic theory. Except the last option, all the other statements are not correct. All the statements are correct, except the 3rd option. As the e/m ratio is same, we can conclude that the cathode ray particles obtained from different gases and vapours have the same charge and mass. Further, the cathode ray particles are similar for any gas and vapour of any matter. Thus, a cathode ray particle is a common constituent of any matter with same mass and charge everywhere. Cathode rays are produced at a pressure of 0.01mm of Hg and a high electrical voltage of 10,000V . The S.I unit of mass is kilogram. Let the number of protons be x.  1 Kg = x × mass of one proton  1 Kg = x × 1.6 × 10–27kg (  mass of a proton = 1.6 × 10–27kg) x

1 1027  1.6  1027 1.6

10  1026  6.3  1026 protons 1.6 Therefore, the mass of 6.3 × 1026 protons is equal to 1 S.I unit of mass (kg). 

7.

8. 9. 10.

11. 12. 13.

Electron Proton Neutron Charge -1.602×1019C +1.602×1019C 0 Mass 0.0005486 amu 1.00727 amu 1.00866 amu

The mass and charge of positive rays change with the change in gas. Hydrogen contains no neutrons. Hence, it is the only gas that produces positive rays with protons. Positive rays comprise ionised atoms whose mass is much higher than that of the particles in cathode rays. Neutron is a neutral (chargeless) particle. Hence, its e/m ratio is zero. The e/m ratio is not constant for positive rays. As to the stability of the atom, Rutherford explained that the revolving electron is under the influence of two types of forces: i) The electrostatic force of attraction between the nucleus and the electron and ii) The centrifugal force directed away from the revolving electron. These two forces are equal and opposite and hence keep the electron in equilibrium in the path. This is

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14.

15. 16. 17. 18. 19.

20. 21. 22. 23. 24. 25. 26.



E 2  E1 nh

nhc E 32. No. of photons (n) = 1000 Wavelength (  ) = 106 A0 = 106  10–10 m 31.



nh c E = =  19.875  10–19 J 33.

1000  6.625  1034  3  108 = 106  1010

E1  2 12000 2    E 2 1 6000 1

Atomic Structure Solutions 34. E = 931.5 Mev = 931.5  106  1.602  10–19 J Applying: E  =

hc : 931.5  106  1.602  10–19 

6.625  1034  3  108 



6.625  1034  3  108 931.5  106  1.602  1019

= 13.334  10 16 m 35. No. of electrons in uni-negative ion = 18 No. of neutrons (n) = 20 Mass number (A) = ? A = Z + n ____________ (1) Z = No. of protons (or) No. of electrons in neutral atom. No. of electrons in neutral atom = 18 – 1 = 17 _________ (2) Substituting 2 in 1, we get, A = 17 + 20 = 37.

36. The atomic number of zinc is 30.  Total no. of protons = 1 + 2 + 3 + ................. + 30. Sum of first ‘n’ natural numbers 

 Total no. of protons 

n  n  1 2

30  31  15  31  465 2

37. Given mass number of lead = APb = 208; atomic number of lead = ZPb = 82  No. of neutrons in lead = nPb = APb – ZPb = 208 – 82 = 126

126  1.53 ___________ (1) 82 Mass number of bismuth = ABi = 209 Atomic number of bismuth = ZBi = 83  No. of neutrons in bismuth = nBi = Abi – ZBi = 209 – 83 = 126   n / P Pb 

126  1.51 ___________ (2) 83 From (1) and (2) the n/p ratio of lead is greater. 38. It is false. Actually, atomic number is the number of protons present in an atom or the number of electrons present in a neutral atom. In a charged atom, the number of electrons are not equal to the number of protons. 39. Hydrogen (protium) atom has no neutrons. Hence, its atomic number and mass number are same.   n / P  Bi 

285 40. False. In hydrogen atom neutrons are absent. 41. Helium (2He4 ), Carbon (6C12 ), Nitrogen (7 N14 ), Oxygen (8O16), etc. 42. Mass number (A) = 23 Atomic number (Z) = 11 No. of protons = No. of electrons = Z = 11 No. of neutrons (n) = A – Z = 23 - 11 = 12 43. During a chemical reaction it is the valence electrons that participate but not the protons. As atomic number is the number of protons present in an atom, it does not change. 44. Mass number (A) = 52 Atomic number (Z) = 24 No. of protons = No. of electrons = Z = 24 No. of neutrons (n) = A – Z = 52 - 24= 28 45. The number of neutrons in calcium are same as that of protons. 46. Addition of neutrons changes mass number but not atomic number. Hence atomic number remains same. 47. Atomic Electronic configuration Element 35 17Cl 12 C 6 39 19K 31 15P

number 17 6 19 15

K 2 2 2 2

L 8 4 8 8

M 7

N

8 5

1

48. From the electronic configuration, it is clear that the element has 20 electrons is calcium. 49. Number of electrons in Mg+2 = 10  E.C. = 2, 8 Number of electrons in Cl– = 18  E.C. = 2, 8, 8 50. Electronic configuration of the given element 2, 8, 2 Total number of electrons = Total no. of protons = 12  The element is Magnesium. 51. Electronic configuration of Al = 2, 8, 3 Total number of electrons in M shell = 3. 52. Carbon i.e., 6C12, 6C13, 6C14 53. Isotopes are the atoms of same element with same atomic number and different mass number.

54. It is hydrogen (1H1, 1H2, 1H3) 55. As isotopes have same number of electrons in their valence shells, their valency remains same. Further, they have number of protons. 56. Isotopes are the atoms of same element with same no. of protons and different number of nucleons.

57. Due to existence of isotopes, they have different atomic masses. 58. A and B are isotopes. 59. As they have same number of electrons, they have similar chemical properties. Due to different atomic masses, they have different chemical properties.

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9th Class Chemistry

286 60. Isotopes are the atoms of same element with same number of protons. 61. 7N14 and 7N15 are isotopes but not isotones 62. Isotones are species having same number of neutrons. The number of neutrons in 32Ge76 are 76 – 32 = 44. The species having 44 neutrons is 33As77 = 77 – 33 = 44 63. The ion isoelectronic with Na+ is Mg+2. Both have 10 electrons. 64. Na+ and Ne both have 10 electrons.

SUMM ATIVE WORKSHEET 1.

2.

3.

The rays deflect towards the +ve plate in an electric field  The rays are negatively charged and hence are cathode rays. (B) The sub-atomic particle present in all the atoms except Hydrogen is neutron. (C) Alpha rays. Let the no. of electrons be x. Given: x ( mass of single electron) = the total mass of neutrons present in one calcium atom.  x ( 0.00054 ) = No. of neutrons in calcium atom  mass of each neutron  x ( 0.00054 ) = 20  1.00866

20×1.00866  x = 0.00054  37358 a) As he was the first person to explain the presence of nucleus in an atom.

b) As he compared his model of an atom with the solar system. 4.

5.

The purpose of lead block is to absorb the radiations emitted by radioactive substance. ZnS screen is used to detect the number and direction of  - rays passing through the gold foil. x = a.m.u = 1.66  10–27 kg y = 1 kg



y 1kg =  6.023×1026 x 1.66×10-27 kg

6.

Oxide ion is O–2

7.

 Total charge = – 2  1.602  10 19 C 1 Coulomb = x ( charge on each electron)



 1C  x 1.602  10 x

19



1C  6.25  1018 19 1.602  10 C

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8.

9.

An atom has mass number A and atomic number Z: (a) No. of protons = Z (b) No. of electrons= No. of protons = Z (c) No. of neutrons = A – Z The electronic configuration of a tripositively charged ion ( X+3 ) is 2,8. Therefore the number of electrons in its neutral atom 10 + 3 = 13 a) Aluminium ( 13 Al27 )  The number of neutrons

in it = 27 - 13 = 14. b) AlCl3 and Al2O3 c) Oxygen d) 3 e) EC of X - 2, 8, 3. Therefore the number of ‘M’ electrons present in it= 3. 10. The no. of electrons in outermost shell of P = 3  Valency of P = 3 The no. of electrons in outermost shell of Q = 7  Valency of Q = 1 Criss-crossing the valencies, we get PQ3 11. During a chemical reaction, it is the number of electrons that change. But atomic number is the number of protons. Hence atomic number does not change during a chemical reaction. 12. 1





19 13. Charge on 8 protons = 8  1.602  10 C

Charge present on 10 neutrons = Zero





19 Charge on 10 electrons = 8  1.602  10 C

14. P (Z = 7 ) – 2, 5 Q (Z=11) – 2, 8, 1 R (Z = 15) – 2, 8, 5 From the above configurations it is clear that P and R have same number of electrons in their outermost shell and hence their valencies are same. Therefore they are likely to have same chemical properties. 15. 1) The elements R, A, J are metals, since they can lose electrons to form cations. 2) The elements A, R, T, I are non-metals, since they can gain electrons to form anions. 3) Hydrogen (1) 4) Helium (2)

HOTS WORKSHEET 1.

From the representation it is clear that Z and A of element is 16 and 32 respectively. i) No. of protons = 16 : No. of electrons = 16 : No. of neutrons = 32 - 16 = 16. ii) No. of nucleons = A = 32

Atomic Structure Solutions

287

iii) Total mass of electrons = Total no. of electrons  mass of each electron. = 16  9.1  10–28 g (  mass of one electron = 9.1  10–28 g )

3.

Element Atomic number Electronic configuration

Total mass of protons = Total no. of protons  mass of each proton. = 16  1.675  10–24 g (  mass of one proton= 1.672  10–24 g ) Total mass of protons = Total no. of neutrons  mass of each neutron. = 16  1.678  10–24 g (  mass of one neutron= 1.672  10–24 g ) iv) No. of atoms of X

=

Total mass of X Mass of single atom of X

Mass of single atom of X= 32 amu 4.

= 32  1.66 1024 g

1a.m.u  1.66 10

24

g

S

4

2,2

A

12

2, 8, 2

I

20

2, 8, 8, 2

i) the number of valence electrons present in S, A and I are 2, 2, 2 respectively. ii) All these elements have similar properties due to presence of same number of valence electrons in their outermost shells. iii) All these are metals as they can lose electrons. iv) All these elements belong to IIA group. Mass of carbon = 0.12 mg = 0.12 103 g  12  105 g .



Substituting the value, we get No. of atoms of X =

iv) Valency of C = 1 and L is also 1 ( 8 - 7 =1) Therefore the formula is CL. v) D2 and L2

32  6.023  1023 32  1.66  10-24

Total no. of electrons/protons/neutrons = no. of atoms  total no. of electrons/protons/neutrons present in one atom of an element. No. of carbon atoms =

(v) Atomic number of X = 16 - Electronic configuration - 1s2 2s2 2p6 3s2 3p4 (vi) No. of valence electrons = 6 (vii) Valency = 8 – 6 = 2 (viii) XO (ix) E.C.- 2, 8, 6  No. of electrons are present in L shell of X = 8. (x) Sulphur 2. Element

Atomic number Electronic configuration

A

3

2,1

D

9

2, 7

C

11

2, 8, 1

L S

17 19

2, 8, 7 2, 8, 8, 1

Atomic numbers of elements A, D, C, L, S are 3, 9, 11, 17, 19 respectively. i) A, C, S and D, L are sets which have same chemical properties. ii) Metals should lose electrons and hence should have 1or 2 or 3 electrons in the outermost shell. Therefore A, C, S are metals. iii) Non-metals should gain electrons and hence should have 5 or 6 or 7 electrons in the outermost shell. D, L are non-metals.

Total mass of carbon Mass of single atom of carbon

Mass of single atom of carbon = 12 amu = 12 × 1.66 1024 g 1a.m.u  1.66  1024 g  Substituting the value, we get No. of carbon atoms =

12 10-5  6.023 1018 12  1.66  10-24

i) No. of electrons in 0.12 mg of carbon = 6.023  1018  no. of electrons in 1 atom of carbon = 6.023  1018  6 ii) No. of protons in 0.12 mg of carbon = 6.023  1018  no. of protons in 1 atom of carbon = 6.023  1018  6 iii) No. of neutrons in 0.12 mg of carbon = 6.023  1018  no. of neutrons in 1 atom of carbon = 6.023  1018  6 iv) Total mass of electrons in 0.12 mg of carbon = Total no. of electrons in 0.12 mg of carbon  mass of each electron. www.betoppers.com

288

= 6.023  1018  9.1  10–28 g (  mass of one electron = 9.1  10–28 g ) v) Total mass of protons in 0.12 mg of carbon = Total no. of protons in 0.12 mg of carbon  mass of each proton. = 6.023  1018  1.675  10–24 g (  mass of one proton= 1.672  10–24 g ) vi) Total mass of neutrons in 0.12 mg of carbon = Total no. of neutrons in 0.12 mg of carbon  mass of each neutron. = 6.023  1018  1.678  10–24 g (  mass of one neutron = 1.678  10–24 g) 5.

6.

7.

8.

9.

Given: No. of electrons in A = No. of protons in A = No. of neutrons in A = x No. of electrons in B = No. of protons in B = No. of neutrons in B = y But y = 2x. EC of A = 2, 8 ( as it contains 8 electrons in L shell) The no. of electrons in A, x = 10  No. of protons in A = No. of neutrons in A = x = 10. Therefore A is Neon. We know that y = 2x = 20.  No. of proton in B = y = 20. Hence, B is calcium. In ‘X’, 5th electron of 3rd shell is its last electron  E.C. = 2, 8, 5 In ‘Y’, 7th electron of 3rd shell is its last electron  E.C. = 2, 8, 7 a) The no. of electrons in X and Y are 15 and 17 respectively. Hence, X and Y are phosphorus and chlorine respectively. b) XY3 c) 4 (P4) and 2 (Cl2) respectively. d) No. of neutrons in X i.e., P = 31-15 = 16 ; No. of neutrons in Y i.e., Cl = 35-17 = 18 Therefore the ratio = 8 : 9 b) Atomic mass is different from mass number. Atomic mass is mass of all the constituents of an atom. Whereas mass number is the total number of nucleons. A) Cl– – 2, 8, 8 +2 D) Ca – 2, 8, 8 C) S2– – 2, 8, 8 L) Ar – 2, 8, 8 The number of electrons, neutrons in A+1 are 10, 12 respectively. Therefore the atomic number (Z) = 11 and mass number = 23. The number of electrons, neutrons in D+2 are 10,

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9th Class Chemistry 12 respectively. Therefore the atomic number (Z) = 12 and mass number = 24. The number of electrons, neutrons in C+3 are 10, 14 respectively. Therefore the atomic number (Z) = 13 and mass number = 27. 10. A) Isotopes are the atoms of same element. B) Isobars are the atoms of different elements. C) Istones are the species containing same number of neutrons. D) Iso-electronic species contain same number of electrons. 11. Element

Z

H Be C O

1 4 6 8

Electronic configuration K L M N 1 ----- --2 2 --- --2 4 --- --2 6 --- ---

Ratio of K:L electrons 1:0 1:1 1:2 1:3

12. The electronic configuration of a tripositively charged ion ( X+3 ) is 2,8. Therefore the number of electrons in its neutral atom 10 + 3 = 13 a) Aluminium ( 13 Al27 )

 The number of neutrons in it = 27 - 13 = 14. b) AlCl3 and Al2O3 c) Oxygen (O2– has 10 electrons) d) 3 e) Electronic configuration of X is 2, 8, 3. Therefore, the number of ‘M’ electrons present in it is 3. 13. In X’ , 5th electron of 3rd shell is its last electron  E.C. = 2, 8, 5 In ‘Y’ , 7 th electron of 3 rd shell is its last electron  E.C. = 2, 8, 7 1) The no. of electrons in X and Y are 15 and 17 respectively. Hence, X and Y are phosphorus and chlorine respectively. 2) XY3 3) 4 (P4) and 2 (Cl2) respectively. 4) No. of neutrons in X i.e., P = 31-15 = 16 ; No. of neutrons in Y i.e., Cl = 35-17 = 18 Therefore, the ratio = Y : X = 9 : 8.

IIT JEE WORKSHEET 1. 6. 11. 16. 21. 26. 31. 35. 40.

2 1 4 2 2 4 2 4 2

2. 1 7. 4 12. 3 17. 3 22. 4 27. 2 32. 2 36. 4 41. 3

3. 3 8. 3 13. 2 18. 1 23. 2 28. 2 33. 4 37. 4 42. 3

4. 4 9. 4 14. 3 19. 2 24. 1 29. 2 34. Z + 6 38. 2

5. 4 10. 3 15. 3 20. 2 25. 2 30. 1 39. 3

4. PERIODIC CLASSIFICATION SOLUTIONS (b) 63

FORMATIVE WORKSHEET Column – I (Classification) Earth, Air, Fire and Water Metal and non-metals Atomic weight table

1. A) B) C)

2.

i) ii) iii)

(c) Germanium

Column – II (Scientist) Greeks’ classification Lavosier’s classification Dalton’s classification

(d) Gallium, germanium (e) Be, Au and Pt. 8.

i – q , ii – r , iii – p

9.

Rare earth elements and actinides.

10. a)

i) False ii) True

A

D

C

L

I

T

F

O

10

10

2

6

0

7

8

4

p If the elements of a triad are arranged in order of increasing atomic mass, the atomic mass of the middle element is the average of the atomic masses of the other two elements.

3.

Thus, At.wt of D 

4.

5.

At.wt of A  At.wt of C 2

137  40 177   88.5  2 2

c)

According to Doberiener, when elements were arranged in order of their increasing atomic mass, the atomic mass of the middle element was approximately the arithmetic mean of the other two elements of the triad.

d)

We know, Newland’s law of octaves states that “When the elements are arranged in the order of their increasing atomic weights, then the properties of the elements were repeated at every eighth element like the eighth note of an octave in music.” X 1

b)

2

3

4

5

6

7

Y 8

e)

e

n

A

–

10

10

10

C

–

2

2

2

L

–

6

6

6

I

–

7

7

7

F

–

8

8

8

I =A– Z = 1 – 1 = 0 A

–

Z

=

n

A

20

–

10

=

10

D

19

–

9

=

10

A, C, L, T, F

11. The electronic configuration of A is 2, 8, 4. Thus, the element belongs to IVA group and 3rd period. And 3rd period can accommodate only 8 electrons.

12. The valencies can be found out from their electronic configuration. Element X Y Z

Z 1 11 17

EC 1 2, 8, 1 2, 8, 7

Valency 1 1 8–7=1

Group number = No. of valence electrons 6 Elements

The number of elements in between ‘X’ and ‘Y’ =6 6.

The discovery of inert gases gave a death-blow to Newlands law of octaves.

7.

a) The physical and the chemical properties of the elements are the periodic functions of their atomic weight.

X and Y belong to I A group. Z belongs to VIIA group. 13. P - an inert gas, Q is a representative element and R is a transition element. 14. (a) Atomic number of element = 36 Therefore its electronic configuration: 2, 8, 18, 8

9th Class Chemistry

290

We know that Group number = Number of valence electrons

The atomic number of B is 1 unit less than C  B belongs to VIIIA group.

In the above case, Number of valence electrons =8

The atomic number of A is 2 unit less than B  A belongs to VIA group.

Hence group number = 8 Period number = Number of shells present In the above case, Number of shells = 4 Hence period number = 4 (b) We know that group number of the element =8 = 8 – group number = 8 – 8 = 0

18. Pnicogens belong to Nitrogen family, Chalcogens belong to Oxygen family, Halogens belong to Fluorine family and aerogens belong to Helium family.

Hence the element belongs to ‘0’ group.

19.

Valency of an element after IVA group

(c) The element is Krypton The neighbouring elements of the group are He, Ne, Ar, Xe and Rn. He, Ne, Ar, Xe and Rn. 15. a) Alkali metals have 1 electron in their outermost shell. Hence, D is an alkali metal. b) Alkaline earth metals have 2 electrons in their outermost shell. Hence, C is an alkaline earth metal. c) Halogens have 7 electrons in their outermost shell. Hence, T is a halogen. d) Noble gases have 8 electrons in their outermost shell. Hence, L is a noble gas. e) Elements that belong to VA group have 5 electrons in their outermost shell. Hence, E is a VA group element. f) The elements that belong to 2nd period contain two shells. Therefore, A, L and E are 2nd period elements. g) Noble gases have zero chemical reactivity. Hence, L has zero reactivity. h) Elements in which the outermost and penultimate shells are incomplete are called transitional elements. Hence, S is a transitional element. 16.

17. 10th element is Neon and it belongs to 2nd period. To get the next period element in the same group, we need to add 8. So, the next element after 10th element is 18th atomic numbered element. For next two elements, we need to add 18 and 18. Therefore, the atomic numbers of the next two elements is 36 and 54 respectively.

Element Atomic number A Z B Z+2 C Z+3 Given that C is an alkali metal.

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Period

Length

No. of elements 1st period Very short period 2 2nd and 3rd periods Short periods 8 4th and 5th periods Long periods 18 6th period Very long period 32 7th period Incomplete period --

Therefore, the right match is: i) - s , ii) - r, iii) - q, iv) - p 20. i) - q, ii) - s, iii) - p 21. Element Atomic number i) Fermium 100 ii) Lawrencium 103 iii) Mendelevium 101 iv) Nobelium 102 22. A) Unnilunium  Z = 101  Mendelevium B) Unnilbium  Z = 102  Nobelium C) Unniltrium  Z = 103  Lawrencium 23. 120  Un + bi + nil = Unbinilium 24. i) – T ii)– F iii) – T iv) – F 25. 1. Li and F belongs to same period. In 2nd period, as we move from IA to VII A, the atomic size decreases. Thus, the size of Li is greater than F. 2. For a given element, as the positive charge increases, the atomic size decreases due to increase in the nuclear pull. 3. Na+ and F– are isoelectronic ions. Since amongst isoelectronic ions, the size of the anion is larger than that of the cation, therefore, Na+ < F–. 4. The atomic size of sulphur is greater than that of oxygen due to more number of shells in sulphur.

Periodic Classification Solutions 26.

Ion K+ Cl S2 Ca+2

Electronic Configuration 2, 8, 8 2, 8, 8 2, 8, 8 2, 8, 8

From the above, it is clear that the given species are isoelectronic in nature. In isoelectronic ions, the lesser that the nuclear charge, larger the size. Hence, the size of sulphide ion is greater than that of any other ion. 27. In, LiI, the anion is largest and the cation is smallest. 28.

Elements

Descriptions These elements belong to same period (3rd) and are arranged in Na, Mg, the increasing atomic numbers. Al, Si Hence, these are arranged in the decreasing order of their atomic radii. These elements belong to same period (2n d) and are arranged in the increasing atomic numbers. C, N, O, F Hence, these are arranged in the decreasing order of their atomic radii. These elements belong to same group (VIA) and are arranged O, S, Se, in the increasing atomic Te numbers. Hence, these are arranged in the increasing order of their atomic radii. These elements belong to same group (VIA) and are arranged in the decreasing atomic I, Br, Cl, F numbers. Hence, these are arranged in the decreasing order of their atomic radii.

29. The size of P is greater than that of S due to lesser nuclear pull in P. The size of C is greater than that of N due to lesser nuclear pull in C. Thus, the right order is : N< C < S < P 30. Among the isoelectronic ions, Mg2+, Na+ and F–, size increases in the order :

291 31. From the above values it is clear that there is large jump in energy from 3rd IP to 4th IP. This is possible only when the element attains stable configuration after the removal of 3 electrons. Hence, the element has 3 electrons in its outermost shell. 32.

Species No. of e’s Ne 10 +3 Al 10 +2 Mg 10 Na+ 10

E. C 2, 8 2, 8 2, 8 2, 8

From the above chart, it is clear that all the species have same electronic configuration. Hence, the IP depends on the extent of nuclear pull. The greater the nuclear pull, the greater the IP. Further, the extent of nuclear pull depends on the number of protons. Therefore, the species with greater number of protons have greater IP and vice-versa. Therefore, the increasing order of IP of the given species is as follows: IP of Ne < Na+2 < Mg+2 < Al+3 33. The IP of cobalt is more because of its small size. 34. A and B belong to second period whereas C and D belong to third period. In general, IE increases along the period and decreases down the group. Therefore, the correct order is B > A > D > C. 35. The reaction, Mg  g    Mg 2  g   2e  occurs in two steps: IE1 Mg  g    Mg   g   e  ;

 H 178 kcal mol1 IE 2 Mg   g   Mg 2   g   e  ;

 H  348 kcalmol 1 Mg  g    Mg 2   g   2e  ;

 H  526 kcal mol 1

36. Cl2O7

Mg2+ < Na+ < F–

37. Na 2 O

Al is the largest due to the presence of more number of shells.

38. This is due to the presence of free electrons. 39. A is a non-metal and B is a metal. www.betoppers.com

9th Class Chemistry

292 40. A  Acidic  Non-metal

10. Carbon, nitrogen and oxygen is a wrong triad as they do not have any similar chemical properties.

B  Amphoteric  Metalloid

11. 1

C  Basic  Metal We know that in a period, as we move from left to right, metallic character decreases and non-metallic character increases. Therefore, their positions in the respective period is as follows: C B A And also, as we move from left to right, the atomic number increases. Therefore, the increasing order of their atomic numbers: C < B < A.

CONCEPTIVE WORKSHEET 1.

Only (1) and (2) are correct. Hydrogen is a nonmetal and chlorine is an electronegative element.

2.

3.

Non-metals are the elements which are brittle, bad conductros of heat and electricity and do not possess metallic lustre .

12. In the above table, the element ‘Na’ shows similar properties with ‘Li’. Sodium is the eighth element after Li. Hence, the table is related to Newland’s law of octaves. 13. Both assertion and reason are correct but reason is not the correct explanation of assertion. 14. I A, VII A

15. 3

16. 4

17. 2

Mendeleev’s periodic table helped in correcting the atomic masses of some of the elements, based on their positions in the periodic table. For e.g., atomic mass of beryllium was corrected from 13.5 to 9. Atomic masses of indium, gold, platinum were also corrected. 18. According to Moseley, the properties of the

elements are the periodic functions of their atomic numbers. 19. Mass number = Number of protons + Number

1) The classification may help to study them better.

of neutrons

2) The classification may lead to correlate the properties of the elements with some fundamental property that is characteristic of all the elements.

Number of neutrons = y - x

3) The classification may further reveal relationship between the different elements

20. Frequency 21. The middle portion of the periodic table includes transition metals.

4.

Elements with same valency are classed together.

22. The inner transition elements constitute both lanthanides (58Ce to 71Lu) and actinides (90Th to Lr). 103

5.

Elements which are malleable and ductile good conductors of heat and electricity and posses characteristic metallic lustre were named as metals.

23. Number of elements in each period = 2 × number of orbitals available in the energy level that is being filled

6.

Both the given statements are correct to reject Doberiener’s triads.

24. The middle portion of the periodic table is occupied by transition elements.

7.

The law of triads was proposed by Dobereiner

25. Atomic number of element M = 20

8.

The law of triads is applicable to chlorine, bromine, iodine.

9.

The mean atomic weight of calcium and barium is equal to the atomic weight of strontium.

Element Calcium Strontium Barium Atomic mass

40.0

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87.5

137

Arithmetic mean 40  137  88.1 2

Number of electrons = 20 Electronic configuration = 2, 8, 8, 2 Number of electronic shells = 4 Number of valence electrons = 2 a) IIA group b) 4th period c) 2

Periodic Classification Solutions d) Loses 2 electrons to attain stable configuration of M2+. Therfore, its oxidation state is +2. e) 2, 8, 8 and 2 f) As it loses electrons to attain stability, it is a metallic element. g) Be, Mg, Sr, Ba, Ra h) Vanadium – V i) Krypton - Kr j) Be k) Ra l) MCl2, MO, M(NO3)2 m) Argon n) 20 o) Potassium – K p) Argon.

293 36. In a group, as we move from top to bottom, ionic radii increases. Therefore, the correct order of their size is : Li+ < Na+ < K+ < Rb+ 37. For isoelectronic species, the greater the nuclear charge, the lesser its size. Thus, the ionic radius of Si+4 is less than O–2. 38. From the question, it is clear that the size of X is very much smaller than that of Y. Therefore, X and Y are fluorine and neon respectively. 39. F < O < C < Cl < Br 40. O2 –>F–>O>F 41. It is defined as the amount of energy required to remove the most loosely bound electron from the isolated gaseous atom. 42. Ionisation potential.

26. Synthetic or Transuranic elements

43. Electron volt / atom or kilo joules / mole

27. 1, 2

44. For a given element, greater is the IP IP n–1

Electronic configuration Unununnium Ununnillium Unnilennium

28.

Z 111 110 109

Element Rontgenium Darmstadtium Meitnerium

29. The concept of orbits was introduced by Neils’ Bohrium. Therefore the element is Bhorium and its atomic number is 107.

Z 104 105 106

30.

Element Rutherfordium Dubnium Seaborgium

Element Hassnium Meitnerium Darmstadtium

31.

Symbol Unq Unp Unh

Z 108 109 110

32. Lanthanides and actinides 33. i) Isotopes ii) Atomic number 34. 3 35.

The size of neon is greater than that of fluorine. Therefore, atomic size of fluorine and neon is 0.72A0 and 1.6A0 respectively.

n+1

> IPn >

45. There is greater shift from 2 nd IP to 3 rd IP. Therefore, the element belongs to IIA group i.e., calcium. 46. VIIIA group elements have highest IP. Further, IP decreases down the group. Therefore, among the given elements, helium has highest IP. 47. In general, IP increases across the period. 48. both increase in size of atom and increase in screening effect 49. As there is greater shift from 2nd IP to 3rd IP, the element belongs to IIA group i.e., alkaline earth metals. 50. Inert gases have highest IP among the given elements. Hence, they occupy the peaks. 51. The third IP for an alkaline earth metal is very high. Hence the value of x will be 154 eV/atom. 52. The second ionization potential value is least for IIA group elements. 53. Cesium 54. decreases, increases 55. decreases , increases 56.

Zero

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9th Class Chemistry

294

SUMM ATIVE WORKSHEET 1.

2.

b) It is a tabular chart, representing systematic arrangement of elements in vertical columns, called groups, and horizontal rows, called periods, in the order of their increasing atomic weights. i) The elements in a periodic table are arranged in vertical columns, known as groups and horizontal rows known as periods. ii) Seven and Eight iii) All of them have similar properties due to presence of same number of valence electrons in their outermost shell. iv) (a) It includes anomalous pairs in which atomic weights of preceding elements are higher than there of the following elements. (b) The position of hydrogen was not fixed. (c) It could not provide a place for rare earth elements and actinides. (d) Isotopes could not find any place in this table.

4.

Proceeding elements Cobalt (58.9) Tellurium (128) Argon (40)

Period No. 1 2 3 4 5 6 7

Starting element H Li Na K Rb Cs Fr

(i) Rubidium (ii) Beryllium (iii) Radioactive halogen - Astatine;

Type of elements 25 element

Manganese

th

50 element

Tin

th

Fermium

th

Rutherfordium

100 element 104 element 6.

7.

.

Name

th

i) In 6th period, elements from Cerium (Ce) with atomic number 58 to Lutetium (Lu) with atomic number 71 are called Lanthanides. ii) In 7th period, elements from Thorium (Th) with atomic number 90 to Lawrencium (Lw) with atomic number 103 are called Actinides. iii) The elements beyond Uranium (Z = 92) are known as trans-uranic elements. iv) The elements beyond Fermium (Z = 100) are known as trans-fermium elements. v) The elements placed in A subgroups, IA and IIA on the left and IIIA through VIIIA (0) on the right are called typical elements. They are also called representative, normal or main group elements. vi) The B-subgroups, i.e., the 5 columns of (III – VII)B, 3 columns of group VIII B and 2 columns of (I – II) V, contain elements which are known as transitional elements.

Atom Valence electrons X 1 Y 3 Z 7

Valency Nature 1 3 1

Metal Metal Non – metal

Compound formation is possible between formula. (1) X and Y - XY 8.

Radioactive inert gas – Radon

9. www.betoppers.com

Name Copper, Silver and gold Ru, Rh, Pd, Os, Ir & Pt

Platinum metals

Following elements Nickel (58.6) Iodine (127) Potassium (39) Ending element He Ne Ar Kr Xe Rn –

Type of elements Coinage metals

a) The physical and chemical properties of the elements are the periodic functions of their atomic weights.

(v)

3.

5.

(2) i) ii) iii)

Y and Z - YZ3 Position of hydrogen is unresolved. Position of helium is not justified. It fails to accommodate lanthanides and actinides in the main body of the table. iv) The arrangement is unable to reflect electronic configuration of many elements in transition group, in lanthanides and actinides. False

Periodic Classification Solutions

295

10. x – Inert gases; y – Representative elements

HOTS WORKSHEET

z – transition elements. 11. A, B, C are three elements with atomic numbers, Z (i) A – VIIA group ; B – IA group

1.

12. Z, Y, X

Strong metals are present on the left hand side, weak metals are present in the middle and nonmetals are present on the right hand side in the periodic table. And metallic nature increases down the group and decreases across the period.

13. 24.5

(i) – D, T, O, T, P, I ;

14. Let the number of protons in M be ‘x’. Nuclear charge (number of protons) is same for all species.

(iii) – S, L ; iv – A

(ii) B (iii) C

(ii) – C, I, P;

2.

G  q, A  s, T  p, E  r

3.

ii and iii are correct

4.

All are incorrect

5.

In the middle of the periodic table are d-block elements which are typical metals and not metalloids.

Number of electrons are 10 in all the three species.

6.

All are correct

7.

i - s, ii - r, iii - q, iv - p

Number of protons are 9, 8 & 7 respectively.

8.

L

So, here, number of electrons  Atomic size. The number of electrons, the greater is the atomic size. Increasing order of atomic size +2

+1

–1

–M 5. The outermost electrons for R, A and M are 2, 7 and 5 respectively. Therefore, the valencies of R, A and M as 2, 1 and 3 respectively. We also know that, group number is equal to the number of outermost electrons of the element. Hence, R, A and M belong to II A, VII A and V A respectively. Elem ent

Z

O

8

P) 2 Q) 2, 4 R) 2, 7 S) 2, 8

E.C

V-e’s

Lewi s symbol  

F

9

2, 7

7

. O. F :

Al

13

2, 8, 3

3

 A 

2, 6

6

24.

25. 27. 28. 30.

31.

 

   



20. 21.

22.

3, IIIA 2, VIA and 0, VIIIA From the Lewis’ symbols, it is clear that the valencies of A, B and C are 2, 3 and 2 respectively. It can also be understood that A is a metal, B and C are non-metals. The different compounds possible by their combinations are : A3B2, AC, B2, C2, BC, B2C, BC2. Change in the options of the question The electronic structure of four elements P, Q, R and S are: P) 2 Q) 2, 4 R) 2, 7 R) 2, 8 The tendency to form electrovalent bond is greatest in a) P b) Q c) R d) S

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32.

35.

43.

44.

– – – –

He C F Ne

In these four elements Fluorine can form greatest electrovalent bond. Since, it has small size it can easily gain one electron to attain stable octet configuration than Helium, Carbon and Neon. Since a strongly electropositive element ‘X’ can form cation, X+ easily and an element ‘Y’ which is strongly electronegative can form anion easily i.e. Y–. Since both are univalent both are held by electrostatic attraction as X+Y–. Ionic nature  Electronegativity difference between the ‘2’ elements in the given compound. Among the given compounds i.e., CaF2, CaBr2, CaCl2 and CaI2 the electronegativity difference between calcium and iodine is least. So, CaI2 is least ionic. N2 < ClF3 < SO2 < LiF < K2O. B < Al < Mg < Ca < K < Cs CsI – Strongest LiI – Weakest Ba+2 has more charge than Ag+ and hence cannot be formed easily. Thus, Ag+ forms easily and the ionic bond is stronger. Therefore, melting point of Ag2O is greater than BaO. The lattice energy of barium sulphate is greater than lattice energy of Sodium sulphate. Hydration energy of barium sulphate is less thanhydration energy of sodium sulphate. Lattice energy of barium sulphate is greater than that of hydration energy of barium sulphate. So, it is sparingly soluble. The greater the dielectric constant, the more the polar nature of the solvent. Hence, in C6H6 (D = 0), CH3COOCH3 (D = 2), CCl4 (D = 0) dielectric constant values are less than CH3OH (D = 32). So, KCl is soluble in CH3OH. AgNO3  HC  AgC  HNO3

Since, HCl is polar, it gives white precipitate with silver nitrate. NaCl < MgCl2 < AlCl3 [If the non-metal is same for all the compounds, then the compound containing least sized metal is most covalent]. a) Y has six electrons in its valence shell. b) 2 c) Covalent

Chemical Bonding Solutions

299

55. d)

Ca + 2, 8, 8, 2

Y 1 2-

Ca2+

Y

or Ca2+ Y2–

2, 8, 8

45.

46.

or CaY

The maximum number of covalent bonds by which 2 atoms bonded to each other are 3 as in case of N2 molecule. N  N Molecule BP LP Tot. e’s CH4

4

0

8

CO2

4

4

16

NH3

3

1

8

BCl3

3

9

24

47.

57. 58.

63.

Green vitriol (FeSO4.7H2O), chile or Indian salt petre (KNO3 or NaNO3) and potash alum (K 2 SO 4.Al2 (SO 4) 3. 24H 2 O) are ionic compounds and dissolve in water to form conducting solutions. CCl4 gives white ppt. with AgNO3 solution is an incorrect statement. Molecules are more stable since, they have lower energy than the respective ions. So, ions react more quickly than molecules. More is the EN difference, more polar is the bond. H |

70.

 H+ .......Cl  a) H  N | H

71. 72.

b) O  S  O A -iv; B - i ; C – ii ; D – iii The structure of Ammonia Boron trifluoride is H3N  BF3.

73.

A : B :  A   B  ionic linkage  . A .  .B  A : B  cov alent linkage 

on Therefore BeCl2 contains no lone pairs on central atom.

A : B  A : B  co  ordinate 

H H

48.

49. 51. 52.

53.

a) H C C H

 7 bond pairs H H b) H N H  3 bond pairs H F F c) F S F  6 bond pairs F F F F Br F d)  5 bond pairs F F

a - (iii), b - (ii), c - (iv), d - (i) Four bond pairs  the molecule should contain 4 combining atoms. Among the given molecules, only XeF4 have 4 combining atoms. a) X-gaint structure, may be graphite X - covalent. b) Y-covalent compound, Z - ionic compound. LiCl would ionise more in water than NaCl is a wrong statement because LiCl has covalent character.

addition

74.

Thus co-ordinate linkage may be suggested as a combination of electrovalent and ionic linkage. a) Structure of hydronium ion (H3O+): Hydronium ion is produced when hydrogen (H+) ion combines with water molecule. In this case (H+) ion accepts a lone pair of electrons present on O atom in H2O forming a coordinate bond. H – O – H + H+

+ +

b)

+

H – O – H or H – O – H H H

Structure of ammonium ion  NH+4 

ion: Ammonium ion is produced when combines with NH3 molecule. In this case hydrogen H+ ion accepts a lone pair of electrons present on Natoms of NH3 molecule forming a coordinate www.betoppers.com

9th Class Chemistry

300

79.

bond.

O  H—O — P — O — H | O | H

H N H + H+ H

+ H N H or H

75. 76. 77. 78.

+

H H–N–H H

10, 12, 12 respectively. No.of valence electrons in N = 5 No.of valence electrons in F = 7 Both ‘a’ and ‘b’ satisfies the stability of CO32–. c

The correct bond structure is

H

80.

81.

H

C

C O

(a)

H C

O

Cl

P Cl

(b)

H

C

N

Cl

(c) (d)

NF3

(iv)

NO

× ×

(iii)

× ×

H2S

×

(ii)

O +

Four bond pairs + no lone pairs = 4 Shape – tetrahedral

× ×

CCl4

× ×

(i)

×

Cl ×× Cl ×Cl × Cl ×× × × × Cl × × ×

× ×

× ×

O N O

× ×

82.

–O

S H

Two bond pairs + two lone pairs = 4 Shape – bent like water

F × N × F ×× × × × F ×

Three bond pairs + no lone pairs = 4 Shape – pyramidal like NH3

×

×

× × × ×

× ×

× ×

× ×

× ×

×

× ×

H

(All diatomic molecules must be linear

× ×

N ××O×× × ×

AlCl3

× ×

× ×

×

× ×

× ×

×

(v)

Cl×× Al×

× ×

Cl Cl××

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Three bond pairs + one lone pairs = 3 Shape – triangular planar like BCl3

H

Chemical Bonding Solutions

301

CONCEPTIVE WORKSHEET 1

2

3

4

5

6

7

8

9

10

a

a

*

b, c

b

c

b

b

d

b

11

12

13

14

15 16

17

18 19 20

b

d

d

b

a

c

a

b

21

22

23

24

25 26

27

28 29 30

*

*

*

*

b

c

*

*

31

32

33

34

35 36

37

38 39 40

*

d

d

d

c

a

c

b

41

42

43

44

45 46

47

48 49 50

a

d

b,c

a

*

b

c

b

51 52 a,b d c,d 61 62

53

54

55 56

57

58 59 60

b

a

c

c

d

c

63 a, b a c, d 72 73

64

65 66

67

68 69 70

c

c

a

c,d

b

74

75 76

77

78 79 80

c

c

b

c

c

d

d

81

82

83

c

d

*

b 71

a

b d c b

d

a

c

* c c

d

c

b

* = Subjective questions

Easily formed



Best donor

+

Best acceptor

More Electropositive

More Electronegative

More size

Less size



All are the members of zero group and are chemically inert, monoatomic and noble gases. ‘He’ and ‘Ar’ are inert and do not form molecules due to their stable complete duplet and octet configuration respectively. Lewisand Kossel Less energy and more stability. When two atoms approach each other and a bond is formed, potential energy is decreased and hence, attains stability. The chemical stability is more for ions than for parent atoms, since ions possess octet configuration in the outermost valence shell. For example, the configuration of Na is 2, 8, 1 and that for Na+ is 2, 8. The maximum valency = The number of valence electrons present in an atom Hence, maximum valency of Sulphur= 6 In ‘MO2’ Valency of the metal with respect to oxygen is 2.

26.



9.

25.



8.

24.



7.

20. 21. 22. 23.



3. 4. 5.

19.

Valency of A = 3 Valency of B = 8 – 6 = 2. Formula : A2B3 Comparatively, sulphur exhibits variable valency. The no. of electrons in the outermost shell is represented in the Lewis symbols. The no. of electrons in the outermost shell is represented in the Lewis symbols. Kossel Octet Theory Valence electrons If the difference in electronegativity of two elements is greater than 1.7, then the two elements can form electrovalent bond easily. Since, here the difference between two elements is 2.5 (3.5 – 1.0) it forms electrovalent bond. In ionic compounds, the cations and anions are easily held together by electrostatic forces of attraction. Since NaCl is an ionic compound, it is held by strong electrostatic force of attraction. In the given compounds CH4, MgCl2, SiCl2, SiCl4 and BF3; MgCl2 can form electrovalent linkage. Since, the elements presents in the remaining compounds are non-metals and can form only covalent linkage.



2.

12. 18.

d

HINTS OR ANSWERS TO THE SELECTED QUESTIONS

1.

11.

K

27.

KCl

Cl

More is the EN difference between the two elements in a compound, more is its ionic character. 

Ionic nature   E.N C  EN m 

 Cl is common in all the elements    IN max   EN Cl  EN m max   EN m min   Size  max  Cs 28.

Cesium chloride has the greatest ionic character. The element with low ionisation potential can form cation easily and the element with high electron affinity can form anion easily. These are the favourable conditions for the formation of ionic bond. www.betoppers.com

9th Class Chemistry

302

29.

30. 31. 33.

34.

35.

36.

37.

38.

Ionic bond is formed between a metal and nonmetal easily. Here in the given compounds, Si, Al, B and Calcium are combining with Nitrogen to give the respective Nitrides. In Si, Al, B and Ca, Calcium is more metallic in nature. So, Ca3N2 is the more ionic compound. MCl3 < MCl2 < MCl, since charge on the metal increases ionic nature decreases. i, v, vi – p ; ii, iii, iv – q Boiling point  Electrostatic force of attraction. More the electrostatic force of attraction more the boiling point. In potassium chloride there is strong electrostaticn force of attraction between the opposite ions is present. So, it has more boiling point. Process of surrounding of water molecules around the oppositely charged ions of crystal is called hydration. This hydration power of water molecule is due to its dipole. (Partial positive charge and negative charge at the ends of hydrogen and oxygen respectively). This hydration power of water molecules decreases the inter ionic attraction in the crystal lattice. Hence, ionic crystals are soluble in water. Most of the ionic substances are soluble in polar solvents like water. Because, dissolving of an ionic solid involves the setting of opposite ions free from the lattice into the solvent. This can happen when the strong electrostatic force of attraction between the opposite ions is weakened. Therefore, solvents having oppositely charged ions, called polar solvents should be used. The best polar solvent is water. Therefore, all ionic compounds are dissolved in water. More electronegativity difference  More polar  More soluble in water. As electronegativity difference between silver and fluorine is more, it is more soluble in water. In ionic solid, ions are tightly packed and these occupy fixed positions in solids. These ions now, cannot move freely to conduct electricity. Hence, these are poor conductors of electricity. Ionic substances are generally soluble in polar solvents solubility of solute  dielectric constant.More the di electric constant, more is the solubility.

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45.

47.

LiF < LiCl < LiBr < LiI [If the metal is same for all the compounds then the compound containing more sized non-metal is most covalent].  4 bond pairs

48.

 6 bond pairs.

50. H

Cl

b) Cl C Cl

a) H C H

51.

Cl

H O c) H H

52.

a) O=C=O c) N  N + a) N =N =O c) H C N

b) O=O d) H–C  N. b) O =C=O d) N  N

53.

a) H P H  1 lone pair H

b)

 2 lone pairs

c)

 no lone pair

d) 56. 57. 58. 60. 71. 74.

 1 lone pair

Covalent bonds are present in diamond. In general the reaction between covalent molecules is slow. Covalent molecules have definite geometry. Electron cloud is unevenly shared in polar covalent bond. NNO

The structure of HNO 3 molecule is HON  O 

O

75. 78. 79. 80.

The structure of N2O5 is O  N  O  N  O   O O In BF3, around B, there are six electrons. c No. of valence electrons in S = 6 No. of valence electrons in 4 ‘O’ = 4×6 = 24 Extra electrons = 2

Chemical Bonding Solutions

83.

b)

NaOH has 1 ionic bond and 1 covalent bond, but no dative bonds. (a) (b)

PH3

H× P× × H H

BCl3

Cl × B × Cl Cl

9.

×

82.

303

SUMM ATIVE WORKSHEET

2.

3. 4. 5.

c)

 2 lone pairs

d)

 2 lone pairs

a)

H N H

b) N  N

H

c) O = O

Octet rule is not obeyed by BCl3.

1.

 3 lone pairs

10.

The electrovalent bond is electrostatic in nature and So, there is specific direction for the bond. Hence, they do not exhibit isomerism. The number of oppositely charged ions surrounding a given ion immediately in a crystal is called coordination number. In sodium chloride crystal, each sodium cation is surrounded by six chloride ions and vice-versa. d d a) Benzene:

11. 12. 13. 14. 15. 16.

H

d)

a) O C O

b)

O

O H N

H

O

c) C N d) O N O Therefore, NO2 contains only one electron. b Covalent compounds are non-electrolytes hence statement (b) is correct. c d c CH4, C2H2  Covalent KCl  Ionic KCN  Ionic + Covalent H |

C H H

17.

C

C

C

C

H H

H

18.

C H

2 bond pairs between two C atoms. b) Acetylene : H  C  C  H 3 bond pairs between two C-atoms

H H | | c)

SO4

NH3

|

1 bond pairs between two C - atoms. |

NH3 Cu

19.

H H

d)

Dative The structure of N2O5 is O  N  O  N  O    O O Covalent + Dative NH3

Ethane: H  C  C  H |

H  N  H+ .......Cl   Ionic + Covalent + |

NH3

 The bonds present between copper and ammonia are dative bonds

|

Ethylene : C|  C|

6. 7.

2 bond pairs between two C - atoms. a c

8.

a) H N H  1 lone pair

O

20.

Cu2+



O

S



O

.5H2O

O

H www.betoppers.com

9th Class Chemistry

304

Molecule N2 CO2 HCl MgCl 2 CCl 4 HCN

HOTS WORKSHEET 1.

a) Y 1

2. 3.

4.

5. 6.

7.

Ionic bode +

X 1

X

-

or Y+ X -

b) Formula of the compound is YX a) Electrovalent or ionic bond b) QP Total numbers of lone pairs = Number of H2O molecules present in 9 grams of water × No. of lone pairs in 1 molecule of water. 9  N  2 [ since 1 molecule of water = 18 contains 2 lone pairs] = N. The total number of valence electrons present in phosphorus is 5. The total number of valence electrons present in O34 is 27. Therefore total number of valence electrons present in PO34 is 32. b a) Both ionic and covalent bonds. b) Both ionic and covalent bonds. c) Covalent bond. d) Covalent bond. e) Covalent bond. a) E lem e nt H C Na Mg

F orm ula of chl oride HCl CC l 4 N aCl M gC l2

Element

b)

C N O F

9. 10. 11.

12.

13. 14.

Nature o f b on d Cov ale nt Cov ale nt Ionic Ionic

Name of the compound with hydrogen Methane Ammonia Water Hydrogen fluoride

15.

a) Ionic bond b) Covalent bond c) Ionic bond d) Covalent bond b LiCl, BeH2 and BCl3 has less numbers of electrons than octet on their central atom. In PCl5, around ‘P’, there are 10 electrons. In SCl4, the outermost shell of sulphur consists of 6 electrons. i.e.,

So, sulphur atom consists of total 10 electrons which is a deviation from octet rule and it also consists of lone pair of electrons. d No. of valence electron in N = 5 No. of valence electron in 4H = 4×1 No.of electrons deficiet = 1 Total = 9–1=8 All the correct.

IITJEE WORKSHEET 1.

NaF is an ionic compound, whereas F2 is a covalent gas.

2.

H

Formula CH 4 NH 3 H 2O

O

O

C N

HF

3.

4K+

8.

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Nature of bond Covalent bond Covalent bond Polar covalent bond Ionic bond Covalent bond Covalent bond

H

C N



C N

Fe C N

C N C N 

4.

Na  C  N  Ag  C  N

5.

a) K+Cl–

Chemical Bonding Solutions

F

F

F

b)

I F +

6. 7. 8. 9. 10. 11.

12.

13.

14.

15.

305

F 2–

c) 2K .O d) K+ –O – H a c d c d a) X is non-polar covalent, Y is ionic, Z is polar covalent b) (X = Br2, Y = NaCl; Z = HCl) 2 A : 1s , 2s2 2p6, 3s2 3p6, 4s1 19 Valency of A = + 1 in AB  B : 1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p5 35 Valency of A = – 1 in B2  C = 1s2, 2s2 2p6, 3s2 3p1 13 Valency of A = + 3 in CB3  AB ionic, B2 covalent, CB3 covalent, DB2 and DB3 ionic. i) If  (electronegativity) = 0 ; Bond is 100% non-polar. ii) If  (electronegativity) = less; Bond is more covalent, less ionic. iii) If  (electronegativity) = more; Bond is more ionic, less covalent. i) X is in VIA group; Y is VIIA group; Z is IVA group. ii) H2X; H Y; ZH4; (Write Lewis dot structure yourself). iii) X2– and Y–. a) Ionic bond. b) Both ionic and covalent bonds. c) Both ionic and covalent bonds. d) Covalent bond. e) Ionic bond. 

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306

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9th Class Chemistry

6. GASEOUS STATE SOLUTIONS

FORMATIVE WORKSHEET 1. 2.

3.

4. 5.

Gases do not have any definite volume. Their volume is equal to the volume of the container. Gases, unlike solids and liquids exert pressure on the walls of the container due to their random motion and continuous collisions. Gases diffuse from one place to another. The intermolecular force of attraction between the gas molecules is negligible.

4 4000   10 3  cm 3 3  V1 = l3 V2 = (2l)3 = 8l3 Vgas =

 V2  % change =  V  1  100  1   8l3  =  l3  1  100 ;  

6.

7. 8.

8  =   1  100 = 700 1  1 litre = 10 cc = 10×(10–2m)3 = 10–3m3  1 m3 = 1000 L  0 C = K . Hence, the temperature variation in kelvin scale would be same i.e., 35°C 1 atm = 105 N/m2 ___________ (1) Let us express N/m2 in dyne/cm2 We know, 1 N = 105 dyne 1m2 = (102 cm2) = 104 cm2 Substituting the above values in (1), we get

105  105  106 dyne / cm 2 104 cm2 Therefore,1atmosphere is equal to 106 dyne /cm2 9. 1 atm = 105 N/m2 = 105 Pascal = 760 torr = 106 dyne/cm2

10. No. of moles =

mass 8 1   4 GMW 32

m GMW  m = n×GMW = 4×28 = 112 g 12. We P = hdg Therefore, for a given pressure, height is inversely proportional to the density of the liquid used in the barometer. As the density of oil is less than that of water, the height of oil barometer would be more than that of water barometer. 13. As the air enters through the tube, the pressure reading will have an error. 14. As P and d are same, height of the liquid column would be inversely proportional to each other. h1g1 = h2g2 11. n =

h1g1 76  g   76  6  456cm g g2 6 The liquid column rises to 456 cm from 76 cm. 15. Atmospheric pressure = Pressure of mercury column of height 76 cm. h = 76 cm = 0.76 m ; d = 13600 kgm–3 g = 10 ms–2 P = hdg = 0.76 × 13600 × 10 P = 1.03 × 105 P Ther efore, the atmospheric pressure is 1.03× 105 P 16. hmercury = 76cm dmercury = 13.6gcm–3 gmercury = g hwater = ? dwater = 1gcm–3 gwater = g We know that P = hdg Pmercury = Pwater hmercury × dmercury= hwater × dwater h2 

76  13.6 1  1033.6cm  10.34m Therefore, the height of the water column is equal to 10.34 m. 17. Atmospheric pressure (P0) = 105 P Depth (h) = 5.1 m Density (d) = 1 g/cc = 1000 kg/m3 g = 10 m/s2 Pdepth = Patmosphere+ Pwater column = P0 + hdg ___ (1) Substituting the above values in (1), we get P = 105 + 5.1 × 103 × 10 Pa = 15.1 × 104 P Therefore, the total pressure at the bottom of the lake is 15.1 × 104 P  h water 

9th Class Chemistry

308 18. h1 = 70 cm = 0.7 m d1 = 13600 kg /m3 h2 = ? d2 = 1000 kg /m3 We know, P = hdg As the pressure is the same, P1 = P2 i.e., h1d1g = h2d2g

 h2 =

P1V1 = P2 V2 , 1× 20 = P2 × 50,

P2 = 20 ×

1 atm. 50

20. P 1V1 = P 2V2

 9.962 ×104 × 95 = 10.13 ×104 × V2

⇒ V2 = 21.

9.962×104 × 95 = 93.424 cm 3 10.13×104

P1 V1 = P2 V2  10 4 × 100 = 10 5 × V2

 or  V2

= 10 cc

V1 = l3 P1 = P 3 3 V2 = (2l ) = 8l P1 = ? Temperature is constant  Boyle’s Law is applicable  P1V1 = P2V2 Þ  P × l3 = P2 × 8l3 P  3 P   P2  8 8 3 P  P2  8 Therefore, the final pressure is P/8 units. 23. V1= V V2 =V + 2V = 3V P2 = ? P1 = P Temperature is constant Boyle’s Law is applicable P1V1 = P 2V2 22. l1 = l l2 = 2l

 P2 

P1V1 P  V P   V2 3V 3

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O

P3 P P2

T3 T2 T1

P1

h1d1 0.7 ×13600 = = 9.52 m. d2 1000

Therefore, the height of the water column is 9.52 m. 19.

24.

P

V

Drop a perpendicular, ‘OP’ on V-axis, such that its volume remains constant. P3, P 2 and P 1 are the respective pressures at T 3 , T2 and T1 respectively. We know, P  T and given P3 > P2 > P1

 T3 > T2 > T1 25. P1 = 760 mm P2 = 740 mm d1 = 2.8 g/cc d2 = ? t1 = t2 = 26°C = constant At constant temperature, the pressure is directly proportional to the density of the gas P1 P2 P P  d 2 = 2 × d1 K  = d1 d 2 P1 d d2 =

740 × 2.8 = 2.726 g/cc. 760

Therefore, the density becomes 2.726 g/cc. 26. Let the initial volume of gas be x. Given standard temperature V1 = x  T = 273 K th V2 = 1/8 V1  V2 = x/8 T2 = ? As the pressure is constant, Charles’ law is applicable.



V1 V2  T1 T2

V2  T1  x  273  34.125 K V1 8 x [OR] = 34.125 –273 = –238.8750C Therefore, the temperature of enclosed gas is – 238.875°C

 T2 

Gaseous State Solutions 27.

309  Volume expelled = 2.5 V – V = 1.5 V

Let the volume of the gas at 0OC = V ml Thus, we have: V1 = V ml; V2 = 2V ml T1 = 0 + 273 = 273 K T2 = ?

1.5 V 3 = Fraction of air expelled = 2.5 V 5

V1 V2 By applying, Charles’ Law, T = T 1 2 Substituting the corresponding values,

we

V 2V have, 273 = T 2 2V ×273 =546K V T2 (in 0C) = 546 – 273 = 273OC. 28. Suppose volume of 200 mg of air at 17OC = V ml As pressure remains constant (being an open vessel), we can apply, T2 =

V1 V2 V V =  = 2 or V2 =1.34 V T1 T2 290 390  Volume of air expelled=1.34V–V=0.34 V Mass of 1.34 V air at 117OC = 200 mg Mass of 0.34 V air at 117OC

200 = × 0.34mg = 50.746 mg 1.34  Mass % of air expelled 200 × 0.34 1 × ×100 = 25.37% 1.34 200 29. V1 = V; V2 = 1.2V T1 = 300 K T2 = ? As the pressure is constant, =

V2 T2 we can apply, V = T 1 1 V2 1.2 V × 300 = 360  T2 = V × T1 = V 1  T2(in0C) = 360 – 273 = 870C 30. Suppose volume of vessel = V cm3 i.e., volume of air in the flask at 27OC = V cm3

V1 V2 V V 4 = i.e., = 2 or V2 = V T1 T2 300 400 3  Volume expelled = 1.33V – V = 0.33 V 31. Suppose volume of vessel = V cm3 i.e., volume of air in the flask at 27 O C = V cm3

V1 V2 V V = i.e., = 2 or V2 = 2.5 V T1 T2 300 750

32. The cylinders explode, if the pressure in the cylinder exceeds the with stading pressure. P1 = 12 atm ;P2 = 14.9 atm; T1 = 300 K ; T1 = ? Cylinder  volume = constant

P2 T2 14.9 × 300 = 372.5K  P = T (or) T2 = 12 1 1  T2 (in0C) = 372.5 – 273 = 99.5°C 34. P1 = 2 atm P2 = 2.5 atm T1 = 273K T2 = ? P2 T2 2.5 =  T2 = × 273 = 341.2 K P1 T1 2 = 68.25°C 35. P1 = 3 atm P2 = ? T1= 300 K T2= 320 K As the volume of tube is constant,Gay Lussaac’s law is applicable.



T P1 T1  P2 = 2 × P1  P2 T2 T1

P2 320 =  P2 = 3.2atm. 3 300 Therefore, the pressure at 47°C is 3.2 atm. 36. Cylinder  Constant Volume System Cylinder blows up if the bursting temperature is less than the melting point (1800 K) of cylinder. P1 = 250 × 105 P T1= 300 K P2 = 1× 106 P T 2= ? Volume is constant Gay-Lussaac’s law is applicable. 



P1 T1   P2 T2

T2 

P2  T1 P1

1×106  T2 = ×300 = 1200 K 0.25×106 As the temperature is less than MP, the cylinder will blow up before it melts. 37. Let the mass of N2 be added to the balloon to expand it. V1 = 6 litres V2 = 10 litres m1= 6 grams m2= ? T and P are constants m = m2 - m1 www.betoppers.com

9th Class Chemistry

310 Temperature and Pressure are constants.  Avogadro’s Law is applicable



V1 m V  1  m2  2 V2 m2 V1

m1

10 6  10 grams 6  Mass of nitrogen added = 10– 6 = 4 grams. 38. Mass (m ) = 11 grams  T = 273K and P = 1 atm Volume (V)= ? CO2  Molecular weight (M) = 44 The formula connecting above terms : PV = nRT  m2 

 PV = Here

m mRT RT  V  WP M

, ‘R’ = 0.0821 lit atm/mol/K

11  0.0821  273 22.4   5.6 litres. 44  1 4 Therefore, the volume occupied by 11 gm of CO2 at STP is 5.6 litres. 39. Let ‘V’ and P be the initial volume and pressure of the gas. Given volume is increased by 40% V

140 7 V  V 100 5 Pressure decreases to 80%  P2 is 80% of P  V2 

 P2 =

80 4P  P2 = 100 5

V1 = V P1= P T1 = –13+273 = 260 K V2 = 7V/5 P2 = 4P/5 T2 = ? Same mass of gas  No. of moles of gas is constant

P1V1 P2 V2 PV T =  T2 = 2 2 × 1 P1 V1 T1 T2 4 7 P × V × 260 5  T2 = 5 PV

 T2 =

4 × 7 × 260 = 291. 2 K 25

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40. Let ‘r’ and ‘T’ be the radius and temperature of the air bubble respectively at the bottom of the pond. r1 = r T 1= T P1= P r2 = 2r T2 = 2T P2 = ? The above terms can be connected as follows: 3 3 P1V1 P2 V2 P1 r1 P2 r2     Vsphere  r3  T1 T2 T1 T2

3

P1 r1 T2 P  r 3 2T P  3   3  P2  T1 r2 8 T 8r The pressure at the top of the pond is P/8. 41. We know that, PV = nRT  P2 

nR T KT T = V P P P Therefore, for V to be constant, both T and P should either be increased or decreased by the same factor. Therefore, volume remains same if we change the temperature and pressure by the same factor. 42. P1 = P V1= 300 ml T 1= T m1= m P2 = P/9 V2 = ? T2 = 3T1 = 3T m2 = m/2 Terms related: P, V, T and m. T  PV = nRT V=

 PV =

PV m =K RT  mT M

( M & R are constant for a given gas)



P1V1 P2 V2  m1T1 m2 T2  V2 =



P1 V1 m 2 T2 × m 1T1 P2

m  3T PV 2 1 27    3 9 V = × 300 P mT 2 2 9

= 4050 ml Therefore, the new volume is 4050 ml. 43. Let P, V, T and m be the initial pressure, volume, temperature and mass of the given gas respectively. m1 = m m2 = m/2 P1= P P2 = 2P T 1= T T2 = ?V1 = V2

Gaseous State Solutions

311

Let us derive the formula connecting the above terms:



PV =

m RT  M

P =K  mT



mT P T2 = 1 1 × 2 P1 m2

P1 P = 2 m1T1 m 2 T2

T2  4T

Therefore, the temperature becomes four times the original. 44. l1 = l l2 = ? T1= 273 K T2 = 546 K P1= 1 atm P2 = 0.5 atm n1 = n2 Let us find the formula connecting these terms. PV = nRT



PV = nR R



PV  K [ n & R are constant] T

P1V1 P2 V2 = T1 T2





3

V2 

2 



M1 r22 42 16 = = = M2 r12 12 1

V1 100 47. r1 = t = 20 = 5 ml / minute 1 V2 40 = ml / minute t2 t M1 = 2 M2 = 32 r2 =

mT 2P T2 = × P m2



M1 : M 2 = ?

P1V1 T2  T1 P2

1 3 546   273 0.5

r1 We know, r = 2



5 = 40/t

M2 r  1 = M1 r2

M2 32

32 = 16 = 4 2

5t 160 =4 t= = 32 minutes. 40 5 Therefore, the time taken by oxygen to diffuse is 32 minutes. 

48. The molecules with same molecular weight have same rate of diffusion. MCO2 = 44, MN2O = 44; MC3H8 = 44 Hence, the right answer is all three: a, b, c 49. Let NH4Cl be formed at a distance of x from HCl.

3

 2  4

  2  3 4  [ V2 = l2 3] 45. V1 = V lit T1= 546 K P1= ?

V2 = 4V T2 = 819 K P2 = P

n1 = n2 The formula connecting the above terms is:

P1V1 P2 V2 =  T1 T2 

P1 =

P2 V2 T1 × T2 V1

P × 4V 546 8 × = P 819 V 3

8  P1  P 3 46. r1 : r2 = 1 : 4

Then OA = x ; OB = 100 - x rHCl MNH3 = = rNH3 MHCl

17 = 0.68 : 1. 36.5

rHCl x x 0.68 = = = But r NH3 100  x 100  x 1 Hence, x + 0.68x = 68

x=

68 = 40.48 m 1.68

r1 M2 r2 M =  12 = 2 r2 M1 r2 M1 www.betoppers.com

9th Class Chemistry

312 50.

Oxygen

Hydrogen

V1 = 1 litre t1 = 1 hr = 60 min M1 = 32

V2= 1 litre t2= ? M2 = 2

The formula connecting these terms is :



V1t 2 M2 1× t 2 2 =  = V2 t1 M1 1× 60 32



t2 = 60

2 1 t   2 32 4 60

60 = 15 min 4 Therefore, the time taken for the diffusion of same volume of hydrogen is 15 min.

 t2 =

CONCEPTIVE WORKSHEET 1. 2. 3.

4.

5.

6. 7. 8. 9.

The term ‘gas’ was derived from the word Chaos. All the given statements are correct. Hence the right option is (d). Among all the states of matter, gases are highly compressible. The high compressibility of the gases is due to the fact that they have large intermolecular spaces. On applying pressure, these molecules simply come close to each other, thereby decreasing the volume of a gas. The intermolecular force of attraction is inversely proportional to the distance between the molecules of states of matter. Thus, the right of intermolecular force of attraction is : Solids > Liquids > Gases.Hence the right option is (a). The intermolecular spaces in a gas are very large. Thus, when two gases are brought in contact with each other, their molecules just move into one another’s intermolecular space, thereby forming a homogeneous mixture Thus, all the given options are correct. All the units given are used to measure volume in different units of systems. We know that T = °C + 273  T = – 273 + 273 = 0K We know that T = °C + 273  T = 273 + 273 = 546 K Torr, Pascal and Bar are the units of pressure. Newton is the unit of force.

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10. a) We know that

Force Area Atmospheric pressure is the force exerted by an air column of 1m2 cross-sectional area on the ground. The force exerted by the air column is nothing but its weight.  Force exerted by the air column = weight of air column = mass of air column × acceleration due to gravity = mg The mass of air column of 1m2 cross sectional is approximately 104 kg and acceleration due to gravity is 9.8 m/s2  10m/s2.  Atmospheric pressure Pressure =

Force m  g 104  10   N / m 2 = 105 N/m2 Area 1 1 = 105 pascal So, atmospheric pressure is 105 N/m2. This means that on an area of 1m2 of the surface of earth, a force of 105 Newton is acting. This is equivalent to the weight of 140 adults of mass, 70kg each. This is also approximately the weight of a body of mass, ten tonnes (10,000 kg). This is also equal to the weight of two elephants, approximately. 

b) Considering the head as a sphere with an average diameter of 20cm the resulting surface area is about 1256cm2. We know that for every one cm2 of area a force of approximately 10 Newton acts due to atmospheric pressure. Therefore the total force of about 1256 10N acts on our head which is equal to the weight of 18 adults weighing 70kg each. 11. The molecules in air beat against a window pane at the rate of two million molecules per square inch per second. Such enormous force of pressure of a gas is enough to power a steam engine or turbojet. This impact would shatter the glass if an equal number of molecular blows were not rained against it from the other side of the pane as well. 12. There is no external pressure in space. Astronauts, while travelling through space, wear heavy space suits so as to create an external pressure. These space suits prevent the bursting of body cells and bleeding of nose and ears.

Gaseous State Solutions 13. At the top of a tall mountain, the pressure is less. This causes our ears to pop in order to balance the pressure between the outer part and the inner part of our ears. 14. Manometer is used to measure the pressure of a gas and barometer is used to measure pressure of atmosphere. 15. To make a barometer compact, the height of the tube which holds the liquid column, should be less. For this, we need a high density liquid. The best suited liquid for this is mercury (Hg) with a high density of 13.6 g/cc. Further, due to its high surface tension, it is non sticky to glass. And is easily viewed through glass because of its shine. Besides, mercury has a high boiling point. 16. At a higher altitude the pressure becomes very much less and the passengers encounter dangers like bursting of body cells, bleeding of noses and ears and also an increase in the blood pressure. To avoid these problems an artificial pressure is created inside the plane. 17. At the top of the mountains and high altitudes, there are a fewer number of molecules of air. Since we could inhale only fewer air molecules each time, we feel suffocated and tend to breathe faster to make up for the deficit. 18. Any gas can be identified from its molecular weight. Let us find the molecular weight from the given data.

m m 66 (or) GMW= = = 64 g GMW n ¼  SO2 19. Let us find the number of moles of each from the given data. n=

We know that no. of moles, n = For A, =

m ; GMW

4 48 = 1 ; For B, = = 1; 4 48

16 =1 16 From the above, it is clear that all the gases possess same number of moles. 20. An air bubble present inside the water is being acted upon by (a) atmospheric pressure and (b) pressure created by the weight of the water column above their air bubble. As the air bubble moves up, the water column above the air bubble gets reduced. The weight of the water column acting on the air bubble reduces, For C, =

313 resulting in the reduction of pressure. As the pressure reduces, the volume of the air bubble increases. Therefore, the pressure and volumes are varying inversely, confirming Boyle’s Law. 21.

4  P1V1 = P2 V2 , 700 × 250 = P2 ×  × 250  , 5 

P2 = 875mm Additional pressure = 875 – 700 =175 mm of Hg. 22. R1 = 2r

4 V1  R13 3 4 4 3 3  V1    2r    8r = P1 = P 3 3 R2 = r 4 4 V2  R 32  V2  r 3 3 3 P2 = ? P1V1 = P 2V2 4 4 P  8r 3  P2  r 3  P2  8P 3 3 23. P1 = 760 mm P2 = 740 mm d1 = 2.8 g/cc d2 = ? t1 = t2 = 26°C = constant P K d



P1 P2 P   d 2  2  d1 d1 d 2 P1

740  2.8  2.726 g / cc. 760 24. P1 = P atm, P2 = 5P atm d1 = d gm/lit., = d2 = ? d2 



P1 P2 P   d 2  2  d1 -----(a) d1 d 2 P1

5P  d  5d gm/litre P 25. Given, V1 = 500 ml T1 = 273 + 27 = 300K Final conditions V2 = ? T2 = 273 + (–5) = 268K By applying Charle’s Law, d2 

V1 V2 500 V2   T1 T2 ; 300 268 V2 

500  268  446.66 ml 300 www.betoppers.com

9th Class Chemistry

314 26. x1 = x

x2 = x +

V1 = ax1 = ax T1 = 300 K T2 = T1 = ?

 V  T  or 

x 5

x 6   V2  a  x    ax 5 5  T2 = ? V2 T2  V1 T1

6 ax V2   T1  5  300 V1 ax

6  T2   300  360K 5 T2 = T1 = 360 – 300 = 60K. 27. V1 = 1 litre V2 = ? t1 = 27°C t2 = 127°C  T1  273  27  300K

293  V  V2  1.035 V2  2V 273 30. t1 = 15°C t2 = – 15°C V2 

 T1  15  273  288K

 T2  273   15   258K

d1 = d d2 = ? d2 – d1 = ?  dT  K  d1T1  d 2 T2

 d2  d2 

d1T1 T2

d1T1 d  288   1.12d T2 258

= d2 – d1 = 1.12d – d = 0.12d = 31. d1 = 1.45kg/m3 d2 = 1.25 kg/m3 t1 = 27°C t2 = ?  T1  27  273  300K

 T2  273  127  400K V2 – V1 = ?

d1T1 = d2T2  T2 

V1 V2 T2 400 4  T  T  V2  T  V1  300  1  3 lts. 1 2 1

 T2 

4 1  1  lts. 3 3 28. t1 = 27°C, t2 = 127°C V1 = 400ml V2 = ? V2  V1 

V  T  V   t  273



V2 t 2  273  V1 t1  273

V 127  273 400  2   V2   400 400 27  273 300  V2  533.3ml. 29. t1 = 10°C t2 = 20°C V1 = V V2 = ?

V  T  V   t  273



V2 t 2  273  V1 t1  273

V 20  273 V2 293  2    V 10  273 V 283

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12 3 d d 100 25

d1T1 d2

d1T1 300  1.45   348K d2 1.25

 T2  348K  t 2  348  273  75C 32. Isochor 33. P T

P1 T1 t   34. a) Pt = PO 1+ b) P = T   273  2 2 35. a) A cylinder b) A cylinder with a movable piston c) Balloon 36. At high temperature and low pressure, a real gas can be converted to an ideal gas. 37. All the given equations are correct. 38. R is Universal Gas Constant in ideal gas equation. 39. System SI CGS In calories

Value 8.314 Joule/ mole/ Kelvin 8.314  107 erg/ mole/ Kelvin 2 calories/ mole/ Kelvin

Gaseous State Solutions

315

40. 1 mole of any gas at S.T.P. occupies 22.4 litres ( Gram Molar Volume)

2

 102  ( MCO = 44 g mol–1); Mx = 44    = 65.97  83.3  2

41. 42.

PM d= RT 50.

PV P/ 2 × V2 =  or  V2 = 4 V T 2T

51.

M 44 = = = 1.96g/L 22.4 22.4

43.

dSTP

44.

P1V1 P2 V2 1 100 1.5× V2  i.e.  T1 T2 273  273 + 91 or V2 = 88.9 cc.

45.

PV = nRT = 

46.

52.

PV 8.2  1010  30 PV = nRT,n =  RT 0.082  300

rO2

=

M O2 MH2

rH2 ; r = O2



5 = 40 / t

r1 r2 =

M2 32

5t 160 =4  t= = 32 minutes. 40 5 49. rx = 83.3 mLs–1

rCO2

M CO2 Mx

;



t 16

4 2 1    VH  16ml 2 VH 2 32 4

V1  2 M2 20  30 32    V2  1 M1 V2  60 64 10 1  (or) V2 = 100 × 2 = 200l. 2 2

1. 2.

PV= Constant  P1V1  P2 V2 (1080 mm of Hg). The final volume will be 60% of initial volume. As the temperature is constant, Boyle’s Law is applicable i.e., P1V1 = P2V2

3.

Volume of a sphere is

–1

M CO 2 83.3 = , 102 Mx

4 3 r and as temperature is 3

constant, Boyle’s Law is applicable  P1V1  P2 V2 4.

nP1V1  P2 V2  n 

5.

V1 V2  T1 T2

 By Graham’s Law of Effusion

=

50 / 20 32  M H2  40 / t 3

SUMM ATIVE WORKSHEET



rx

1 1 1 : : 64 32 16

MO 2

M H2 V1 M2 Vo 2    V2 M1 V 2 Mo2



32 = 16  4 2

rco2 = 102 mLs

rO2



54.

55.



M

; rCO2 = rO2 = rCH2 =

rx Mo 2 r 128   x  2 ro 2 Mx ro2 32

32 = 16 = 4 ; rH 2 : rO2 2

V2 40 r2 = t = ml / minute t 2 M1 = 2 ; M2 = 32 M2 M1

1

M SO2

rSO2 32 1  r  64 = 2 O2

53.

V1 100 t1 = 20 = 5ml / minute;

r1 We know, r = 2

rH2



=4:1 48. r1 =

r

M O2

= F; t = 64 min

w wRT RT,M = M PV

9  0.0821 300  27. 1  8.2

rH2

rO2



= 1: 2 : 2

= 109 mol = 6.02 ×1014 molecules 47.

rSO2

P2 V2 P1V1

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9th Class Chemistry

316

6.

V1 V2  T1 T2

7.

P1V1  P2 V2

HOTS WORKSHEET 1.

 V2  Percentage change in volume   V  1  100  1  9. No. of molecules = No. of moles  6.023  1023 and ‘n’ can be obtained by applying PV = nRT 10. PV= nRT 8.

 PV 

Pressure of air = Pressure of Hg h1 d1 g = h2d2g h1 × 1.2 × g = 0.7 × 13600 × g

 h1 = 2.

m mRT RT  M  M PV

11. 10 K

0.7  13600  g = 7933m = 7.933km. 1.2  g

V1 = 2.5L P1 = 740mm P2 = 780mm V2 = ? PV

1 1 P1V1 = P2 V2  V2 = P  2

12. P = hdg 13. P = hdg = ( l2  l1 )dg 14.

P1V1  P2 V2

 Q2   1  100 = 15. % Change    Q1  16.

3.

 m2   1  100   m1 

17. No.of molecules = no. of moles  6.023  10

r1 d2 M 2 u1 s1t 2     r2 d1 M1 u 2 s 2 t1

19.

r1 V1t 2 M2 d    2 r2 V2 t1 M1 d1

V2 =

22.

4 3 4 10  r = P2 × 9   r 3  P2 = atm. 3 3 9 Isothermal condition means that the temperature is constant P1 = 1atm P2 = ? V1 = 105L V2 = 35L P1V1 = P 2V2

 10 ×

4.

PV

105  1

1 1  P2 = V  35 2

5.

PM [ Ans: P4] RT Mass of helium in gaseous mixture  100 Total mass of gaseous mixture

4 4  (r3 + 8r3) = 9   r3 3 3

P2 = ? P 1V 1= P 2V 2

23

r1 V1t 2 M2 d    2 r2 V2 t1 M1 d1

Mass % 

4 4  r3 +  (2r)3 3 3

 V2 =

20. Any gas can be identified from its molecular weight.

21. d gas 

4 3 r 3

P1 = 10 atm

P1V1 P2 V2  T1 T2

18.

r1 = r  V1 =

2.5  740 = 2.371 L 780

.

Ans: 19.95

 V2  23. %change    1  100 .  V1  24. 1 mole of any gas at S.T.P occupies 22.4 litres. 25. The weight of 1 mole of any gas at S.T.P is equal to its gram molecular weight.

= 3atm.

Isobarical condition means that the pressure is constant and therefore,

V1 T1  V2 T2 is applicable. V1 = 52 L T1 = 25° C = 25 + 273 = 298K V2 = 104 L T2 = ?

V1 T1 54 298 Applying V  T , we get, 104  T 2 2 2

 T2 = 574K Temperature change = 574 – 273 = 301K

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Gaseous State Solutions 6.

317

P1 = 5 atm T1 = 250K P2 = ? T2 = 300K

10. d = 1.56 g/L T = 65 + 273 = 338 K

745 atm 760 P = 0.0821 litre - atom - mole-1.Kelvin-1 P=

P1 T1  P2 T2 ( as volume is constant )

We know : PV = nRT  P ×

300 T2  5 = 6 atm. × P1 = 250 T1 m1 = 7g m2 = 21g P2 =

7.

 V2  % change in volume=  V  1 × 100  1   m2   V m  =  m  1 × 100  since 2  2   1  V1 m1  

8.

 21  =   1 × 100 = 200 %,  7  V = 18L T = 27+273 = 300K P = 70cm = 70/76 atm R = 0.0821 litre - atm mole–1 K–1 m = 1.35g PV = nRT

 PV =

m  RT MW

mRT 1.35  0.0821 300  = 2.015 PV 70 / 76  18 V = 380ml = 0.38L T = 27 + 273 = 300K R = 0.0821 lit-atm mole–1.Kelvin–1

 MW=

9.

P = 800 mm =

800 atm. 760

m = 0.455 g molecular weight, M = ? PV = nRT  PV =

m RT M

mRT 0.455  0.0821  300  = 28 800 PV  0.38 760  C2H4 or N2

 M=

 d=

m m = . RT T d M

PM dRT or M = RT P

1.56  0.0821  338 = 44.2 745 / 760 11. T1 = 27 + 273 = 300k P1 = 1 atm d1 = d T2 = ? d2 = 0.75 d P2 = 1 atm

 M=

PM 1 = K (  P, M and R  RT dT

We know : d =

are constants). d1T1 = d2T2 d × 300 = 0.75d × T2 T2 =

300d = 400 K. 0.75d

V1 100 12. r1 = t = = 5ml / minute 20 1 V2 40 r2 = t = ml / minute t 2 M1 = 2 M2 = 32 r1 We know, r = 2



5 = 40 / t

r M2  1 = M1 r2

M2 32

32 = 16  4 2

5t 160 =4  t= = 32 minutes. 40 5 13. Let A and B be the ends through which HCl and NH4Cl enters the tube.



A

O HCl x

B NH3 100 – x

NH4Cl

Then OA = x ; OB = 100 – x www.betoppers.com

9th Class Chemistry

318 rHCl rNH3 =

M NH3 M HCl

=

17 = 0.68 : 1 36.5

rHCl x x 0.68 But = =  rNH3 100  x 100  x 1

Hence, x + 0.68x = 68

68 = 40.48m 1.68 14. ‘9’ carbon atoms weigh = 9×12 = 108a.m.u. ‘13’ hydrogen atoms weigh = 13×1 = 13a.m.u. The weight of 2.33×10–23g of other component = x=

2.33  1023  14a.m.u. 1.66  1024  The mass of one molecule = 108+13+14 = 135

 V2  15. % change in volume =  V  1 ×100 = 10  1   P2  % change in pressure =  P  1 ×100 = ?  1  We know : P1V1 = P2V2 

P2 V1 = P1 V2

 V1  % change in pressure =  V  1 ×100  2 

V2 V2 1 1 11  V = +1 = Given ; V –1 = 10 10 10 1 1

 M=

dRT P

1.5  0.0821  300 = 40.06 700 / 760  Molecular weight of CO and CO2 mixture is 40.06 Let ‘x’ be the % of moles of CO in the mixture. Average mol.wt =

 28x   100  x  44

100  4006 = (28x) + (100-x) 44  4006 = –16x + 4400 or 16x = 394 or x = 24.75  Mole % of CO = 24.75 Mole % of CO2 = 100–24.75 = 75.25 17. Mass of empty vessel = 50.0g Mass of vessel + liquid = 148.0g  Mass of liquid = 98.0g  Density of liquid = 0.98g/mL

m 98 = = 100mL d 0.98  Volume of vessel = 100mL = 0.1L Mass of empty vessel = 50.0g Mass of vessel + mass of gas = 50.5g  Mass of gas = 0.5g Now, we have : m = 0.5g V = 0.1L R = 0.0821 lit-atm/mole/kelvin T = 300 K P = 760mm = 1 atm Mol. wt = M = ? We know, PV = nRT

 Volume of liquid =

m RT T M

 M=

mRT PV

0.5  0.0821 300 = 123. 1 0.1 18. Pressure of H2 after reaction is over, has been given (0.35bar) indicating O2 has completely reacted.

 M=

 10



 change in pressure =  11  1 ×100   1  10  11  =   100    100 11  11  Pressure decreases by 9.09%. 16. d = 1.5g/L T = 300K 700 atm 760 R = 0.0821lit-atm/mole/kelvin M=?

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PM RT

=

 PV =

V1 10  V = 11 2

P=

We know d =

 PV  Thus, number of moles of H2   and that O2  RT  can be determined. Let the no.of moles of H2 remaining be nH . 2

 nH2 = PV = 0.35V RT RT Let the total no.of moles be nT  PV  V nT =   =  RT  total RT

Gaseous State Solutions

319

 Total no.of moles of H2 and O2 reacted



= n H 2  n O2



reacted

=

V 0.35V 0.65V  = RT RT RT

We know : 2H2+O2  2H2O. From the reaction it is clear that, H2 and O2 react in 2 : 1 molar ratio.

 

 n H2

reacted

 

= 2 n O2



We have, n H 2  n O2

 2n O2 + n O2 =  3 n O2 =

0.65V RT

 n O2 =

0.65V 3RT



reacted

reacted

0.65V = RT

0.65V RT

0.65V 1.3V ×2 = 3RT 3RT Let us find the total no.of moles of H2 pressure in the mixture.

 n H2 = 2 n O2 =

n 

H2 total

 

= n H2

unreacted

 

+ n H2

reacted

0.35V 1.3V 0.78V + = RT RT RT  Fraction of H2 present in the mixture =

No.of moles of H 2 = Total no. of moles of mixture =

0.78V RT V RT

0.78V RT × = 0.78 RT V  of H2 in the mixture = 0.78×100 = 78% The fraction of O2 present in the mixture = 1– 0.78 = 22 % The % of O2 present in the mixture = 0.22×100 = 22%.  composition of original mixture = 78% of H2 and 22% of O2. 19. As pressure is hdg, by the comparison we can determine fall in Hg level per hour. From the given condition, volume of oxygen consumed per hour can be determined. h1d1g = h2d2g  h1d1 = h2d2 =

 h1 =

h 2 (drop in with given fluid) × h 2 (density of fluid) d1 (density of Hg)

37  1.034 = 2.81mm Hg drop in 30 min. 13.6 = 5.62mm Hg drop/hr. P1 = 5.62mm P2 = 760mm V1 = 16.0cm3 V2 = ? T1 = 310.5 K T2 = 273.15 K =

P1V1 P2 V2 = T1 T2 P1V1 T2 5.62  16 273.15  V2 = T × P = × 310.15 760 1 2 3 2 3 –1 = 0.1042 cm = 1.042×10 mm hour 20. m = 100g M = 40 P = 2 atm T = 27+273 = 300 K i) Volume of the cylinder before dent may be calculated as : PV = nRT

 V=

nRT P

100 0.0821 300 × = 30.79L. 4 2 Mass of gas after dent = 100 – 10 = 90g

 V=

nRT 90 0.0821 300 = × = 27.71L P 40 2 ii) Since the temperature remains constant, Boyle’s law may be applied. P1V1 = P2V2  2×30.79 = P2×27.71 So, V =

2  30.79 = 2.22 atm. 27.71 21. We have been given total moles of O2 present in both flasks and since volume and pressure are same, from the ratio of temperatures, number of moles of O2 in each flask can be calculated and hence, the volume of each flask is determined. We calculate final pressure from the unknown volume of flask. For flask1, P1V1 = n1RT1 For flask 2, P2V2 = n2RT2 Since both flasks are of equal volume and have been joined by a narrow tube, pressure is also same in both flasks. P1V1 = P2V2 T1 = n2RT2  n1RT

 P2 =

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9th Class Chemistry

320

n1 T2 600 2 4 = = ;  n = T  300 1 2 2 1 n1 + n2 = 0.6 mol  n1 = 0.4 mol at 300 K  n2 = 0.2 mol at 600 K Volume of the flask containing 0.3 mol O2 in each, V=

nRT P

 V=

0.3  0.0821  300 L 0.5

Hence, final pressure Pf =

(mol.wt) 2

2

(mol.wt)B = 4(mol.wt)A 24. V1 = 1000 ml T1 = – 73 + 273 = 200 K P1 = 2atm P2 = 0.5 atm T2 = 127 + 273 = 400 K V2 = ?

P1V1 P2 V2 P1V1 T2 = V = ×  2 T1 T2 T1 V2

n 2 RT2  n1RT1   or  V  V 

=

0.2  0.0821 600  0.5 = 0.66 atm 0.3  0.0821 300 22. As we go up to higher altitudes, there is decrease in temperatures; simultaneously there is decrease in pressure when volume increases. Thus, we use gas equation. =

2  1000 400 × = 8000ml. 200 0.5

IIT JEE WORKSHEET KEY

P1V1 P2 V2 = T1 T2

1

2

3

4

5

6

7

8

9

10

1

2

1

4

*

1

*

4

*

*

11 12 13 14 15 16 17 18 19 20

P1 = 1 atm P2 = ? V1 =

P1

 (mol.wt) = 2 × P = 2 × = 4 1 1 2  (mol.wt)2 = 4(mol.wt)1 = or

*

*

*

1

1

*

4

3

1

21 22 23 24 25 26 27 28 29 30

4 4  (1)3 =  m3 3 3

2

4 4  (3)3 =9   m3 3 3 T1 = 20+273 = 293 K T2 = – 20+273 = 253 K V2 =

3

2

2

4

3

3

1

1

nRT 0.5  0.0821 546   22.4litres P 1 V1 = 0.2 litres V2 = ? P1 = 1atm P2 = 1atm T1 = 273k T2 = 546k Pressure is constant  Charles’ Law V

4 1  3  253 293 9  4  3

2.

= 0.032 atm. 23. We know, PV = nRT

P1 n1  P  n (Since V, R, T are constants) 2 2

P

m /(mol.wt)

1 1 1  P  m /(mol.wt) 2 2 2

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2

HINTS/SOLUTIONS TO THE SELECTED QUESTIONS 1. n = 0.5, P = 1atm, T = 273 + 273 = 546 k v=? Pv = nRT

P1V1 P2 V2 = T1 T2 P1V1 T2  P2 = T × V  P2 = 1 2

3

V2 T2 546   V2   0.2  0.4 litres V1 T1 273 3.

For a given mass of gas:

P1V1 PV  2 2 T1 T2

Gaseous State Solutions

PV T PV 2T  V2  1 1  2  V2   V T1 P2 T 2P

4. 5.

Hence, the new volume is same as that of the original. Same rate of diffusion  same mol.wt  NO, C2H6 as both have same molecular weight. Let the mol.weight of unknown gas

rH 2  6 (given) rx  6.

7.

8.

rH 2 x   6   x  36  2  72 rx x 2

The value of Ra is: 1.987 cal/mol/kelvin (or) 8.314 Joule/mol/kelvin (or) 0.0821 lit - atm/mol/kelvin 2gm of H2 (or) 32 gm of O2 is 1 mole of each gas. We know that 1 mole of gas at NTP occupies 22.4litres. rA = 2rB

PV = nRT

nRT 1 0.0821 546   44.8litres P 1 14. According to Charles’ Law, at constant pressure, for a given mass of gas, volume is directly proportional to its absolute temperature. 15. Volume of H2 = 2 litres Volume of Cl2 = 11.2 litres V

 2 HCl H2 + Cl2  1 vol + 1vol   2 vol  2 lit . 2 lit   2 lit 11.2 litres of Cl2 available, only 2 litres reacts and the rest 9.2 litres remain unreacted, and 2 litres of HCl is formed. Therefore, the final composition of mixture is: 9.2 litres of Cl2 and 2 litres of O2. 16. V = 22.4 litres

n1 

2 1 2

 A  B  ?

n2 

8 1  32 4

r1 2 r    or  A  B r2 1 rB A

22 1  44 2 nT = n1 + n2 + n3 = 1.75 T = 273k , P = ?

2 9.

321

B A

B A 1 (or) 4   (or)   4  0.25 A B

We know that

R  0.0821

lit  atm mol  kelvin

 0.0821  1000  82.1

cc  atm mole  kelvin

cc  atm mole  kelvin

0.5 1  moles of H2 2 4 At STP, 1 mole of H2 occupies 22.4 litres

10. 0.5g of H 2 

1 22.4 litres (or) moles of H 2 occupies 4 4 5.6 litres 11. As n and R are constants, for the volume to same, both T and P should be changed by same factor. 12. The value of R = 8.314 joule/mole/kelvin 13. P = 1atm, T = 273 + 273 = 546 k n = 1mole, V = litre 

n3 

nRT 1.75  0.0821 273   1.75atm V 22.4 17. Constant pressure  Charles’ law P



V2 T2  V1 T1 (or) T2 = 2 × 273 = 546 K (or) 273°C

18. According to Charles’ Law of constant pressures, for a given mass of gas, volume is directly proportional to its absolute temperature. 19. V1 = 100 ccm, P1 = 104 Pa, P2 = 105 Pa, V2 = ? At constant temperature : P1V1 = P2V2  V2 

P1V1 104  100   10cc P2 105

20. P1 = atm V1 = V V2 = V/4 P2 = ? At constant temperature : P1V1 = P2V2

 P2 

1 V  4atm V/4 www.betoppers.com

9th Class Chemistry

322 21. Density of a gas =

PM P (or) d  RT T

P  For d1  T  should be minimum.   22. Given mass of gas and volume are constant As temperature varies  Pressure increases  Rate of collision decreases  Energy increases 23.

m O2 m N2 N O2 N N2



?

N N2 N O2 N N2



n O2



m O2

m n n N2 ; But M

m N2



M N2 M O2

1 28 7    4 32 32

24. According to Avogadro, equal values of gases contain same number of molecules. 25. As the masses are same, the gas with less mol.wt contains more no. of moles or molecules.  Nitrogen contains more no. of molecules.  Greater pressure is exerted by nitrogen. 26. Wt. of 1cc of HC= wt of 4cc of O2 at NTP. Let us final the wt. of O2. 22400cc of O2 at NTP weighs 32g  4cc of O2 at NTP weighs

4  32  5.7 103 g 22400 wt. of 1cc og HC = 5.7 × 10–3g wt. of 2240cc (2.24 litres) = x 

2240  5.7  103  5.7  2.24  12.76g 1 27. 22400 mol of O2 at NTP weighs 32g 112 mol of O2 at NTP weighs = xg x

28.

rH4 rx

VA MB 50 MB   (or) (or) VB MA 40 64

5  5  64  MB = 100 4 4 30. Lighter gas differs fast and hence ammonium chloride ring is formed near the heavier gas. HCl is heavier than NH3. MB 

N O2

x

r1 V1  2 M2   r2 V2  1 M1 

1  4

Number of molecules × No. of moles. 

29. VA = 50ml VB = 40ml tA = t B MA = 64 MB = ? We know that

112  32  16 102 g  0.16g 22400  2 (given)

2

Mx  M x  16  4  64 16

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

7. MOLE CONCEPT SOLUTIONS

FORM ATIVE WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS

1. 2.

(A) (C) Sodium Carbonate + Ethaoic acid   Sodium ethanoate +Carbon dioxide + Water  

  8.  Reactants

Products

Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid. 5.3 g + 6g = 11.3g Mass of products = Mass of sodium ethanoate + Mass of carbondioxide + Mass of water 8.2 g + 2.2 g + 0.9g = 11.3g 3.

=

(B) The balanced chemical equation is : 5. 

2KClO3  2KCl

(2×122.5)g ‘x’g

x  4.



3O2

(2×74.5)g (3×32)g 1.92 g

2.96 g

2.96  2  122.5  4.88 g 2  74.5 X

6  56.3 = 7.730 atoms 43.7

Also 56.3 parts of X in I correspond to = 4X atoms. 43.7 parts of X in II will correspond to =  4  43.7 = 3.1 atoms 56.3

Now, the atomic ratio X : O in the second oxide

0.277  100 = 20.14. 1.375

Second experiment : Copper taken = 1.179 g Copper oxide formed = 1.476 g Oxygen present = 1.476 – 1.179 g = 0.297 g Hence % of oxygen of CuO

I oxide : 43.7 parts 56.3 parts II oxide : 56.3 parts 43.7 parts Now 43.7 parts of oxygen corresponds to = 6 oxygen atoms  56.3 parts of oxygen in II correspond to =

Hence, the formula of 2nd oxide is X2O5. (A) First experiment: Copper oxide = 1.375 g Copper left = 1.098 g Oxygen present = 1.375 – 1.098 = 0.277 g Hence % of oxygen in CuO =

(C) Oxygen

3.1 7.73 : or 1 : 2.5 or 2:5 3.1 3.1

=

6.

0.297  100 = 20.12. 1.476

Percentage of oxygen is the same in both the above cases, so the law of constant composition is proved. (B) In the first compound Hydrogen = 5.93% Oxygen = (100 – 5.93)% = 94.07%. In the second compound Hydrogen = 11.2% Oxygen = (100 - 11.2)% = 88.8%. In the first compound the number of parts by

9th Class Chemistry

324

mass of oxygen that combine with one part by 94.07  15.86 parts. mass of hydrogen  5.93

10.

In the second compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen 

7.

88.8  7.9 parts. 11.2

The ratio of masses of oxygen that combine with fixed mass (1 part) by mass of hydrogen is 15.86 : 7.9 or 2 : 1 (C) i) Mass of copperoxide = 3.672 g Mass of copper = 2.938 g Mass of oxygen = Mass of copperoxide – mass of coppper = 3.672 - 2.938 = 0.734 g Ratio of mass of copper : Oxygen = 2.938 : 0.734 

8.

(C) Mass of copper oxide = 1.288 g Mass of copper = 1.03 g Mass of copperoxide – Mass of copper = Mass of oxygen. 1.288 g – 1.03 g = 0.258 g Mass of copper : Mass of oxygen = 1.03 : 0.258



xA1  yA 2 98.892  12  1.108  13.0034  = xy 98.892  1.108

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x  62.929   100  x  64.928 100  63.54 = 62.929x + 6492.8 – 64.928x  x = 69.40  Percentage abundance of Cu 63 = x1 = 69.40% and Percentage abundance of Cu65 = x2 = 100 – 69.40 = 30.60% (A, B) Atomic weight 

12

Weight of one atom of that element . 1 weight of th of C 12atom 12

Total weight

(C) Number of atoms  weight of singleatom Weight of single atom = (Atomic weight) amu = 39 amu = 39 × 1.66 × 10–24g. Total weight= Gram atomic weight = 39g

1.03 :1 =4:1 0.258

Given, Relative abundance of C12 isotope = x = 98.892% Atomic weight of C12 isotope = A1 = 12 a.m.u Relative abundance of C13 = y =1 .108% Atomic weight of C13 = A2 = 13.0034 a.m.u Average atomic weight of C = ? We know, Average atomic weight

xA1  yA2 x y

63.54 

11.

= 9.

We know, Average atomic weight 

2.938 :1 0.734

=4:1

12 a.m.u Given, Atomic weight of Cu63 = A1 = 62.929 Atomic weight of Cu65 = A2 = 64.928 Atomic weight of Cu = 63.54 Percentage abundance of Cu63 = x1 = x = ? Percentage abundance of Cu65 = x2 = 100 – x = ?

39g  Number of atoms  39  1.66  1024 g 

13.

1024  6.023  1023 1.66

(C) Smallest unit of mass = amu Mass of one electron = 0.00054 amu 1amu

No. of electrons = 0.00054amu = 1852.3 14. 15.

(A) 1gram atom of any element contains 6.023×1023 atoms. (D) We know, Number of gram atoms (na) =

Mole Concept Solutions

325

Given weight  m  Gramatomic weight  GAW  –––––– (1)



– 12 atom = x Molecular weight =

(i) Applying (1), we get, number of gram atoms (na) =

4 4 1

Weight of one molecule 1 Weight of th of C-12 atom  12

(ii) Applying (1), we get, number of gram atoms(na) =

16.

1  0.25 4

Weight of one molecule Its molecular weight

Therefore, the number of gram atoms present in 1g of helium is equal to 0.25. (A) We know, number of gram atoms (na) = n m  H n He GAW

mH  GAW  H m H  GAW  He    m He m He  GAW H  GAW He

= Weight of

17.

19.

20.

(C) The heaviest naturally occurring element is uranium i.e., 92 U 238 The lightest atom is hydrogen i.e., 1 H1. Given, weight of hydrogen = weight of 1 gram atom of uranium. Let’s find the weight of 1 gram atom of uranium. Number 

of

gram

atoms

n  a

Given weight  m  Gram atomic weight  GAW  –––(1)

1

m 238

  GAW of Uraniumis 238g 

(B) Weight of one molecule = (molecular weight) amu Molecular weight of CO2 = 12 + (2 × 16) = 44.  weight of one molecule of CO2 = 44 amu (C) To find out the molecular weight, we need to find the weight of one molecule in terms of amu. Weight of one molecule in terms of amu =

21.

 na   18.

238  238 1



xg 1024   6.023  1023 x 1.66  1024 g 1.66

 Number of atoms of B



yg 1024   6.023  1023 y 1.66  1024 g 1.66

 Number of atoms of A = Number of atom of

B.

(D) Atomic weight 

weight in grams 1.66  1022 = = 100 amu. 1.66×1024 1.66×1024

 molecular weight of the compound = 100. (C) Let us consider two elements, A and B, whose atomic weights are ‘x’ and ‘y’ respectively.  Gram atomic weight of A = ‘x’ g Gram atomic weight of B = ‘y’ g weight of one atom of A = ‘x’ amu = x × 1.66 × 10–24g ; weight of one atom of B = ‘y’ amu = y × 1.66 × 10–24g  Number of atoms of A

 m  238g

Therefore, we need to find the number of gram atoms of hydrogen in 238 g of it. Applying 1, we get, Number of gram atoms

1 th of C - 12 atom = y 12

 x = y..

Substituting the values in the above equation, we nH 1 4 2 get n  2  1  1 He

1 weight of one atom = weight of th of C 12 Its atomic weight

weight of one atom 1 weight of th of C  12atom 12

 Similarly, the number of atom of any element in its gram atomic weight = 6.023 × 1023. Now, consider any two compounds, say ‘C’ and ‘D’, whose molecular weight are ‘p’ and ‘q’ respectively.

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9th Class Chemistry

326

1 mole of atoms = Gram atomic massSo, 1 mole of calcium atoms = Gram atomic mass of calcium= 40g Now, 40 g of calcium = 1 mole of calcium

 Gram molecular weight of C = ‘p’ g ; Gram

molecular weight of D = ‘q’ g weight of one molecule of C = ‘p’ amu = p × 1.66 × 10–24g weight of one molecule of D = ‘q’ amu = q × 1.66 × 10–24g Number of molecules of C 

So,

22.

= 0.125 mole Thus, there are 0.125 mole in 5 grams of calcium. The above problem can also be solved directly by using the formula : Number of moles of atom

'q 'g 1024   6.023  1023 24 'q ' 1.66  10 g 1.66

(A) We know, Number of gram molecules

n 



m

Given weight  m  Gram molecular weight  GMW  ––––

(1) Applying (1) to the cases, we get, i) Number of gram molecules of Neon (Ne) = 5 1   0.25 GMW of Ne  20  20 4 7 1   0.25 GMW of N 2  28  28 4 (B)1 gram mole of methane weigh 16g. 

(N2) =

Weight of methane = 1 × 16 = 16g  Weight of hydrogen in 16g of methane = 4 g (as 4 hydrogen atoms are present) We know, Number of gram molecules of Weight of thesubs tan ce  m 

hydrogen = Gram molecular weight GMW   

24.

25.

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Mass of element in grams Gram atomic mass of element



5 1  = 0.125 mole 40 8



26.

4 2 2

(D) Weight of compound gas = Number of moles × Gram molecular weight.  Weight of compound gas  Gram molecular weight ( Number of moles are constant ie., 1) For the weight of compound gas to be minimum, its molecular weight should be minimum. The compound gas with minimum molecular weight is CH4. Hence, it weighs the least. We know that :



Thus, 5 grams of calcium constitute 0.125 mole of calcium. A yet another way of writing the above formula is by using the term ‘molar mass’ in place of ‘gram atomic mass’. That is : Number of moles of atom

ii) Number of gram molecules of nitrogen

23.

1  5 mole 40

1  mole 8

'p'g 1024   6.023  1023 'p' 1.66  1024 g 1.66

Number of molecules of D 

5 g of calcium 

27.

Mass of element Molar mass of element

In the above case, mass of element is 5 grams and molar mass of element is 40 g/mol. The atomic mass of aluminium is given to be 27 u. This means that 1 mole of aluminium atoms has a mass of 27 grams. Now, 1 mole of aluminium atoms = 27 g So, 4 moles of aluminium atoms = 27 × 4 g = 108 g Thus, the mass of 4 moles of aluminium atoms is 108 grams. We know that 1 mole of atoms contains 6.022 × 1023 atoms. Now, 1 mole of sodium atoms = 6.022 × 1023 atoms So, 0.2 mole of sodium atoms = 6.022 × 1023 × 0.2 atoms

Mole Concept Solutions

28.

327

= 12.044 × 1022 atoms Thus, 0.2 mole of sodium element has 12.044 × 1022 atoms in it. The ratio of atoms in carbon and magnesium will be the same as the ratio of their moles. So, first of all we should find out : (i) moles of carbon in 1 gram of carbon, and (ii) moles of magnesium in 1 gram of magnesium This can be done as follows. (a) 1 mole of carbon = Gram atomic mass of carbon = 12 grams Now, 12, g of carbon = 1 mole So, 1 g of carbon = mole _____________ (1) Thus, we have J2 mole of carbon element and it contains x atoms of carbon. Now, since an equal number of moles of all the elements contain an equal number of atoms, so moles of magnesium will also contain x atoms of magnesium. We will now calculate the moles of magnesium in 1 gram of magnesium. (b) 1 mole of magnesium = Gram atomic mass of magnesium = 24 grams Now, 24 g of magnesium = 1 mole So, 1 g of magnesium = Thus, we have

So, 1 g of magnesium  We know that,

1 mole of magnesium 12

contains = x atoms So, 

1 mole of magnesium contains 24

x 12 atoms 24

x  atoms 2

Thus, if 1 gram of carbon contains x atoms, then 1 gram of magnesium will have

29.

and it contains x atoms of carbon. Now, since an equal number of moles of all the elements contain an equal number of atoms, so

1 12

moles of magnesium will also contain x atoms of magnesium. We will now calculate the moles of magnesium in 1 gram of magnesium. (b) 1 mole of magnesium = Gram atomic mass of magnesium = 24 grams Now, 24 g of magnesium = 1 mole

x atoms 2

in it. To solve such problems we should remember that “equal number of moles of all the elements contain equal number of atoms”. (i) Let us first convert 4 grams of calcium into moles. We have been given that the atomic mass of calcium is 40 u, so 1 mole of calcium is 40 grams. Now, 40 g of calcium = 1 mole So,

4 g of calcium =

1 mole ____ (2) 12

1 mole of carbon element 12

1 mole 24

1  4mole 40 

Now,

1 mole 10

1 mole of calcium will have the same 10

number of atoms as

1 mole of neon. 10

So, we should now convert —mole of neon into mass in grams. (ii) The atomic mass of neon is 20 u, so 1 mole of neon will be equal to 20 grams. Now, 1 mole of neon = 20 g So,

1 1 mole of neon  20  g 10 10 www.betoppers.com

9th Class Chemistry

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30.

31.

Thus, 2 grams of neon will contain the same number of atoms as 4 grams of calcium. The atomic mass of an element is numerically equal to the masS of 1 mole of its atoms. Since 1 mole of atoms is 6.022 × 1023 atoms, so it means that the atomic mass of an element is numerically equal to the mass of its 6.022 × 1023 atoms. Now, 1 atom of element X has mass = 2.65×10~23 g So, 6.022 × 1023 atoms of element X have mass = 2.65 × 10–23 × 6.022 × 1023 = 15.96 g = 16 g Thus, the atomic mass of the element X is 16 u. The element of atomic mass 16 u is oxygen, having the symbol O. Here we have been given that the atomic mass of carbon (C) is 12u and the atomic mass of oxygen (O) is 16 u. So, first of all we will calculate the mass of 1 mole of carbon dioxide by using these values of atomic masses. The mass of 1 mole of carbon dioxide (CO2) will be equal to its molecular mass expressed in grams. That is: 1 mole of CO2 = Molecular mass of CO2 in grams = Mass of C + Mass of O × 2 = 12 + 6 × 2 = 12 + 32 = 44 grams So, the mass of 1 mole of carbon dioxide is 44 grams. Now, 44 g of carbon dioxide = 1 mole So, 22 g of carbon dioxide  

1  22mole 44

1 2

= 0.5 mole (or 0.5 mol) Thus, 22 grams carbon dioxide are equal to 0.5 mole. www.betoppers.com

The above problem can also be solved directly by using the formula: Number of moles of molecules =

Mass of substance in grams 22 1   Gram molecular mass of substance 44 2

= 0.5 mole Thus, 22 grams of carbon dioxide constitute 0.5 mole of carbon dioxide. A yet another way of writing the above formula is by using the term ‘molar mass’ in place of ‘gram molecular mass’. That is : Number of moles of molecules =

Mass of substance Molar mass of substance

32.

33.

In the above case, mass of substance is 22 grams and molar mass of the substance is 44 g/ mol. In order to solve this problem, we should know the mass of 1 mole of water. This can be obtained by using the given values of the atomic masses of hydrogen and oxygen as follows : 1 mole of water (H2O) = Molecular mass of H2O in grams = Mass of 2H atoms + Mass of O atom = 2 × 1 + 16 = 2 + 16 = 18 grams Thus, the mass of 1 mole of water is 18 grams. Now, Mass of 1 mole of water = 18 g So, Mass of 0.5 mole of water = 18 × 0.5 g = 9g Thus, the mass of 0.5 mole of water (H20) is 9 grams. We know that : 1 mole of oxygen contains = 6.022 × 1023 molecules So, 0.25 mole of oxygen contains = 6.022 × 1023 × 0.25 = 1.505 × 1023 molecules

Mole Concept Solutions

34.

Thus, 0.25 mole of oxygen contains 1.505 × 1023 molecules. In this problem, we should convert 4 g of methane into moles and 4 g of oxygen should also be converted into moles by using the respective gram molecular masses. The substance having more moles will contain more molecules in it. (i) Here, Mass of methane (CH4) = 4 grams Gram molecular mass of CH4 = Mass of C + Mass of 4H = 12 + 4 × 1 = 16 grams Now, Number of moles of CH4 

Mass of CH 4 in grams Gram molecular mass of CH 4 4  16 

1 4

= 0.250 mole ____________ (1) (ii) Here, Mass of oxygen (O2) = 4 grams Gram molecular mass of O2 = Mass of 2 ‘O’ atoms = 2 × 16 = 32 grams Now, Number of moles of O2 

Mass of O2 in grams Gram molecular mass of O 2



4 32



1 8

= 0.125 mole ________________ (2) We find that 4 g of CH4 contains more moles of molecules (0.250 moles), whereas 4 g of O2 contains less moles of molecules (0.125 moles). Since 4 g of methane has more moles,

329

35.

it contains more molecules (than 4 g of oxygen). The ratio of molecules in sulphur dioxide and methane will be the same as the ratio of their moles. So, first of all we should find out the number of moles of sulphur dioxide in 1 gram of sulphur dioxide, and the number of moles of methane in 1 gram of methane. This can be done as follows: (i) The molecular formula of sulphur dioxide is S02. So, 1 mole of S02 = Mass of S + Mass of 2 ‘O’ = 32 + 2 x 16 = 64 grams Now, 64 g of sulphur dioxide = 1 mole So, 1 g of sulphur dioxide  Thus, we have

1 64

___

(1)

1 mole of sulphur dioxide and 64

it contains x molecules in it. Now, since equal moles of all the substances contain equal number of molecules, therefore, mole of methane will also contain x molecules of methane. We will now calculate the number of moles in 1 gram of methane. (ii) Molecular formula of methane is CH4. So, 1 mole of CH4 = Mass of C + Mass of 4H = 12 + 4 × 1 = 16 grams Now, 16 g of methane = 1 mole So,

1 g of methane =

1 mole ___ (2) 16

We know that : 1 mole of methane contains = x molecules 64

So. 

1 mole of methane will contain 16

x  64 molecules 16

Thus, if 1 g of sulphur dioxide contains x

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9th Class Chemistry

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36.

molecules, then 1 g of methane contains Ax molecules. This problem will be solved by using the fact that equal moles of all the gases contain equal number of molecules. Let us first convert 16 grams of sulphur dioxide into moles. 1 mole of sulphur dioxide, S02 = Mass of S + Mass of 2 ‘O’ = 32 + 2 × 16 = 64 grams Now, 64 g of sulphur dioxide = 1 mole So, 16 g of sulphur dioxide 



Now,

_______________

(1)

1 mole of sulphur dioxide will have the 4

same number of molecules as

54  0.051g Al 102

= 0.027 g Al __________________ (3) The atomic mass of aluminium is given to be 27 u. This mean that 1 mole of aluminium atoms (or aluminium ions) has a mass of 27 grams, and it contains 6.022 × 1023 aluminium ions. Now, 27 g of aluminium has ions = 6.022 × 1023 So, 0.027 g of aluminium has ions

1  16 mole 64 1  mole 4

1 mole of 4

6.022 1023  0.027 27

1 oxygen. So, we should now convert mole 4



of oxygen into mass in grams. 1 mole of oxygen, O2 = Mass of 2 ‘O’ atoms = 2 × 16 = 32 grams Now, 1 mole of oxygen = 32 grams

= 6.022 × 1020 Thus, the number of aluminium ions (Al3+) in 0.051 gram of aluminium oxide is 6.022 × 1020.

So,

37.

= 54 + 48 = 102 grams ____________ (1) Now, 1 mole of Al2O3 contains 2 moles of Al. So, Mass of Al in 1 mole of Al2O3 = Mass of Al × 2 = 27 × 2 = 54 grams __________________ (2) Now, 102 g aluminium oxide contains= 54 g Al So, 0.051 g aluminium oxide contains

38.

1 1 mole of oxygen  32  grams 4 4

= 8 grams ______________ (2) This problem involves aluminium ions. Please note that the mass of an aluminium ion is the same as that of an aluminium atom. In order to solve this problem, first of all we have to find out the mass of aluminium atoms in 0.051 g of aluminium oxide (which will give us the mass of aluminium ions) This can be done as follows : 1 mole of Al2O3 = Formula mass of Al2O3 in grams = Mass of Al × 2 + Mass of O × 3 = 27 × 2 + 16 × 3

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 32 g O2 has mole =1 64  1  2moles  64 g O2 has mole = 32 6.023  1023  64 32 g O contain  2 32 23 molecules = 12.04×10 N molecules of O2 weigh 32 g 1 molecule of O2 weights

32 = 5.313×10-23 g 6.023×10 23 39.

 6.023 × 1023 molecules of CO2  44 g  1021 molecules of

44  1021 g 6.023  1023 = 7.31 × 10–2 g = 73.1 mg  CO2 left = 200 – 73.1 = 126.9 mg Also Mole of CO2 left

CO2 

Mole Concept Solutions

331 The weight of 1 mole of CaCO3 = 100 gm  100gm CaCO3 = 6.023 × 1023 molecules

wt. 126.9  103   2.88×103 = m.wt. 44 40.

 10gm CaCO3 

22 2   2.5  10  8.20  1610.7 g 7 Weight of cobalt in alloy =

1022 molecules  6.023 × 1022 molecules of CaCO3 = 50 × 6.023 × 1022 = 3. 0115 × 1024. (A) 1 mole of sucrose contains

=  r 2h × d =

1610.7  12  193.3g 100  58.9 g cobalt has atoms = 6.023 × 1023  193.3 g of cobalt has atoms =

41.

6.023  1023  193.3 = 19.8 × 1023 58.9 (mol. wt. CaCl2 = 111 g)  111 g CaCl2 = N ions of Ca2+  222 g CaCl2 =

45.

6.023 × 1023 molecules  1 molecule of sucrose has 45 atoms  6.023 × 1023 molecules of sucrose has 45 × 6.023 × 1023 atom/mole.

46.

N  222  2 N ions of Ca 2+ 111  111 g CaCl2  2 N ions of Cl–

 222 g CaCl2 

2  N  222 ions of Cl = 4 111

N ions of Cl Mass of carbon in dot = 1 × 10–6 g 23

 No. of atoms =

6.023  10  1 10 12

6

47.

= 5×1016 atoms of C 43.

Molecule has C, H and other components  Wt. of 9 C atoms = 12 × 9 = 108 amu  Wt.of 13 H atoms = 13 × l = 13 amu Wt. of 2.33 × 10–23 g of other atom 2.33  1023  14.04amu 1.66  1024  Total weight of one molecule = 108 + 13 + 14.04 = 135.04 amu Mol. weight = 135.04 (C) Total no. of protons in given amount of CaCO3 = No. of molecules in the given amount of CaCO3 (Nm) × No. of protons in 1 molecule of CaCO3 . In 1 molecule of CaCO3, the no. of protons = 20 + 6 + 24 = 50  Total no. of protons = 50×Nm Let us find the no. of molecules. We know, 1 mole of CaCO3 contains 6.023×1023 molecules.

Terms in the problem are moles and volume at S.T.P. The standard relation connecting the above two terms is: occupies 1 mole of any gas at S.T.P.   22.4 litres  0.125 moles of any gas at S.T.P.. occupies   x litres  x

–

42.

 x

48.

0.125  22.4  2.8 litres 1

Therefore, 0.125 moles of any gas at S.T.P occupies 2.8 litres. Terms in the problem are volume at S.T.P. and number of moles. The standard relation connecting the above two terms is: make 22.4 litres of any gas at S.T.P.  1 mole make  2.8 litres   x moles

=

44.

6.023  1023  10 = 6.023 × 100

Weight of alloy cylinder = Volume × density

2.8  1 1   0.125 22.4 8

Therefore, 2.8 litres of any gas at S.T.P make 0.125 moles. Terms in the problem are volume of gas at S.T.P. and weight. The standard relation connecting the above two terms is: occupies GMW of any gas   22.4 litres at S.T.P. occupies  71 g of Cl2   22.4 litres at S.T.P.. occupies 17.75 g of Cl2   x litres www.betoppers.com

9th Class Chemistry

332

 x

49.

Therefore, 17.75 g of Cl2 at S.T.P. occupies 5.6 litres. Terms in the problem are mass and volume of gas at S.T.P. The standard relation connecting the above two terms is: weighs 22400 cc of any gas at S.T.P    GMW weighs 22400 cc of hydrogen at S.T.P.  2 g weighs 500 cc of hydrogen at S.T.P.   xg  x

50.

51.

17.75  22.4 22.4   5.6 litres 71 4

52.

500  2  0.045 g 22400

Therefore, the mass of 500 cc of hydrogen at S.T.P. is 0.045 g or 45 mg. Any gas can be identified from its molecular weight. Let us find the same from the given data. Volume occupied by the gas at S.T.P. = 2.24 litres Mass of the gas = 1.6 g Molecular weight = ? Terms involved in the problem are weight and volume of gas at S.T.P. The standard relation connecting the above two terms is: occupies GMW of any gas   22.4 litres at S.T.P. 2.24 litres of the given gas at S.T.P. weigh   1.6 g 22.4 litres of any gas at S.T.P. weigh   GMW = 16g  Molecular weight = 16g Therefore, the gas with molecular weight 16 is methane (CH4). Mass of the volatile compound = m = 0.2 g Volume occupied by the volatile compound at S.T.P. = VS.T.P. = 56 ml Gram Molecular weight = GMW = ? Terms in the problem are m, VS.T.P and GMW Formula connecting these terms is, VS.T.P = n × 22.4 =

 GMW 

m  22.4 GMW

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 Gram molecular weight (GMW) = 80 g Molecular weight = 80  Therefore, the molecular weight of the volatile compound is 80. Terms involved in the problem are volume of gas at S.T.P and number of molecules. The standard relation connecting the above two terms is: 22.4 litres of the given gas at S.T.P. 23 contains   6.023 ×10 molecules 22.4 litres of hydrogen at S.T.P.  23 contains   6.023 ×10 molecules 1.12 litres of hydrogen at S.T.P. contains   x molecules  x

53.

m 0.2  22.4   22400  80g VSTP 56

1.12  6.023  1023 = 3.0115 × 1023 22.4

molecules Therefore, the number of molecules present in 1.12 litres of hydrogen at S.T.P. is 3.0115 ×1022 molecules. We know that ‘N’ molecules of any gas at S.T.P. occupy 22.4 litres.  x  22.4  litres N  

 x molecules occupy 

54.

 Volume occupied by any gas at S.T.P = f (number of molecules (x)) As the number of molecules are same, the volume occupied by them is same. Therefore, the given statement is True. Percentage weight of iron = 0.25; Molecular weight of haemoglobin = 89600 No. of atoms of iron = n = ?; Atomic weight of iron = 56 We know, Percentage composition of Fe = n × At. wt. of Fe 100 Molecular weight

 0.25 =

n × 56 100  n = 89600

Mole Concept Solutions

333

=

0.25  89600 4 100  56

55.

40+144 + 22 + 224 = 430  % composition of Ca in Ca(C6H11O7)2 =

Therefore, the number of iron atoms per molecule of haemoglobin is 4.13 We know, percentage composition of an element =

n × At. wt. 100 mol.wt

__________

1 40  100  9.3 430

d) In calcium citrate, Ca3(C6H5O7)2: Molecular weight of Ca3(C6H5O7)2 = (3 × 40)+ (12 × 12) + (l0 × l) + (14 × 16) = 120+144+10 + 224 = 498  % composition of Ca+2 in Ca3(C6H5O7)2

(1)

Molecular weight of glucose = 180 Applying (1), we get, Percentage composition of carbon = 6  12 72  100   100  40 180 180

=

Percentage composition of hydrogen =

CONCEPTIVE WORKSHEET

12  1 12 100  100  6.7 180 180

HINTS AND SOLUTIONS TO THE

Percentage composition of oxygen =

SELECTED QUESTIONS

6  16 96 100  100  53.3 180 180

56. 57.

(B) 2 g-atom Fe give 1 mole Fe2O3 (or) 2×56 g Fe give 160 g Fe2O3.  5 × 106 g Fe give

58.

5  106  160 = 7.142 × l06 112

g or 7.142 tons We know, percentage composition of an element =

n × At. wt 100 mol. wt

3  40  100  24.48 490

__________

(1)

a) In calcium carbonate, CaCO3: Molecular weight of CaCO3 = 40 + 1 2 + 48 = 1 00  % composition of Ca in CaCO3 = 1 40 100  40 100

b) In calcium lactate, Ca(C3H5O3)2: Molecular weight of Ca(C3H5O3)2 = 40 + (6×12)+(10×l) + (6×16)= 40+72+10+96 = 218  % composition of Ca in Ca(C3H5O3)2: = 1 40 100  18.34 218

c) In calcium gluconate, Ca(C6H11O7)2: Molecular weight of Ca(C6H11O7)2 = 40 + (12 × 12) + (22 × l) + (14 × 16)

1. 2. 3. 4. 5. 6. 7. 8. 9.

(C) (C) (C) (C) (B) (D) (A) (C) (B) Carbon dioxide and Oxygen combine in the fixed proportion of 3 : 8 by mass to produce 11g of Carbon dioxide. Therefore the same mass of Carbon dioxide (11g) will be obtained if we have 3 g of Carbon in 50 g of Oxygen. The extra Oxygen 50 – 8 = 42 g of Oxygen) will remain unreacted. 10. (C) Hydrogen and Oxygen combine in a fixed ratio of 1 : 8 by mass to form water. So, 1g of hydrogen gas requires 9 g of oxygen 3g of hydrogen gas requires 8×3g of oxygen = 24 g. 11. (C) 12. (C) 13. (C) 14. We know, Average atomic weight

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9th Class Chemistry

334

of the mass of single molecule of a compound

99.75 ×15.9949 + 0.25 ×17.9992 = 100

=

to the mass of 26.

1595.491275 + 4.4998 100

27. 1599.99 = = 16 100 1 th of C 12 . 12

(A)

16.

(D) Weight of one atom of an element = Atomic weight × amu.



18. 19. 20. 21.

40  1.66 10 1.66  1024

 40

 Mass number = 40 and the element is either calcium or Argon.  Number of protons = 20 (or) 18. (C) Weight of He atom = 4 amu = 4 × 1.66 × 10–24g (B) Weight of an element = Atomic weight × 1.66 × 10–24g (D) All the given statements are correct. (A) We know, number of gram atoms =

28.

GMW 

1 weight is cons tan t  GAW

22. 23. 24.

25.

maximum, the gram atomic weight of the element should be minimum. And the gram atomic weight is minimum for hydrogen. Therefore, hydrogen can generate maximum number of gram atoms from a given amount. (B) The smallest particle of matter that can exist independently is a molecule. (a,b,c,d) H2, N2, F2, O2. (A) i) Sodium  monoatomic (Na) ii) Helium  monoatomic (He) iii) Oxygen  diatomic (O2) iv) Ozone  triatomic (O3) v) sulphur  polyatomic (S8) (D) Molecular weight has no unit as it is a ratio

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gases) From the above equation, it is clear that the gas with the least molecular weight contains the maximum number of gram molecules. The gas with the least molecular weight is hydrogen. Therefore, it contains the maximum number of gram molecules. (C) Any gas can identified from its molecular weight. m

 Number of gram atoms

 For the number of gram atoms to be

1 ( mass is same for all the GMW

We know, GMW  n ––––––(1) m

Given weight of thesubs tan ce  m  Gram atomic weight  GAW 



Given weight  m  Gram molecular weight  GWM 

 nm 

24

(B) Atomic weight 

(C) The weight of ammonia molecule = 17 amu. = 17 × 1.66 × 10–24g (D) Given that the weight of the gas is 100 g We know, Number of gram molecules  nm 

15.

17.

1 th of C – 12 isotopic atom 12

32  16g  Molecular weight 2

Weight of thesubs tan ce  m 

4

= Gram molecular weight GMW   4   1 29.

We know that : 6.022 × 1023 atoms of helium = 1 mole So, 9.033 × 1024 atoms of helium 

30.

1  9.033  1024 6.022 1023

= 15 moles (or 15 mol) Thus, 9.033 × 1024 atoms of helium are 15 moles of atoms. 1 mole of iron = Gram atomic mass of iron = 56 grams We know that 1 mole of iron element contains 6.022 × 1023 atoms of iron.

Mole Concept Solutions

335

Now, 56 g of iron contains = 6.022 × 1023 atoms So, 2.8 g of iron contains 



31.

6.022 1023  2.8 56

= 3.011 × 1022 atoms Thus, a piece of iron metal having a mass of 2.8 grams contains 3.011 × 1022 atoms of iron. 1 mole of carbon atoms means 6.022 ×1023 carbon atoms. In this case 1 mole of carbon atoms weighs 12 grams. This means that the mass of 6.022 × 1023 atoms of carbon is 12 grams. Now, 6.022 × 1023 atoms of carbon have mass = 12 g So, 1 atom of carbon has mass

= 1.99 × 10–23 g Thus, the absolute mass of 1 atom of carbon is 1.99 x 10~23 gram. In order to solve this problem, we should convert 100 grams of sodium into moles of sodium, and also 100 grams of iron into moles of iron. The element having more moles will have more atoms. Please note that since the atomic mass of sodium is 23 u, the molar mass of sodium will be 23 g/mol. Similarly, since the atomic mass of iron is 56 u, the molar mass of iron will be 56 g/mol. We will now calculate the moles of sodium atoms (Na) and iron atoms (Fe), one by one. (i) Mole of sodium  



6.022  1022 2

12  g 6.022 1023

32.

(ii) Mole of iron 

Mass of sodium Molar mass of sodium 100 = 4.34 23

33.

100 = 1.78 56

We find that 100 grams of sodium contain 4.34 moles of atoms whereas 100 grams of iron contain 1.78 moles of atoms. Since 100 grams of sodium has more moles, it contains more atoms than 100 grams of iron. We know that : 6.022 × 1023 molecules of sulphur dioxide = 1 mole So, 12.044 × 1022 molecules of sulphur dioxide



34.

Mass of iron Molar mass of iron



1  12.044  1022 23 6.022 10

2 = 0.2 mole 10

Thus, 12.044 × 1022 molecules of sulphur dioxide are 0.2 mole. We know that 1 mole of water contains 6.022 x 1023 water molecules. So, in order to calculate the number of molecules in 0.06 gram of water, we should first calculate the mass of 1 mole of water in grams. Now, Molecular mass of water, H20 = 2 + 16 = 18 u So, Gram molecular mass of water =18 grams (or 1 mole of water) Thus, 1 mole of water has a mass of 18 grams and it contains 6.022 x 1023 molecules of water. Now, 18 g water contains = 6.022 x 1023 molecules So, 0.06 g water contains 

6.022 1023  0.06 18

= 2.007 × 1021 molecules www.betoppers.com

9th Class Chemistry

336

35.

Thus, a drop of water weighing 0.06 gram contains 2.007 × 1021 molecules of water in it. First of all we should find out the mass of 1 mole of nitrogen molecules (N2). Now, 1 mole of N2 = Molecular mass of N2 in grams = Mass of 2 N atoms in grams = 2 × 14 grams = 28 grams Thus, the mass of 1 mole of nitrogen molecules is 28 grams. Now, 1 mole of nitrogen molecules contains 6.022 × 1023 molecules. This means that the mass of 6.022 x 1023 molecules of nitrogen is 28 grams. Now, Mass of 6.022 × 1023 molecules of N2 = 28 g So, Mass of 3.011 × 1024 molecules of N2 



36.

molecular mass of 32 u is oxygen, having the formula 02. 37.

Thus, the mass of 3.011 x 1024 molecules of nitrogen is 140 grams. The molecular mass of a substance is numerically equal to the mass of 1 mole of its molecules. Since 1 mole of molecules is equal to 6.022 x 1023 molecules, it means that the molecular mass of a substance is numerically equal to the mass of its 6.022 x 1023 molecules. In this problem we have been given the absolute mass of 1 molecule of the substance, so all that we have to do is to find out the mass of its 6.022 x 1023 molecules. Now, Mass of 1 molecule of substance = 5.32 × 10 –23 g So, Mass of 6.022 x 1023 molecules of substance = 5.32 × 10"23 × 6.022 × 1023 = 32g Thus, the molecular mass of the given substance is 32 u. The substance having a

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6.023  1023 atoms 4 = 1.506 × 1023 atoms Standard molar volume is the volume occupied by 1 mole of gas at NTP.  1.429 g of O2  1 litre at NTP 32 litre at NTP = 22.39 litre  32 g of O2  1.429 at NTP Molar volume = 22.39 litre mol–1  1g He has

38.

39.

Given that, 3M + 2N  M3N2 Let a is at. wt. of metal  (3a + 28) g M3N2 has metal = 3a 3a  100  100 g M3N2 has metal =  3a  28 

28  3.011 1024 6.022 1023 28 10 = 140 g 2

 4 g He has 6.023 × 1023



40.

3a  100  72  3a  28

 a = 24

44 =1 44 1 mole of CO2 contain 1 mole of ‘C’ atoms and 2 moles of oxygen atoms.  44g of CO2 has 2 × 6 × 1023 atoms of oxygen.

(A) No. of moles of CO2 =

4.4g of CO2 has 

12  1023  4.4 = 1.2 × 1023 44

atoms. 41.

(A) Density 

Mass ; Density of water = 1 Volume

gm/ml  1 ml of water weighs 1 gm  0.0018 ml = 0.0018gm Number of moles



42.

weight 0.0018   1 104 Molecular weight 18

 Number of water molecules = 6.023 × 1023 × 1 × 10–4 = 6.023 × 1019 (D) Density of water = 1gm/ml  1 ml of water weighs 1 gm

 1 litre (1000ml) of water weighs 1000g  No. of moles =

1000 = 55.55 18

Mole Concept Solutions

337

No. of molecules = No. of moles ×6.023×1023 = 55.55×6.023×1023 = 55.55 NA

43.

Number of moles of laughing gas (N2O) = 0.25 V(occupied) = ? 1 mole of any ga at S.T.P occupies 22.4 litres. Using the above relation, we get, occupies 1 mole of N2O    22.4 litres. occupies 0.25 moles of N2O   x

x

0.25  22.4  5.6litre 1

 0.25 mole of laughing gas at S.T.P occupies 44.

a volume of 5.6 litres. 22.4 litres of any gas at S.T.P make 1 mole Let us apply the above relation to each of the given cases. (a) 22.4 litres of marsh gas (CH4) at S.T.P makes   1 mole 2.24 litres of marsh gas (CH4) at S.T.P makes   x mole

x

2.24  1  0.1 22.4

 2.24 litres of marsh gas at S.T.P makes 0.1 mole. (b) 22.4 litres of tear gas (CCl3 NO2) at makes S.T.P   1 mole makes 1.12 litres of tear gas at S.T.P  x

x

1.12  1  0.05 22.4

 1.12 litres of marsh gas at S.T.P makes 0.05 mole. makes (c) 22.4 litres of N2 at S.T.P   1 mole makes 11.2 litres of N2 at S.T.P  x

x

11.2  1  0.5 mole 22.4

 11.2 litres of N2 at S.T.P makes 0.5 mole. 45.

(i) Two moles of HC1 can be written as 2HC1. We can see that the two moles of HC1 contain 2 moles of H atoms (or hydrogen atoms). (ii) One mole of NH3 contains 3 moles of H atoms (or hydrogen atoms).

46.

Now, two moles of HC1 contain 2 moles of hydrogen atoms whereas one mole of NH3 contains 3 moles of hydrogen atoms. It is obvious that 1 mole of NH3 contains more hydrogen atoms. a) Molecular weight of potassium dichromate[K2Cr2O7]= 78+ 104+ 112 = 294 Number of atoms of potassium in 1 molecule of potassium dichromate = 2 We know, percentage composition of an element =

n × At. wt  100 mol. wt

 Percentage composition of potassium in

K2Cr2O7 

2  39 78  100  100   26.53% 294 294

b) Molecular weight of calcium phosphate [Ca3(PO4)2 =3×40+2×31 + 8 × 16 = 310 Number of atoms of phosphorus in 1molecule of calcium phosphate = 2  Percentage composition of phosphorus in Ca3(PO4)2 =

2  31 62  100  100   20% 4. 310 610

Weight of each FeSO4 tablet = 325 mg = 0.325 g Amount of Fe+2 that is fatal to the child = 550 mg = 0.55 g Let the number of tablets that constitute a legal dose to the child be x.

x =

0.55g Weight of Fe contributed by 1 FeSO4 tablet +2

-------------- (1) Calculation of the amount of Fe+2 in 1 FeSO4 tablet: 1 mole of FeSO4 contain 1 gram atom of Fe  152 g of FeSO4 contain 56 g of Fe  0.325 g of FeSO4 contain

0.325  56 = 152

0.11 g of Fe Substituting the above value in (1), we get, x = 0.55 5 0.11

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9th Class Chemistry

338

47.

Therefore, the molecular weight of insulin is 941.17.

Therefore, 5 tablets of FeSO4 turn to be lethal to the child. Molecular weight of urea [CO(NH2)2] = 12 + 16 + 28 + 4 = 60  60 g of urea contains 28 g of nitrogen  5 × 1000 g of urea contains ‘x’g of nitrogen =

28  5  1000  2333.3g 60

SUMMATIVE WORKSHEET HINTS AND SOLUTIONS TO THE SELECTED QUESTIONS 1.

 Mass of nitrogen supplied to the soil by 5

48.

kg of urea is 2333.3 g. a) AgCl  % of Ag = 108 108  100   100  75.26 143.5 108  35.5

b) Ferric oxide = Fe2O3  % of Fe = 2  56  100  70  2  56   48

c) K2SO4  % of K

 2  39 

78

 100   100  44.82 78  32  64  2 =39   32  64

49.

The formula shows that 1 mole H2O contains 2 g atom of hydrogen and 1 g atom of oxygen. The weight of 1 mole is 18 g. 18 g of water contain 2 g of hydrogen and 16 g of oxygen.  Percentage composition of hydrogen = 2  100 = 11.1 % 18

The percentage composition of oxygen = 16  100 = 88.9% 18

50.

2.

Minimum molecular weight of insulin can be found out, assuming that 1 molecule of it contains 1 sulphur atom. Percentage weight of sulphur = 3.4; Molecular weight of insulin = x = ? No. of atoms of sulphur = n = 1; Atomic weight of sulphur = 32 We know, Percentage composition of S n × At. wt of S 100 Molecular weight

 3.4 =

1 32 1 32 100 100   941.17 x 3.4

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3.

(B) Calcium carbonate Calcium oxide + Carbon dioxide Total mass of reactants = Mass of calcium carbonate Total mass of products = Mass of calcium oxide + mass of carbon dioxide = 5.6g + 4.4g = 10g But according to law of conservation of mass. The total mass of reactants is equal total mass of products in a chemical reaction. So the mass of calcium carbonate is 10g. (A) Mass or boron in the compound = 0.096 g. Mass of compound = 0.24 g % of boron = Mass of boron in compound × 100 Mass of compound 0.096   100 = 40 %. 0.24 % of oxygen = 100 – 40 = 60% (D) We have learnt according tothe law of constant composition that the percentage of the elements present in different samples of a pure substance always remains the same. This means that in the second sample of calcium carbonate, the ratio of the different elements present will remain the same or will remain unchanged. Thus, Mass of calcium in 1.5 g of the sample 40 = × (1.5 g) = 0.6 g 100 Mass of carbon in 1.5 g of the sample 12 = × (1.5 g) = 0.18 g 100 Mass of oxygen in 1.5 g of the sample =

Mole Concept Solutions

4.

18 × (1.5 g) = 0.72 g 100 (B) Let us find out the ratio by mass of silver and chlorine in the two samples of silver chloride which are formed. First expirement : Mass of silver = 0.50 g Mass of silver chloride = 0.68 g Percentage of silver in the sample Mass of silver

= Mass of silver chloride × 100

 0.50g 

= 0.68g × 100   = 73.53 Percentage of chlorine in the sample = 100 – 73.53 = 26.47 Second experiment : Mass of silver = 1.0 g Mass of silver chloride = 1.33 g Percentage of silver in the sample Mass of silver

= Mass of silver chloride × 100

1.0g 

= 1.33g × 100  

5.

= 75.19 Percentage of chlorine in the sample = 100 – 75.19 = 24.81 The ratio by mass of silver and chlorine in the two samples is nearly the same. This illustrates the law of constant composition. All that we have to do in this problem is to calculate the mass of reactants and products separately, and then compare the two. If the two masses are equal, then the law of conservation of mass gets verified. The given reaction can be written as :

339

(i) Sodium carbonate and ethanoic acid are reactants. So, Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid = 5.3 + 6 = 11.3 g _____________ (1) (a) Sodium ethanoate, carbon dioxide and water are products. So, Mass of products = Mass of sodium ethanote + Mass of carbon dioxide + Mass of water = 8.2 + 2.2 + 0.9 = 11.3 g ______________ (2) We find that the mass of reactants is 11.3 g and the mass of products is also 11.3 g. Since the mass of products is equal to the mass of reactants, the given data verifies the law of conservation of mass. 6. This problem is to be solved by using the law of conservation of mass in chemical reactions. In this reaction, calcium carbonate is the reactant whereas calcium oxide and carbon dioxide are products. Now, from the law of conservation of mass we have : Mass of products = Mass of reactants or Mass of calcium oxide + Mass of carbon dioxide = Mass of calcium carbonate So, 5.6 + Mass of carbon dioxide = 10 And, Mass of carbon dioxide= 10-5.6 = 4.4 g. Thus, the mass of carbon dioxide formed is 4.4 g. 7. In order to solve this problem we have to calculate the ratio (or proportion) of copper and oxygen in the two samples of copper oxide compound. Now : (a) In the first experiment : Mass of copper = 1.03 g __________ (1) www.betoppers.com

9th Class Chemistry

340

And, Mass of copper oxide = 1.288 g So, Mass of oxygen = Mass of copper oxide - Mass of copper = 1.288 – 1.03 = 0.258 g __________ (2) Now, in the first sample of copper oxide compound : Mass of copper : Mass of oxygen = 1.03 : 0.258 

10.

1.03 :1 0.258

= 4:1 __________ (3) (b) In the second experiment : __________ Mass of copper = 2.938 g (4) And, Mass of copper oxide = 3.672 g So, Mass of oxygen = Mass of copper oxide – Mass of copper = 3.672 – 2.938 = 0.734 g _____________ (5) Now, in the second sample of copper oxide compound : Mass of copper : Mass of oxygen = 2.938 : 0.734 

8.

9.



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Mass of boron in compound 100 Mass of compound



0.096 100 0.24

= 40 _____________________ (1) (ii) Mass of oxygen in compound = 0.144 g And, Mass of compound = 0.24g So, Percentage of oxygen 

Mass of oxygen in compound  100 Mass of compound



2.938 :1 0734

________________ = 4:1 (6) From the above calculations we can see that the ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same 4 : 1. So, the given figures verify the law of constant proportions. Here we have been given that hydrogen and oxygen always combine in the fixed ratio of 1 : 8 by mass. This means that : 1 g of hydrogen gas requires = 8 g of oxygen gas So, 3 g of hydrogen gas requires = 8 × 3 g of oxygen gas = 24 g of oxygen gas Thus, 24 grams of oxygen gas would be required to react completely with 3 grams of hydrogen gas. Our answer will be governed by the law of

constant proportions. Now, since carbon and oxygen combine in the fixed proportion of 3 : 8 by mass to produce 11 g of carbon dioxide, therefore, the same mass of carbon dioxide (11 g) will be obtained even if we burn 3 g of carbon in 50 g of oxygen. The extra oxygen (50 - 8 = 42 g oxygen) will remain unreacted. (i) Mass of boron in compound = 0.096 g And, Mass of compound = 0.24 g So, Percentage of oxygen

11.

0.144 100 0.24

= 60 ______________ (2) Thus, the percentage composition of the compound is : Boron = 40%; Oxygen = 60% We know, number of gram atoms nA = m GAM

mH  GAM H m H  GAM He n  H    m He n He m He  GAM H  GAM He Substituting the values in the above equation, nH 1 4 2 we get n  2  1  1 He

12.

The heaviest naturally occurring element is uranium i.e., 92 U 238 The lightest atom is hydrogen i.e., 1 H1. Given, mass of hydrogen = mass of 1 gram atom of uranium.

Mole Concept Solutions

341

Let’s find the mass of 1 gram atom of uranium. 

Number of gram atoms  na  

Given mass Gram atomic mass –––(1)

Applying (1) to the cases, we get, i) Number of gram molecules of Neon (Ne) = 5 1   0.25 GMW of Ne  20  20 4

m 1   GAM of Uraniumis 238g  238

ii) Number of gram molecules of nitrogen

 m  238g

Therefore, we need to find the number of gram atoms of hydrogen in 238 g of it. Applying 1, we get, Number of gram atoms

 na   13.

17.

238  238 1

weight of one atom 1 weight of th of C  12 atom 12

weight of one atom 1  th of = mass of atomicmass 12

Weight of the subs tan ce hydrogen = Gram molecular mass 

18.

C – 12 atom = x Molecular mass = Weight of one molecule 1 Weight of th of C-12 atom 12



Weight of one molecule Molecular mass

1 = Mass of th of C - 12 atom = y  x = 12

14.

15.

19.

16.

We know, Number of gram molecules  nm 

Mass of compound gas = Number of moles × Gram molecular mass.  Mass of compound gas  Gram molecular mass ( Number of moles are constant ) For the mass of compound gas to be minimum, its molecular mass should be minimum. The compound gas with minimum molecular mass is CH4. Hence, it weighs the least.  32 g O2 has mole =1 64  1  2moles 32

6.023  1023  64  32 g O2 contain 32 = 12.04×1023 molecules N molecules of O2 weigh 32 g 1 molecule of O2 weighs

22

weight in grams 1.66  10 = = 100 1.66×1024 1.66×1024 amu.  molecular mass of the compound = 100.

4 2 2

 64 g O2 has mole =

y. Mass of one molecule = (molecular mass) amu Molecular mass of CO2 = 12 + (2 × 16) = 44.  mass of one molecule of CO2 = 44 amu To find out the molecular mass, we need to find the mass of one molecule in terms of amu.  Mass of one molecule in terms of amu =

7 1   0.25 GMW of N 2  28  28 4 1 gram mole of methane weigh 16g.  Mass of methane = 1 × 16 = 16g  Mass of hydrogen in 16g of methane = 4 g (as 4 hydrogen atoms are present) We know, Number of gram molecules of

(N2) =

Atomic mass 

Given mass Gram molecular mass ––––(1)

20.

32 = 5.313×10-23 g 23 6.023×10 23  6.023 × 10 molecules of CO2  44 g 21  10 molecules of

44 1021 g = 7.31 × 10–2 g = CO2  6.023  1023 www.betoppers.com

9th Class Chemistry

342

73.1 mg  CO2 left = 200 – 73.1 = 126.9 mg Also Mole of CO2 left =

21.

of CaCO3. In 1 molecule of CaCO3, the no. of protons = 20 + 6 + 24 = 50  Total no. of protons = 50×Nm Let us find the no. of molecules. We know, 1 mole of CaCO3 contains 6.023×1023 molecules. The mass of 1 mole of CaCO3 = 100 gm  100gm CaCO3 = 6.023 × 1023 molecules

wt. 126.9×10-3 = = 2.88×10-3 m.wt. 44 Mass of alloy cylinder = Volume × density =  r2h × d = 22 2   2.5   10  8.20  1610.7 g 7 Mass of cobalt in alloy =

1610.7  12  193.3g 100 23  58.9 g cobalt has atoms = 6.023 × 10  193.3 g of cobalt has atoms =

22.

6.023  1023  193.3 = 19.8 × 1023 58.9 Mol. mass of CaCl2 = 111 2+  111 g CaCl2 = N ions of Ca  222 g CaCl2 =

 10gm CaCO3 

26.

27.

N × 222 = 2 N ions of Ca 2+ 111 –  111 g CaCl2  2 N ions of Cl  222 g CaCl2 

23.

2  N  222 ions of Cl  111

25.

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0.635  6.023  1023  0.06023  10 23 63.5 Number of atoms of Cu in 0.635 g of it is  0.06023 × 1023. Given, mass of laughing gas (N2O) = 11 g number of molecules = ? GMM of any substance contains 6.023 × 1023 molecules. By, using the above relation, we get, contains 44 of N2O  6.023 × 1023 molecules Na 

= 5×1016 atoms of C

2.33  1023  14.04 amu = 1.66  1024  Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu Mol. mass = 135.04 Total no. of protons in given amount of CaCO3 = No. of molecules in the given amount of CaCO3 (Nm) × No. of protons in 1 molecule

m N GAM

(where, N = 6.032 × 1023) Substituting the given values in the above equation, we get,

= 4 N ions of Cl– Mass of carbon in dot = 1 × 10–6 g

Molecule has C, H and other components  Wt. of 9 C atoms = 12 × 9 = 108 amu  Wt.of 13 H atoms = 13 × l = 13 amu Wt. of 2.33 × 10–23 g of other atom

× 1022 molecules 22 6.023 × 10 molecules of CaCO3 = 50 × 6.023 × 1022 = 3. 0115 × 1024. 1 mole of sucrose contains 6.023 × 1023 molecules  1 molecule of sucrose has 45 atoms 23  6.023 × 10 molecules of sucrose has 45 × 6.023 × 1023 atom/mole. Given, Mass of copper (m) = 0.635 g Number of atoms of copper (Na) = ? Atomic mass of copper (GAM) = 63.5 We know, number of atoms  N a  

6.023  1023  1  106 No. of atoms =  12 24.

6.023  10 23  10 = 6.023 100

28.

contains 11 g of N2O  x

Mole Concept Solutions

29.

343

11  6.023  1023  x = 1.51 × 1023 44  11 g of laughing gas contains 1.51 × 1023 molecules. Cost of 11 g of gold = Rs. 5000 Cost of each atom of gold = ? Number of atoms of gold



mass  Gold  GAM  Gold 

(7) (8) (9) (10) (11) 12.

 6.023  1023 

1  6.023  1023  34  10 21 197 cos t Given, 34 × 1021 atoms of gold   Rs. 5000 cos t 1 atom of gold   Rs.x

 x

1 5000  Rs.147  10 21 23 0.336  10  Cost of 1 atom of gold is Rs. 147 ×10–21. Given, Volume of the gas = 560 cc = 0.56 litre Mass of the gas = 1.1 g Molecular mass = ? 22.4 litres of any gas at S.T.P weigh its gram molecular mass. Applying the above relation, we get, weigh 22.4 litres of the gas   GMW

 x

96  6.023  1023 = 12.046 × 1023 48

molecules We know that, 1 molecule of ozone contains 3 oxygen atoms   12.046 ×1023 molecules of ozone contain, 3 × 12.046 × 1023 = 36.138 × 1023 atoms contains (c) 256 g of S8 molecule  6.023 × 23 10 molecules contains 64 g of S8 molecule  x

weigh 0.56 litre   1.1g

1.1  22.4  44 g 0.56  The molecular mass of the gas is 44.

 GMW 

HOTS WORKSHEET HINTS AND SOLUTIONS TO THE SELECTED QUESTIONS Hints / Answers to the selected questions (1) 5.816  10–3 mol of CaSO4, 11.6  10–3 mol of H2O, CaSO4.2H2O (2) C21H 30O 2 (3) 1.79  10–5 g boron. (4) V = 1.23  10-23 cm3, 1.432 A0. (5) 921.6 g (6) % abundance of C13 is 1.109% ; % abundance of C12 is 98.891%

6  6.023  1023 = 18.669 × 1023 2

molecules We know that, 1 molecule of hydrogen contains gas  2 atoms 23  18.069 ×10 molecules of hydrogen contain, 2 × 18.069 × 1023 atoms = 36.38 × 1023 atoms contains (b) 48 g of ozone  6.023 × 1023 molecules contains 96 g of ozone  x molecules

 x

30.

Mg - 27.7 %, P – 23.6 % & O – 48.7 % 617.93 (a) No (b) Yes (c) No (d) No (a) 3200 g, 32.65 g (b) 24 (a) 24.092  1023 atoms of oxygen; 112 g of CO (b) 5  1016 atoms. GAM (or) GMM of any substance contains Avogadro number of atoms / molecules i.e., 6.023 × 1023 atoms/ molecules. By applying the abhove relation to each of the given cases, we get, contains (a) 2 g of Hydrogen gas  6.023× 23 10 molecules contains 6 g of Hydrogen gas  x

 x

13.

64  6.023  1023 = 1.5 × 1023 256

molecules We know that, 1 molecule of S8 contains 8 sulphur atoms   1.5 × 1023 molecules of S8 contain, 8 × 1.5 × 1023 = 12 × 1023 atoms The problem is all about finding molecular mass. www.betoppers.com

9th Class Chemistry

344

6.023 × 1023 atoms or molecules of a substance weigh GAM or GMM. Let us apply the above relation to each of the given cases. a) Given, 3.0115 × 1022 molecules weigh 2.4 g weigh 6.023 × 1023 molecules   GMM weigh 3.0115 × 1022 molecules   2.4 g  GMM 

b)

x

2.4  6.023  1023 = 2.4 × 20 3.0115  10 22

19.92  10 18  6.023  10 23  120 105

x

16.

1 mole or GMM of any gaseous substance at S.T.P occupies 22.4 litres occupies a) 38 g of F2 at S.T.P    22.4 litres occupies 19 g of F2 at S.T.P  x

x

19  22.4  11.2litres 38

 19 g of F2 at S.T.P occupy 11.2 litres. b)

occupies 40 g of Argon at S.T.P    22.4 litres occupies 5 g of Argon at S.T.P    x litres

 x

22.4  5  2.8litres 40

 5 g of Argon at S.T.P occupy 2.33 litres. c)

occupies 17 g of NH3 at S.T.P    22.4 litres occupies 8.5 g of NH3 at S.T.P    x litres

x

8.5  22.4  11.2 litres 17

 8.5 g of Argon at S.T.P occupies 11.2 15.

litres. Given, Volume occupied by 10 grams of CO2 = Volume occupied by 10 grams of O3 Let us calculate of the volume occupied by 10 grams of CO2 gas. 1 mole or GMM of any gas at S.T.P occupies 22.4 litres.

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22.4  10  5.09 litres –––––––– (1) 44

Let us now calculate the volume occupied by 10 grams of O3 gas. occupies 48g of O3 at S.T.P    22.4 litres occupies 10 g of O3 at S.T.P   x

= 48 Given, 1 lakh molecules of the substance weigh 19.92 × 10–18 g weigh 6.023 × 1023 molecules   GMM 5 –18 weigh 10 molecules   19.92 × 10 g

 GMM  14.

Applying the above relation, we get occupies 44 g of CO2    22.4 litres. occupies 10 g of CO2    x

17.

22.4  10  4.67 litres –––––––– (2) 48

From (1) and (2), Volume occupied by 10 grams of CO2  volume occupied by 10 grams of O3 gas  The given statement is false. GMM of any gas at S.T.P occupies 22.4 litres. Applying the standard relation, we get, 64g of SO2 at S.T.P occupies 22.4 litres. ––––––––– (1) 80g of SO3 at S.T.P occupies 22.4 litres ––––––––– (2) From (1) and (2), we get, Volume occupied by the gram molecular mass of SO2 = Volume occupied by gram molecular mass of SO3  The given statement is true. 22.4 litre or 22400 ml of any gas at S.T.P weighs GMM. Let us apply the above relation to each of the given cases. i) Calculation of the mass of 5.6ml of O2: weigh 22400 ml of O2 at S.T.P   32 g weigh 5.6 ml of O2 at S.T.P   xg 5.6  32  0.01 g 22400 Calculation of mass of 2.24 litres of CH4: weigh 22.4 litres of CH4 at S.T.P   16 g

x

ii)

weigh 2.24 litres of CH4 at S.T.P   xg

x

2.24  16  1.6 g 22.4

iii) Calculation of mass of 112cc of brown gas: Brown gas  NO2 weigh 22.4 litres of NO2 at S.T.P   46 g

Mole Concept Solutions

345

 Total number of atoms of ‘C’ present in 22

weigh 0.112 litres of NO2 at S.T.P   xg

g of CO2

0.112  46 x  0.23 g 22.4

18.

We know that at STP 22400 ml of any gas weighs GMM  100 ml weighs

19.

21.

GMM  100 22400

 m  GMM Therefore, unless and until GMM of different gases is, the above statement is false. Given, Mass of the volatile compound at S.T.P. = m = 0.7 g Volume of the volatile compound at S.T.P. = V = 140 ml = 0.14 litre GMM = ? 22.4 litres of any gas at S.T.P. weighs ts gram molecular mass. Applying the standard relation, we get, weigh 22.4 litres at S.T.P.   GMM weigh 0.14 litres at S.T.P   0.7 g  GMM 

20.

= Total number of molecules of CO2(Nm) × 1 –––––– (1) Total number of atoms of ‘O’ present in 22 g of CO2 = Total number of molecules of CO2(Nm) × s2 –––––––– (2) Let us first calculate the number of molecules of CO2. Number of molecules of

CO 2 

GMM CO2

 N (where N = 6.023 ×

1023) Substituting the given values in the above equation we get, Number of molecules of 22  6.0213  1023  3.0115  1023 –– 44 –––––––– (3) Substituting the value of (3) in (1) and (2), we get, Total number of atoms of C = 3.0115 × 1023 × 1 = 3.0115 × 1023 Total number of atoms of O = 3.0115 × 1023 × 2 = 6.023 × 1023 23  22 g of CO2 contains 3.0115 × 10 atoms 23 of ‘C’ and 6.023 × 10 atoms of ‘O’. Mass of sugar (C12 H22 O11) = 342 g GMM of C12 H22 O11 = 342 g Total number of atoms of different elements present in 342g of sugar = ? We know that, Total number of atoms of each element = Total number of molecules × Number of atoms of each element present in 1 molecule of CO2. 1 molecule of sugar (C12 H22 O11) contains 12 ‘C’ atoms, 22 ‘H’ atoms and 11 ‘O’ atoms. Number of molecules of sugar (C12 H22 CO 2 

0.7  22.4  112 g 0.14

 Molecular mass of the volatile compound is 112. We know that, 6.023 × 1023 molecules of any occupies gas at S.T.P    22.4 litres.  Equal number of molecules of any gas at occupies S.TP.    Equal volumes.  In the given problem, both the gases CO2 and CH4 contain equal number of molecules (i.e., 1 billion), they occupy same volume at S.T.P  The given statement is True. Given, Mass of CO2 = 22 g GMM of CO2 = 44 g Number of atoms of each element present in 22 grams of CO2 = ? We know that, Total number of atoms of each element = Total number of molecules × Number of atoms of each element present in 1 molecule of CO2. 1 molecule of CO2 contains 1 atom of carbon and 2 atoms of oxygen.

m CO2

22.

O11) 

342  6.023  1023  6.023  1023 342

(Where N = 6.023 × 1023) –––– (1)  Total number of carbon atoms = 6.023 × 1023 × 12 = 7.22 × 1024  Total number of hydrogen atoms = 6.023 ×

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9th Class Chemistry

346

23.

1023 × 22 = 1.32 × 1025  Total number of oxygen atoms = 6.023 × 1023 × 11 = 6.625 × 1024 Given, Mass of CaCO3 = 50 g GMM of CaCO3 = 100 g Total number of protons, electrons, neutrons in 50 g of CaCO3 = ? We know that, Total number of electrons/protons/neutrons present in a given amount of substance = Total number of molecules (Nm) × Total number of electrons/protons/neutrons present in 1 molecule of that substance. 1 molecule of CaCO3 contains 50 electrons, 50 protons and 50 neutrons. i.e., in 1 molecule of CaCO3, No.of electrons = No. of protons = No. of neutrons = 50 In 50 g of CaCO3, total no.of electrons = total no.of protons = total no. of neutrons = x But, x = Total number of molecules present in 50 g of CaCO3(Nm) × 50 –––––– (1) Let us calculate the value of Nm 50  6.023  1023  3.0115  10 23 100 Substituting the above value in (1), we get, x = 3.0115 × 1023 × 50 = 15.0575 × 1024 Therefore, the total number of electrons = protons = neutrons = 15.05× 1024 Mass of king of chemicals (H2SO4) = 24.5 g Number of nucleons (protons + neutrons) = ? We know that, Total number of nucleons in 24.5 g of H2SO4 = Total number of molecules (Nm) × Number of nucleons present in 1 molecule of H2SO4 –––––––––––– (1) 1 molecule of H2SO4 contains 50 protons and 48 neutrons or 98 nucleons ––––– (2) Number of molecules in 24.5 g of H2SO4 = Nm 

24.

24.5  6.023  10 23  1.505  10 23 ––––– (3) 98 Substituting the values from (3) and (2) in (1), we get www.betoppers.com

25.

Total number of nucleons in 24.5 g of H2SO4 = 1.505 × 1023 × 98 = 1.474 × 1025. a) Calculation of the amount of SO2 present in 0.25 mole weigh 1 mole of SO2   64g weigh 0.25 moles of SO2  x  x

b)

0.25  64 1

Calculation of the number of molecules 1 mole of any substance contains Avogadro number of molecules, i.e., 6.023 × 1023 molecules Applying the relation, we get, contains 1 mole of SO2  6.023 × 1023 molecules contains 0.25 moles of SO2  x x

0.25  6.023  1023 1

 1.50575  1023 molecules.

c)



d)



e)

Calculation of number gram atoms of ‘S’ present in 0.25 moles of ‘SO2’: 1 mole of atoms = 1 gram atom 1 mole of SO2 contain 1 mole or 1 gram atom of S. 0.25 moles of SO2 contain 0.25 moles or 0.25 gram atoms of S. Calculation of number of gram atoms of ‘O’ 1 mole of atoms = 1 gram atom 1 mole of SO2 contain 2 moles or 2 gram atoms of S. 0.25 moles of SO2 contain 0.5 moles or 0.5 gram atoms of S. Calculation of the volume occupied SO2 at STP: 1 mole of any gas at S.T.P occupies 22.4 litres. Applying the above relation, we get, 1 mole of SO2 at S.T.P occupies 22.4 litres.

Mole Concept Solutions

347

occupies 0.25 moles of SO2 at S.T.P   x

0.25  22.4  5.6 litres. 1 Calculation of the total number of protons Total No.of protons = Number of protons in 1 molecule of SO2 × No. of molecules Number of protons in SO2 = 32 × 0.25 × 6.023 × 1023 = 48.184 × 1023 protons. Calculation of the amount of CO2: 6.023 × 1023 molecules of any substance weigh GMW. weigh 6.023 × 1023 molecules of CO2   44 g weigh 1.5 × 1022 molecules of CO2  x

x

f)



26.

a)



b)

1.5  1022  44 x  1.1g 6.023  1023 Calculation of the number of moles of CO2: 6.023 × 1023 molecules of any substance make 1 mole. 6.023 × 1023 molecules of CO2 weigh   1 mole. weigh 1.505 ×1022 molecules of CO2   x x

c)

C

d)

Let us find the mass ‘O’ present in 1.1g of CO 2. We know that, 44g of CO2 contains 32g of O  1.1g of CO2  ? contains

1.1  32  0.8g ––––––– (2) 44 Number of gram atoms of weight  o  0.8 O   0.05. GMM  o  16  x

e)



Calculation of the volume occupied by 1.5× 1022 molecules of CO2 of S.T.P: Avogadro number of molecules of any gas occupies at S.T.P    22.4 litres.

 6.023 × 1023 molecules of CO2 at occupies S.T.P    22.4 litres 22 1.5 × 10 molecules of CO2 at S.T.P occupies    x



1.1  12 12 3    0.3 g 44 40 10 Number of gram atoms of

Given mass ––––––––– (1) Gram atomic mass

We know that, from the above calculation, 1.5 × 1022 molecules of CO2 mass 1.1g.

Calculation of number of gram atoms of C: We know that, number of gram atoms

 x

Number of gram atoms of O = ? Number of gram atoms 

1.5  1022  1  0.025 moles 6.023  1023

Given mass Gram atomic mass Let us find the mass of ‘C’ present in 1.5 × 1022 molecules of CO2. The mass of CO2 contains the above number of molecules i.e., 1.1 g. We know that, 44g of CO2 contain 12g of C  1.1g of CO2 contain xg of C

0.3  0.025 12

1.5  10 22  22.4  0.56 litres 6.023  10 23 Calculation of total number of protons Total number of protons = Number of molecules of CO2 × Number of protons present in CO2 1 molecule of CO2  contain 22 protons Total number of protons = 1.5 × 1022 × 22 = 33.12 × 1022.  x 

f)

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9th Class Chemistry

348

27.

Given, Mass of substance = ‘x’ g Let us assume, gram molecular mass of the substance (GMM) = W g Number of atoms in one molecule of that substance = K We know that, total number of atoms in a substance = Total number of molecules (Nm) × Number of atoms in each molecule = Nm × K –––––– (1) Let us first calculate the total number of molecules. Number of molecules  N m  

m  N –––– GMM

––– (2) Where ‘N’ is Avogadro number with a value of 6.023 × 1023 Substituting the given values in the above equation, we get, x 23 Number of molecules  N m    6.023  10 M

––––––––– (3) Substituting the value of equation (3) in equation (1), we get, Total number of atoms in ‘x’ grams of a compound  28.

Given, Volume of H2S = 2.8 litres Mass of H2S = ? 22.4 litres of any gas at S.T.P weighs GMM Weighs 22.4 litres of H2S gas at S.T.P   34g Weighs 2.8 litres of H2S gas at S.T.P   xg  x

29.

x  6.023  1023  K. M

.8  34  4.24 g 22.4

 2.8 litres of H2S at S.T.P weighs 4.25 g.

Given, Mass of SO2 = 8 g Volume of SO2 = ? occupies 64 g of SO2 at S.T.P    22.4 litres. occupies 8 g of SO2 at S.T.P   x 22.4  8  2.8 litres 64 8 g of SO2 at S.T.P occupy 2.8 litres. x

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30.

Given, Volume of H2SO4 = 2 Mass of H2SO4 = ? 1 mole of any substance weighs GMW. Weighs 1 mole of H2SO4   98g Weighs 2 mole of H2SO4  x x

31.

98  2  196g 1

 2 moles of H2SO4 weigh 196 g. Given, Mass of sodium = 0.23 g Number of moles = ? 1 mole of any substance weighs its GAM. Weighs 1 mole of sodium   23g Weighs x mole   0.23 g

0.23  1  0.01 moles 23  0.23 g of sodium makes 0.01 mole. x

IIT JEE WORKSHEET KEY 1) (D) 2) (A) 3) (C) 4) (C) 5) (B) 6) (A) 7) (C) 8) (C) 9) (B) 10) (C) 11) (D) 12) (A) 13) (B) 14) 4.159g 15) Pressure and temperature are same 16) 6.02  1024 17) (C) 18) (B) 19) (A) 20) (D) 21) (D) 22) (D) 23) 7.525  10 22 24) (C) 25) 0.5 26) 5 27) C12 28) (D) 29) (C) 30) (A) 

8. SOLUTIONS SOLUTIONS them is that clouds are formed in the upper

FORM ATIVE WORKSHEET

1. 2.

3.

HINTS / ANSWERS TO SELECTED QUESTIONS (C) a) – v ; b) – iv; c) – i ; d) – ix ; e) – ii ; f) – viii ; g) – vii ; h) – vi ; i) – iii A solution is a homogeneous mixture of two or

atmosphere while fog gets formed in the region 11. 12.

more substances. A homogeneous mixture means that the mixture is just the same throughout. A colloid is a kind of solution in which the size of solute particles is intermediate between those in true solutions and those in suspensions. 4.

True solutions: Salt solution, sugar solution;

13. 14.

Colloidal solutions: Ink, Starch solution, Blood 5.

The path of the light beam and the colloidal particles become visible.

6.

Soap solution. This is because the particles of solute are bigger in soap solution than in sugar solution.

7.

Starch solution shows the Tyndall effect as it is a colloidal solution.

8.

The decrease in density of solution – B o n addition of solute indicates that, some of the solute particles present in solution got precipitated. This indicates that solution is supersaturated. As the density of solution – C remains unchanged on addition of solute, solution cannot dissolve anymore solute. Hence, it is saturated. (C) The solubility of a substance at a given temperature is defined as the quantity of that substance, that can be dissolved in 100 grams of the solvent, at that temperature, to get a saturated solution. Hence to know the solubility we should know the amount of solute in 100g of solvent. 1st method: Given: 20g of H2O contains 5g of the salt. 100g of H2O contains

Solutions: Brine ; Suspensions: Chalk water

5  100  25g of the 20

salt Solubility of the salt at 600C is 25. 2nd method: Given: Amount of solute = w1 = 5g Amount of water (solvent) = w2 = 20g Solubility = S = ?

mixture, Muddy river water, Shaving cream Colloidal solution: Soda-water, Milk, Smoke in air. 9.

close to earth. (B) The increases in density of solution - A on addition of solute indicates that, it has dissolved the solute added to it. Therefore, solution – A is unsaturated.

In dirty clothes, the dust particles are present on oil drops sticking to them. Simple water cannot remove these oil drops from the clothes because

We know that, Solubility (S) 

emulsion. Soap plays the role of emulsifier and

––– (1) Substituting the above values, in equation (1), we

helps in forming a stable emulsion between the two. This means that soap helps in removing these dirty clothes get washed by soap solution. Both fog and cloud are the examples in which liquid is the dispersed phase and gas (air) is the dispersion medium. The only difference between

5  100  25 20 Weight of solution = 100g Solid left over on evaporation is 50g get S 

oil drops along with the dust sticking to them. The 10.

wsolute  100 –– w solvent

water and oil as such do not form a stable

15.

wsolute  100 ––––– (1) Solubility (S)  w solvent

9th Class Chemistry

350 How to get weight of solute and solvent? Given 50g of solid is left over on evaporation  Amount of solute = 50g ––––– (2) We also know that, wsolution  wsolute + wsolvent  100 = 50 + wsolvent  wsolvent = 50g ––––– (3) Substituting (2) and (3) in (1), we get

50  100  100 50 Solubility = S = x Weight of solution = Wsolution = ? We know that Wsolution = Wsolute (w1) + Wsolvent (w2 )  Wsolution = w1 + w2 ––––– (1) How to get weight of solvent (w2)? We know that, Solubility

w1 We know that solubility (S)  w  100 2

 w2 = Amount of water

Solubility (S)  16.

S

w1  100  w 2  100w1 ––––– (2) w2 S

Substituting the above values in equation (2), we get

100y ––––– (3) x Substituting the value from equation (3) in (1), we get Amount of solute  w 2 

Wsolution  y 

17.

20.

21. 22. 23. 24. 25. 26.

100y xy  100y  x x

 y(100  x)   Wsolution    gm x  Therefore, the weight of

w solute  100 We know that solubility = S  w solvent –––––––(1)

solution is

 y(100  x)    gm x Solubility = S = K Amount of solute (w1) = ‘L’ g Amount of solvent (w2) = ? w1 We know that solubility  S  w  100 2

10  100  S  10 100 Therefore, solubility is 10. We know percentage by weight = w % S

Weight of solute  w1  w1   100 = Weight of solution  w1  w 2  w1  w 2

18. 19.

100w1  100  L   g S  K 

(C) Solubility (S) = 40 Amount of solute = w1 = 10g Amount of water = Amount of solvent = w2 = ?

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10 × 100 = 9.09 110 Therefore, percentage by weight of the solution = 9.09. Before dilution After dilution Initial weight % = w1% = 30% Final weight % = w2% = 6% Mass of initial solution = m1 = 200 g Mass of final solution m2 = m Mass of water added = m2 – m1 = m – 200 How to get m? Percentage by weight =

 w2 = weight of solvent 

100 w1 100  10   25g S 40 Therefore the amount of water requires to dissolve 10g of solute is 25g. From the above information it is clear that : Solute –A forms an exothermic solution. Therefore, its solubility decreases with increase in temperature. Solute – B forms an endothermic solution. Therefore, its solubility increases with increase in temperature. no effect.(in the case of solute – C) (C) (D) (A) (C) (B), (C), (D) Weight of the solute (NaOH) = wsolute = w1 = 10 g Weight of the solvent (H2O) = wsolvent = w2 = 100 g Solubility (S) = ? Percentage by weight = ? 

27.

Solutions

351

w solute We know, w% = w × 100  w% solution

29.

1



Vsolute We know, v / v% = V × 100 solution

w solution

Note: Before and after dilution, the mass of the solute remains constant.

 %w 

1 w solution



 20

w1 % m2  w 2 % m1

30 m2   m2 = 1000 g 6 200 Mass of water added = 1000 – 200 = 800  g. Mass of the solvent = wsolvent = 300 g Mass of the solute = wsolute = A = ? Percentage by weight = 10 We know that percentage by = w % =

30.

w solute  100 w solution 

70  100 x  70

 20 (x + 70) = 7000

5600  280 ml 20 Therefore, the volume of solvent required to prepare 20% (v/v) solution is 280 ml. Weight of the solute = 10 g Weight of the solvent = 90 g Density of the solution = 2 g/ml Weight volume percentage = ? The total weight of the solution = weight of the solute + weight of the solvent = 10 + 90 = 100 g  The volume of the solution = 20x = 5600  x =



28.

Volume percentage = v/v% = 20; Volume of H2SO4 =Vsolute= 70ml Volume of solvent = Vsolvent = x = ?

Mass of the solution 100   50ml. density of the solution 2

w solute  100 ––––––––––(1) w solute  w solvent

We know, weight – volume percentage = Substituting the above values in equation (1), we get

Weight of solute  100 Volume of solution By substituting the values, we get,

w solute 10   100 w solute  300

28.

10  100 = 20 50 Therefore, the weight – volume % of the solution is equal to 20 g/ml. Weight – volume % =

On simplifying, we get mass of solute = w solute = 33.33 grams Therefore, the mass of the solute = 33.33 grams. Solubility is S  In the saturated solution, ‘S’ g of solute is present in 100 g of solvent. w solute  w1  S g wsolvent = w2 = 100 g We know that percentage by weight





Weight of solute  100 Weight of solution

Weight of solute   100 Weight of solute  Weight of solvent

CONCEPTIVE WORKSHEET 1. 3. 4.

HINTS / ANSWERS TO SELECTED QUESTIONS All 2. (C) (A) A suspension is a heterogeneous mixture in which the small particles of a solid are spread throughout a liquid without dissolving in it. Examples: Chalk-water mixture, Muddy water, Milk of magnesia, Sand particles suspended in water, and Flour in water.



w1 S  100   100 w1  w 2  S  100  www.betoppers.com

9th Class Chemistry

352 5.

6.

A colloid is a kind of solution in which the size

7.

A colloid is a kind of solution in which the size

of solute particles is intermediate between those

of solute particles is intermediate between those

in true solutions and those in suspensions. Some

in true solutions and those in suspensions. Some

of the examples of colloids (or colloidal

of the examples of colloids (or colloidal

solutions) are : Soap solution, Starch solution,

solutions) are : Soap solution, Starch solution,

Milk, Ink, Blood, Jelly and Solutions of synthetic

Milk, Ink, Blood, Jelly and Solutions of synthetic

detergents.

detergents. A suspension is a heterogeneous

In a colloidal solution, the particle size is such (1

mixture in which the small particles of a solid

nm to 100 nm), that these particles scatter the

are spread throughout a liquid without dissolving

light rays as they fail on them. Because of

in it. Examples: Chalk-water mixture, Muddy

scattering, the path of the light as well as the

water, Milk of magnesia, Sand particles

particles become visible. But in a true solution, the particle size so small (less than lnm) that these particles are not in a position to scatter the light. Therefore, true solution does not show any

8. 9. 10. 11.

suspended in water, and Flour in water. (A) (B) (C), (D) (C)

Tyndall effect. 12.

Method–I

Method–II

15g of water contains 5g of a substance. 100g of water contains

5  100  33.4 g of 15

salt.

Weight of solute (w1 ) = 5g Weight of solvent (w2) = 15g Solubility (S) = ? We know

0

Solubility of the substance at 35 C is 33.4

Solubility (S) 

w1 5 100   100 w2 15

= 33.34 13.

Solubility (S) = 50 Amount of salt (solute) = w1 = 90g Amount of water (solvent) = w2 = ? We know that,

14.

Solubility = S = 40 Amount of solute (w1) = ? Amount of water = amount of solvent (w2) = 50g

w1 Solubility (S)  w  100 –––––– (1) 2

w1 We know that solubility  S  w  100 2

Substituting the above values in equation (1), we get Weight of water (solvent)

S  w2 ––––– (1) 100 Substituting the above values in equation (1), we get Amount of solute

100w1 100  90   180g. S 50 Therefore, the weight of water required to prepare the solution is 180g.  w2 

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 Amount of solute w1 

40  50  5  4  20 100 The amount of solute required to prepare the saturated solution is 20g.  w1 

Solutions 15.

353 Weight of solvent (w2) = 100 g and also Weight of solution = Weight of solute (w1) + weight of solvent (w2) = (100 + S) g

30g of solution on evaporation gave 5g of solute  Amount of solution = 30g Amount of solute (w1) = 5g and amount of solvent (w2) = 30 – 5 = 25g Solubility (S) = ? We know that, S 

w1 We know that, % by weight  w  w  100 1 2

w1  100 –––––– (1) w2

S  100 100  S –––––––––(2) Substituting (2) and (1), we get  % by weight 

Substituting the above values in the equation (1),

5  100  20 25 Therefore, the solubility of solute is 20. Solubility curves : The graphs which indicate the effect of temperature on solubility of a solute are called solubility curves. Utility of solubility curves : (a) The solubility of a given salt at any temperature can be read directly from its solubility curve. (b) The nature of curve gives an idea about the effect of temperature on the change of solubility. A steep curve indicates a rapid change, while a flat one indicates rather a slow change in solubility with rise of temperature. (c) Substances whose solubility decreases with rise in temperature are : NaOH, Na2CO3, Na2SO4, etc,. (d) Substances whose solubility increases with rise in temperature are: NaNO3, KNO3, KCl, NH4Cl, etc,. (e) Substances whose solubility fairly increases or no change with rise in temperature are: NaCl, Li2SO4. Increase of temperature, increases the volume of a gas resulting in escape of a gas molecules from the solution. This decreases the solubility. False Due to the formation of hydrogen bonding. (D) (C), (D) we get S 

16.

17.

18. 19. 20. 22. 23. 24. 25. 26.

(C) (C) (A), (B), (C), (D) Given percentage by weight =

1 (Solubility) — 2

——— (1) Let ‘S’ be the solubility of the solute.  S g of solute is present in 100 g of the solvent  Weight of solute (w1) = ‘S’ g and

27.

S 1 100 1  100 S   100  S 2 100  S 2  200  S  S  100 The volume of solute = 50 ml; The volume of solvent = 500 ml Volume of the solution = 500 + 50 = 550 ml The percentage by volume = v/v % = ? We know,

Vsolute 50  100   100  9.09 v/v%  V 550 solution 28.

29.

So, volume percentage of solution is 9.09. Weight – volume percentage of solution = 15 Density of solution = 1.06 g/ml Weight of solution = ? Weight of solute = ? We know that, 15% (w/v)  15g of solute is present in 100 ml of solution. a) Weight of the solution = (Volume × density)solution = 100 × 1.06 = 106g b) Weight of solvent = Total weight of solution – weight of solute = 106 – 15 = 91g Therefore, the solution contains 15 g of solute and 91 g of solvent. Aqueous solution of glucose = 10%(w/v)  10 g of glucose is present in 100ml of solution Volume of solution containing 1 gram mole of glucose = ? Weight of glucose (w) = 1 gram mole = Gram molecular weight of glucose expressed in grams = 180 g (72 +12 + 96 = 180) We know that percentage by volume =

w  100 x 100 V 180 180  100  V   100  1800 ml V 10 Therefore, the volume of solution in which 1 g mole of glucose dissolved, will be 1800 ml. www.betoppers.com  10 

9th Class Chemistry

354 30.

Weight of NaCl = wNaCl = x? Weight of water = wwater = 54 g Weight percentage =10

2.

w(COOH)2  2.2g

Volume of solution = V = 500 ml Density of solution = d = 1.1 g/ml Weight of solution = wsolution = 500 × 1.1 = 550 g Weight percentage = w% =?

w  w%  solute  100 w solution  w% 

w solute  100 w solute  w solvent

x  100  x  6g x  54 Therefore, the weight of NaCl is 6 g.

w% 

 10 

HINTS / ANSWERS TO SELECTED QUESTIONS Weight percentage of sugar solution = w% = 18; Weight of solution = 180 g Let the weight of glucose and water be ‘x’ and ‘y’ respectively.

w% 

(or)

18  180  32.4g 100 But, weight of water = weight of solution – weight of sugar  y = 180 – x = 180 – 32.4 = 147.6 g Therefore, 32.4 g of glucose and 147.6 g of water are present in the solution. x

w solute  100 w solution

 w% 

w solute  100 w solute  w solvent

15 1500  100   20 15  60 75 Therefore, the weight % of the resulting solution is 20. Given, solubility = 50  50 g of solute is present in 100 g of the solvent.  Weight of solute (w1) = 50 g and weight of solvent (w2) = 100 g and also Weight of solution = Weight of solute (w1) + Weight of solvent (w2) = 100 + 50 = 150g We know that, Weight % 

w solute  100 We know, w%  w solution x  100 180

Weight of K2CO3  w K 2CO3  15g Weight of water = wwater = 60 g Weight percentage = w% = ?

SUMMATIVE WORKSHEET

 18 

w solute 2.2  100  w%   100  0.4. w solution 550

Therefore, weight percent of oxalic acid is 0.4. 3.

1.

Weight of oxalic acid, (COOH)2 =

4.



w solute 50  100   100  33.33 w solution 150

Therefore, the weight percentage of the solution is 33.33. 5.

Case–I (At 800C) Solubility at 800C = S1 = 50

Case–II (At 300C) Solubility of 300C = S2 = 30

Amount of solute dissolved in solution at 800C = m1 = 50 g

Amount of solute dissolved in solution at 300C = m2 = ?

Amount of solvent at 800C = Amount of solvent at 300C = 50g = Constant Weight of solute that comes out = Weight of solute left undissolved (m) = Initial amount of dissolved solute in solution (m1) – Final amount of dissolved solute in solution (m2) www.betoppers.com

Solutions

355

 mleft = m1 – m2 –––––– (1); How to get m2? We know that, Solubility (S) 

9.

Amount of solute  100 Amount of solvent



 S  Amount of solute ( amount of solvent is same in both the cases)



S1 m1 S 30   m2  2  m1   50  30g S2 m 2 S1 50

10.

Volume of solute 100 Volume of solution Here, Volume of solute (alcohol) = 50 mL And, Volume of solvent (water) = 150 mL So, Volume of solution = Volume of solute + Volume of solvent = 50 + 150 = 200 mL Now, putting these values of ‘volume of solute’ and ‘volume of solution’ in the above formula we get:

Solubility = S = 25 Weight of solute = w1 = 10g Weight of solvent (w2) = ?

w1 We know that solubility (S)  w  100 2

Concentration of solution 

100 w1 S –––– (1) Substituting the above values in equation (1), we get  Weight of solvent

 Weight of solvent  w 2 

100  10  40 gm 25 Therefore, weight of solvent that dissolve 10g of solute is 40g. Here, Mass of solute (sugar) = 30 g And, Mass of solvent (water) = 370 g So, Mass of solution = Mass of solute + Mass of solvent = 30 + 370 = 400 g

8.

Concentration of solution =

50 100 200

50 2 = 25 per cent (by volume) Thus, the concentration of this alcohol solution is 25 per cent or that it is a 25% alcohol solution (by volume) This solution also contains a liquid solute (acetone) mixed with a liquid solvent water (because it is an aqueous solution), so we have to calculate the concentration of this solution in terms of volume percentage of solute (acetone) Here, Volume of solute (acetone) = 2 mL And, Volume of solution = 45 mL Concentration of solution 

11.

30 30 100 = = 400 4

7.5 percent (or 7.5 %) Thus, the concentration of this sugar solution is 7.5 per cent (or that it is a 7.5% sugar solution)

110 100  100  = 20 per cent (or 20%) 550 5 Thus, the concentration of this salt solution is 20 per cent (or it is a 20% salt solution) Concentration of solution 

(A)

 w2 

Mass of solute  100 Mass of solution



 At 300C only 30g of solid is the dissolved solute. Substituting the values of m1 and m2 in (1), we get  50 – 30 = 20g is in undissolved solute Therefore the amount of solute that comes out = 20g 6. 7.

Here Mass of solute (salt) = 110 g And, Mass of solution= 550 g Concentration of solution

12.



Volume of solute 100 Volume of solution



2 200  100  = 4.4 per cent 45 45

Hint: Solubility =

wt.of solute ×100 wt.of solvent

[Ans: 19 gram ]

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9th Class Chemistry

356 13.

14.

Hint:Solubility is the amount of solute dissolved in 100g solvent at a given temperature.

weight of solute ×100 Hint: % wt = weight of solution

7. 8.

[Ans: 20 ] 15.

Hint: % wt =

wt. of solute ×100 wt. of solution

[ Ans: 23.33]

HOTS WORKSHEET

9.

HINTS / ANSWERS TO SELECTED QUESTIONS 1.

2. 3.

4.

5. 6.

Substances whose solubility decreases with rise in temperature are : NaOH, Na2CO3, Na2SO4, etc,. Substances whose solubility increases with rise in temperature are: NaNO3, KNO3, KCl, NH4Cl, etc,. Substances whose solubility fairly increases or no change with rise in temperature are: NaCl, Li2SO4. P is unsaturated and Q is super saturated solution. A saturated solution on heating becomes unsaturated and on cooling become super saturated solution. (a) Petrol, kerosene or ammonia solution (b) Turpentine (c) borax solution Refer Novel points The amount of heat change during the formation of a solution depends mainly on two factors: Lattice energy: It is the amount of heat required to separate a mole of the ionic substances into its component positive and negative ions. This is an endothermic process. Heat of hydration: These ions get hydrated. The ions hold the water molecules by ion dipole bonds. Heat is liberated during the process of hydration. Thus it is an exothermic process. (a) If lattice energy > hydration energy, the system cools down.

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(b) If lattice energy < hydration energy, the system heats up. Solution - I is endothermic in nature and solution - II is exothermic in nature. Graph I indicates that solubility increases rapidly with increase of temperature, Graph-II indicates that there is slight increases in solubility due to increase of temperature and graph - III indicates that there is no change in solubility with change in temperature. Graph - I – KNO3, Pb (NO3)2 Graph - II - KCl Graph -III - NaCl

wt.of solute ×100 wt.of solvent

10.

Solubility =

11.

[Ans: 19 gram ] Solubility is the amount of solute dissolved in 100g solvent at a given temperature.

12.

Solubility =

wt.of solute ×100 [Ans: 20 ] wt.of solvent

13.

Solubility =

wt.of solute ×100 [Ans: 20 g] wt.of solvent

14.

This due to escaping away of the solute molecules from the solution. For NaCl, there is no change and for CaSO4 it increases.

15.

IIT JEE WORKSHEET HINTS / ANSWERS TO SELECTED QUESTIONS (1) (D) (3) (A) (5) (C) (7) (B) (9) no change

(2) (A), (B) (4) (D) (6) Henry’s law (8) Solubility (10) Gas in water



9. COMPOUNDS OF CARBON SOLUTIONS

FORMATIVE WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS 1. 2.

3. 4. 5.

Statements ‘A’ and ‘B’ are correct. The correct answer is (B) Both assertion and reason are correct and reason is the correct explanation of assertion. The correct answer is (A) (1) – q (2)– s (3)– p (4) – r The correct answer is (D) Distance between ‘C–C’ bonds in diamond is more. The correct answer is (C) 3500ºC SiO2 + 3C (s)   SiC(s) + 2CO (g)

3500ºC SiC(s)   Si + C (graphite) The correct answer is (C) 6. ‘A’ and ‘C’ are allotropes. 7. Graphite can withstand very high temperatures. So, ‘A’ is false, ‘B’ is true. The correct answer is (B) 8. The correct answer is (B) 9. Diamond is inert, hard and stable. The correct answer is (D) 10. Both assertion and reason are correct and reason is the correct explanation of assertion. THe correct answer is (A). 11. Carbon monoxide was first prepared by Lasson

in 1776, by heating zinc oxide with wood charcoal. The correct answer is (A)

 formic acid 

conc

0

100 C   CO  g   H 2 SO4 .H 2 O  aq 

(ii)

 COOH 2  s   H 2 SO4

  H 2 SO4 .H 2 O  aq   CO2  g   CO  g 

C  s   CO 2  g    2CO  g   t

(iii)  Red hot 

So, A = CO, B = CO2, C = CO

The correct answer is (B) 15. It is collected by downward displacement of water. The correct answer is (B) 16. All the given statements are correct. So, the correct answer is (D) 17. The most poisonous gas which causes death by affecting respiratory problems is CO. 18.

19.

in 1776, by heating zinc oxide with wood charcoal. Therfore, CO is prepared reduction of a metallic oxide. The correct answer is (A) 12. It is very dangerous to sleep in a room where coal or wood is burning and the doors and the windows are closed. Owing to the limited supply of air in such a room, carbon monoxide is produced. Since the gas is colourless and has a barely detectable no smell, people sleeping in the room do not feel its presence and run the risk of CO poisoning. So, Both assertion and reason are correct and reason is the correct explanation of assertion. The correct answer is (A) 13. Carbon monoxide was first prepared by Lasson

HCOOH  H 2 SO4  aq 

14. (i)

20. 21.

22.

23.

C  s   CO 2  g    2CO  g   t  Red hot 

So, it is an endothermic reaction. The correct answer is (A). Carbon dioxide is formed and is absorbed by Caustic potash. The correct answer is (B) Assertion is correct and reason is incorrect. So, the correct answer is (C). SiO2 is a compound of oxygen and silicon, the two most abundant elements of earth’s crust and is used in building construction. 2 SiO +2C  Si+2CO (poisonous gas and a stable diatomic molecule). The correct answer is (B). Both assertion and reason are correct and reason is the correct explanation of assertion. The correct answer is (A). When a filter paper soaked in palladium chloride or platinum chloride is taken in the jar of carbon monoxide, first it turns pink, then green and finally black. A = Pink, B = Green, C = Green The correct answer is (B)

9th Class Chemistry

358 24. Carbon monoxide is obtained in the form of water gas and producer gas when air ( moist) is passed over a bed of white hot coke. The following reactions occur. 2C  O2  2CO C  O2  CO2

CuCl     CO  g   H 2 O   CuCl.CO.2 H 2 O  aq 

C  H2O  CO  H2  water gas 

25. 26.

27.

28.

29. 30.

The correct answer is (D) The correct answer is (C) The unconscious victim of carbon monoxide poisoning should be given artificial respiration with carbogen (a mixture of 95% oxygen and 5% carbon dioxide), till normal breathing is restored. The correct answer is (B) In the gas masks, the air is breathed in through hopcalite (a mixture of 50% MnO2, 30% CuO, 15% Co2O3 and 5% Ag2O), when carbon monoxide is oxidised to carbon dioxide. The correct answer is (A) Both assertion and reason are correct and reason is the correct explanation of assertion. The correct answer is A The correct answer is (C)   Na SO  H O  CO Na CO  H SO  2

3

2

4

2

4

2

The correct answer is A 38. CO2 dissolves forming carbonic acid.

CO2  H 2O  H 2CO3 CO remains undissolved and is collected. Recovery of CO2 from H2CO3 can be done by warming. The correct answer is (B). 39. Carbon monoxide gets oxidised to carbon dioxide. The correct answer is (B) 40.  COOH 2  s   H 2 SO4   H 2 SO4 .H 2 O  aq   CO2  g   CO  g  The correct answer is (B) 41. When burning magnesium ribbon is taken in the jar of carbon dioxide, it continues burning, because the heat of burning magnesium decomposes it to carbon and oxygen. The oxygen so liberated sustains the burning of magnesium,

2Mg  CO 2   2MgO  C

2

Ca  HCO 3 2  s  + 2HNO 3  aq   Ca NO3 2 +2H2O  +2CO2  g Δ

The correct answer is (B). 31. Carbon dioxide gas is neither combustible nor does it support combustion. It can also be poured out of a gas jar like a liquid as it is heavier than air. The correct answer is (B)   CaCO3  H 2O  CO 2 32. Ca  HCO3  2  The correct answer is (A) 33. Dry ice has nothing in common with ice formed from water, except that it is snow white. The correct answer is C. 34. Both are correct The correct answer is (C) 35. A = CO; B = CO2 The correct answer is (C) 36. CO – forms metal carbonyls, almost insoluble in water even at high pressure CO2 – forms metal carbonates, farly soluble in water www.betoppers.com

Dry ice – It is used as a refregerant. So, the right match is : p – (ii), (iv) ; q – (iii), (v); q – (i) The correct answer is (B) 37. Carbon monoxide dissolves in ammonical cuprous chloride solution forming addition compound.

The correct answer is (B). 42. All the given statements are true.

CONCEPTIVE WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS 1. 4. 7. 10. 12.

(B) (D) (A), (B) (C) (A)

2. (A), (C) 5. (B) 8. (B) 11. (C)

3. (A) 6. (A) 9. (A)

100 0 C HCOOH (aq) + H2SO4 (aq)   CO(g) + H2SO4 .H2O(aq). 13. (A) 14. (A) 15. (D) ]16. (D) 17. (C) 18. (B) Sun light CO (g) + C 2 (g)  

19. (D) 22. (D) 25. (C)

20. (A) 23. (C) 26. (C)

COC 2 (g)

 Carbonyl chloride or Phosgene 

21. (B) 24. (B) 27. (A)

Compounds of Carbon Solutions 28. (C) 31. (C) 34. (C) 36. (D)

29. (A) 32. (A) 35. (A)

359

30. (C) 33. (C)

A) CaCO3  s  + 2HCl  aq 

4.

   CaC  2  aq  + H 2 O    + CO2  g 

B) Mg  HCO3 2  aq 

5.

  MgCO3  s  + H 2 O    + CO2  g  

C) Fe2  CO3 3  s     Fe2 O3  s  + 3CO2  g 

37. (A)

6.

CO2  H2O  H2CO3  carbonic acid 

38. (B) Pure CO2 has a faint sour taste. 39. (C) The level of water increases showing that CO2 is soluble in water. One volume of water can dissolve one volume of CO2. 40. (C) 41. (B) 42. (C) 43. (A) 44. (C) 45. (A) 46. (A) 47. (A) 48. (C) 49. (C) 50. (B) 51. (A) 52. (B) 53. (C)

SUMMATIVE WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS

1.

2.

3.

Diamond is hardest because of a) Strength and uniformity of C-C bonds in diamond. b) Stable and rigid crystal lattice Graphite is soft because: parallel layers of carbon atoms in graphite are held by weak forces of attraction. Diamond: High density is due to compactness of crystal structure. ( distance between C-C bond is less). Graphite: low density is due to open crystal structure. ( distance between carbon atoms in high.) Diamond: In this unit, a carbon atom is linked to four other carbon atoms by covalent bonds ( of equal lengths).

7. 8. 9. 10. 11.

12.

13. 14.

15. 16.

17.

Graphite: In this unit a carbon atom is linked covalently to three other carbon atoms by covalent bonds in the same layer forming regular hexagons. Diamond is non conductor due to absence of free electrons in the crystal where as graphite has free mobile electrons in the crystal to conduct. Diamond is very inert, does not react with most chemicals . It is less prone to chemical attack due to its compact structure. Graphite is more prone to chemical attack due to its more open structure. Graphite lubricants are less affected due to its high M.P (35000C). Because of their strength and uniformity of C-C bonds, stable and rigid crystal lattice. Clay controls the hardness of the lead. Because it is a product of burning of carbon or its compounds in a limited supply of air or oxygen. Incomplete combustion of carbon fuels releases carbon monoxide. A pale blue flame is at the top because of following reaction: 2CO  O2  2CO2 . Where as in the middle following reaction takes place: CO2  C  2CO When excess CO2 is passed through KOH potassium bicarbonate is formed. It is completely soluble and hence leaves no precipitate. Because it is almost as heavy as air. Oxy haemoglobin dissociates releasing oxygen to tissues where as carboxyl - haemoglobin prevents haemoglobin from taking up oxygen. A person thus dies due to lack of oxygen. To prevent them from ‘CO’ poisoning. In both the mentioned types, the resultant solution contains sodium sulphate and sodium bicarbonate which being electrolytes themselves, may conduct electricity resulting in an electric shock, or may enhance short circuiting which may cause another fire. CO2 does not support combustion.

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9th Class Chemistry

360

Dry ice produces a temperature of -400c, lasts longer, freezes rapidly, sublimes and is non poisonous- Hence advantageous when compared to normal ice. 19. % of lead in lead pencil is zero. 20. It traps gases on its porous surface. 21 . 1-E; 2-A; 3-B; 4-F; 5-C; 6-D; 7-G; 22. 1) Coal 2) Lampblack 3 ) Coke 4) Coke 5) Lamp black 23. 1) C  O2  CO2   18.

2) C  2H 2  CH 4

4.

 3) C  2S   CS2

5.

0

1000 C 4) C  H 2O   CO  H 2

24.

5) C  2H2SO4  CO2  2H 2O  2SO2 1) b 2) c 3)b 4)b 5)c

25.

1. CuCO 3  C uO  CO 2

6.

 2. Ca(HCO3 ) 2   CaCO3  H 2O  CO 2

3. Na 2 CO3  H 2SO 4  Na 2SO 4  CO 2  H 2 O  4. Mg(HCO3 ) 2   MgCO 3  CO 2  H 2O or

 Ca(HCO 3 ) 2   CaCO 3  H 2O  CO 2 5

.

Ca(HCO3 ) 2  2HNO3  Ca(NO3 ) 2  2CO2  2H2 O

7.

8.

HOTS WORKSHEET

1.

HINTS/ANSWERS TO THE SELECTED QUESTONS When equal weights of allotropes in pure form are separately burnt and the carbon dioxide formed is absorbed in previously weighed potash bulbs(KOH), if the same quantity of K2CO3 is formed from each of the allotropes, it proves that all allotropes are chemically identical. C + O2  CO2

2. 3.

CO2 + 2KOH  K2CO3 + H2O Diamond can scratch iron, but the reverse is not true because iron is less harder than diamond. Diamond has a 3 dimensional network of tetrahedral units in which each carbon is strongly bonded

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9. 10.

11.

12.

to four adjacent carbon atoms. This produces a rigid three dimensional structure which makes diamond the hardest substance. Whereas, in graphite the flat parallel layers of carbon atoms are held by weak Van der Waal’s forces. Therefore they slide over one another on a slight application of forces which makes graphite soft. Wood charcoal is porous. These pores are filled with gases which make the wood charcoal to float on water. When the water is heated, gases present in the pores escape. As a result, the pores are occupied by water making it to sink in water. The ignition temperature of charcoal is very less. This makes charcoal to catch fire easily. Hence, it is used in gunpowder along with potassium nitrate (KNO3) and sulphur. Octane (C8H18) a carbon compound, present in petrol releases carbon monoxide (which is a poisonous gas), on incomplete combustion. This carbon monoxide is released as exhaust fumes from vehicles. Hence, it is hazardous to health. Activated charcoal has more adsorption capacity than wood charcoal. So, activated charcoal is preferred to wood charcoal. Reactants are able to penetrate between the hexagonal layers of carbon atoms in graphite. But it is not possible in diamond due to its three dimensional structure, in which the atoms are held firmly by strong covalent bonds. So, diamond is more inert than graphite. Diamond does not possess free electrons while graphite possesses free electrons. Graphite is composed of flat two dimensional sheets of carbon atoms. Weak vander Waal’s forces hold these layers together, and this makes graphite soft. The carbonmonoxide burns with a pale blue flame at the top of the charcoal oven for ming carbondioxide. 2CO + O2  2CO2 KOH leaves no precipitate whereas other alkalies such as NaOH form bicarbonates which are sparingly soluble and may get precipitated.

Compounds of Carbon Solutions 13.

14.

15.

It is almost-insoluble in water and almost as heavy as air. Coal burns in the presence of insufficient oxygen and forms CO. Carbon monoxide is 300 times more soluble in blood than oxygen. Thus, when it comes in contact with blood, it reacts with haemoglobin (red blood corpuscles). It binds itself with haemoglobin to form carboxy haemoglobin, a compound which cannot absorb oxygen. Thus, red blood corpuscles lose the capacity to carry oxygen to the cells and tissues Thus, the person dies for want of oxygen. This because both CO2 and SO2 turn lime water milky.

16.

Lead carbonate leaves a black residue i.e., PbO.

17.

Calcium carbonate reacts with dil. HCl to form CaCL2 which is a soluble chloride and hence do not precipitate in test tube to form hard solid which cannot be removed. While calcium sulphate formed by (CaCO3 + H2SO4) is hard and settles down at the bottom of the test as a precipitate that cannot be removed and may even crack the test tube.

18.

19.

20.

Mg reduces CO2 to carbon. 2Mg + CO2  2MgO + C and therefore continues to burn in CO 2 .

361 21.

Fe2O3 + 3C  2Fe + 3CO

22.

2H 2SO 4 + C  CO 2 + 2H 2O + 2SO 2 [conc.] 4HNO3 + CCO2 + 2H2O +4NO2

23.

 C  Na 2CO3  Na 2O  2CO

24.

 C + 2S   CS2

 Ca + 2C   CaC2

25. 26.

 Ni  s   4CO  g    Ni  CO 4   

 Nickel carbonyl 

27.

  CaCl  H O  CO CaCO 3  2HCl  2 2 2

28.

100 C HCOOH  H 2SO 4  aq    CO  g  

0

 formic acid 

conc

H 2SO 4 .H 2O  aq  29.

C 2 H 4  3O 2   2CO 2  2H 2 O  t

30.

(a) NaOH  CO 2   NaHCO 3

In case of dil.H2SO4 , CaSO4 is formed. This is deposited on the surface of marble and acts as a protective layer. This layer stops the further reaction to liberate CO2. When limited amount of carbon dioxide gas is assed through excess of KOH, it reacts to form potassium carbonate and water. The carbonate so formed is soluble in water. Hence, no milkiness is seen. Whereas when limited amount of carbon dioxide is passed through limeweater, it turns milky, on the account of the formation of insoluble calcium carbonate, which is white in colour.

1000C C + H2O   CO + H2

(b) 2NaOH  excess   CO 2  less    Na 2 CO 3  H 2 O 31.

C(A) = CO

B) = CO2

C) = H2O

D) = COCl2 E) = NH2CONH2 32.

conc.H 2SO4 A COOH  CO  CO 2  H 2O |

COOH Cl2 KOH  CO  CO 2 is absorbed



COCl 2

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362 33.

9th Class Chemistry  CaCO3  CaO + CO2

(A)

 CaO + C  CaC2 + CO

(B)

(C)

CaC2 + 2H2O  Ca(OH)2 + C2H2 34.

A) = CaC2O4;

B) = CaO ;

(D)

(E)

C) = CO ;

D) = CO2

IIT JEE WORKSHEET KEY

(1) (A) (5) (A) (9) (B) (13) (C) (17) (D)

(2) (B) (6) (C) (10) (B) (14) (D) (18) (A)

(3) (C) (7) (C) (11) (B) (15) (A) (19) (D) 

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(4) (C) (8) (B) (12) (C) (16) (C) (20) (A)

10. COMPOUNDS OF NITROGEN SOLUTIONS

1. 4. 7. 10. 13. 14.

FORMATIVE WORKSHEET

CONCEPTIVE WORKSHEET

HINTS/ANSWERS TO THE SELECTED QUESTONS

HINTS/ANSWERS TO THE SELECTED QUESTONS

(C) 2. (B) 3. (A) (C) 5. (C) 6. (C) (C) 8. (D) 9. (A, B, C, D) (B) 11. (C) 12. (C) (D) (B) Ammonia cannot be dried by conc. H2SO4, CaCl2 or P2O5. Since it reacts with them to given certain compounds. 2NH3  H2SO4   NH4 2 SO4

 Ammonium sulphate 

CaC 2  4NH3  CaC 2 .4NH3

 Additive compound 

P2O5  6NH3  3H2O  2  NH4 3 PO4

1. 4. 7. 8.

16. (B) Mg3N2 + 6H2O  3Mg(OH)2 + 2NH3 AlN + 3H2O  Al(OH)3 + NH3  17. (A) 18. (A) 19. (A) 20. (C) 21. (A) 22. (B) 23. (A) 24. (D) 25. (C) 26. (C)

3. 6.

(A, C) (C)

0

9. 12. 15. 18. 21. 22.

(A) (A) (B) (A) (C) (B)

10. 13. 16. 19.

(C) (B) (B) (C)

11. 14. 17. 20.

(B) (C) (B) (B)

2NH 3  H 2SO4    NH 4  2 SO4 23. 26. 29. 32. 35. 38. 41. 44. 47. 50.

NH 3  HNO3   NH 4 NO3 (C) 24. (A) 25. (D) 27. (A) 28. (B) 30. (B) 31. (B) 33. (D) 34. (A) 36. (A) 37. (D) 39. (D) 40. (A) 42. (A) 43. (D) 45. (A) 46. (A) 48. (C) 49. (A) 51. (B) 52.

(D) (C) (A) (B) (B) (B) (D) (A) (A) (B)

NH3  3C 2  NC 3  3HC

SUMMATIVE WORKSHEET

8NH3  3C 2  6NH4C  N2

HINTS/ANSWERS TO THE SELECTED QUESTONS

 A

 Excess 

27. 30. 33. 36. 39. 42. 45. 48.

(D) (D)

100 C CaC2  s   N 2  g    CaNCN  s   C  s 

15. (A) 2NO2  7H2  2NH3  4H2O

2. 5.

 NH 4 Cl  aq   NaNO 2  aq    NH 4 NO2  NaCl  aq 

 Ammonium phosphate 

2NO  5H2  2NH3  2H2O

(A) (C) (A) (A)

(A) (C) (C)

(B) (D) (A) (C) (B)

 Excess 

C

D

 B

28. 31. 34. 37. 40. 43. 46. 49.

(A) (B) (C) (A) (C) (A) (B) (A)

29. 32. 35. 38. 41. 44. 47. 50.

(D) (C) (A) (B) (A) (A) (A) (C)

1.

2. 3.

The complete apparatus is made up of glass. No rubber tube, bung or cork is used. HNO3 corrodes all these. The glass retort is glass stoppered.  KHSO4 + HNO3 KNO3 + H2SO4   2KNO3   2KNO2 + O2  . Pure O2 is obtained by this method.

9th Class Chemistry

364

4. (i)  2NH3  + CaCl2 + 2H2O 2NH4Cl + Ca(OH)2   NH4Cl (ii) NH3 + HCl   3Cu + N2 + 3H2O (iii) 3CuO + 2NH3   5Cu + 2NO + 3H2O 5CuO + 2NH3  5. CaCl2, P2O5, H2SO4 react with moist NH3, hence they are not used.  NH4HSO4 NH3 + H2SO4   CaCl2 . 8NH3 8NH3 + CaCl2   2(NH4)3 PO4 6NH3 + P2O5 + 3H2O   3MgO + 2NH3 6. Mg3N2 + 3H2O  7. Being lighter than air, the gas is collected by the downward displacement of air. 8. Because of its high solubility. 9. CaO  4Na2ZnO2 + 10. NaNO3 + 4Zn + 7NaOH  NH3 + 2H2O NH3 turns Red litmus blue being a basic gas. 0

20.

21. 22.

23.

24. 25. 26. 27.

0

750 C  900 C 11. 4NH3 + 5O2   4NO + 6H2O +  Pt

12. 13.

Finely divided iron powder. Urea (NH 2 CONH 2), ammonium sulphate 28. (NH4)2SO4.

14.

 Cu(OH) 2 + CuSO4 + 2NH3 + 2H2O  blue

15.

16.

17.

18. 19.

(NH2)2 SO4  [Cu(NH3)4](OH)2 Cu(OH)2 + 4NH3  (a) Excess NH3 reacts with Cl2 : 8NH3 +  N2 + 6NH4Cl 3Cl2  (b) Excess Cl2 reacts with NH3 : NH3 + 3Cl2   NCl3 + 3HCl Once heated, the catalyst maintains its temperature due to the heat released in the reaction, and no further heating is required.  Cu(NO3)2 + 2H2O (i) Cu + 4HNO3  + 2NO2  4Cu(NO3)2 + (ii) 4Cu + 10HNO3  5H2O + N2O KNO3 or NaNO3 with conc. H2SO4.  Cu(NO3)2 + 2H2O (i) Cu + 4HNO3  + 2NO2 0

170 C (ii) N 2 O . (NH 4 NO 3   N 2O + 2H2O)

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29.

30.

 2NH3 + (iii) (NH4)2SO4 + 2NaOH  Na2SO4 + 2H2O Fuming HNO3 contains dissolved oxides of nitrogen. It is prepared by distilling NaNO3 / KNO3 with conc. H2SO4 in the presence of starch. It is a mixture of conc. HCl and HNO3 in the ratio of 3 : 1. Mg and Mn.  Mg(NO3)2 + H2 Mg + 2HNO3   Mn(NO3)2 + H2 Mn + 2HNO3  NaNO3 and KNO3.  2NaNO2 + O2 2NaNO3   2KNO2 + O2 2KNO3  NO, NO2, N2O. HNO3 being powerful oxidising agent, oxidises H2 to H2O. N2O.  2NO N2 + O2   2NO2 2NO + O2   4HNO3 4NO2 + O2 + 2H2O   Ca(NO3)2 + H2O + CaCO3 + 2HNO3  CO 2  Mg(NO3)2 + H2O MgO + 2HNO3   NaNO3 + H2O + NaHCO3 + HNO3  CO 2  2H2O + 4NO2 + CO2 C + 4HNO3   H2SO4 + 6NO2 + 2H2O S + 6HNO3   Cu(NO3)2 + 2H2O + Cu + 4HNO3  2NO 2 FeSO4 . NO.

HOTS WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS

1.

2. 3.

A higher ratio by weight of the alkali may counteract the loss by sublimation of ammonium chloride on heating in the solid state. Quick lime being basic in nature does not react with basic ammonia gas. NH4NO3 is not used since it is explosive in nature and may itself decompose forming nitrous oxide and water vapour.

Compounds of Nitrogen Solutions

4.

5.

6.

7. 10. 11.

12.

13.

 CaSO4 + (a) (NH4)2SO4 + Ca(OH)2  2H2O + 2NH3.  Na2SO4 + (b) (NH4)2SO4 + 2NaOH  2H2O + 2NH3.  Al(OH)3 + NH3. (c) AlN + 3H2O   3Mg(OH) 2 + (d) Mg3N 2 + 6H2O  2NH3. A higher ratio of hydrogen or an increase in concentration of the reactants favours the forward reaction to give optimum yield of ammonia. The reactants should be dry, pure and free from impurities, since impurities tend to poison the catalyst and reduce its effectiveness. P 2O 5. 8. AlN. 9. CaCl2. Fe2O3.  NH4Cl. (a) NH3 + HCl   (NH4)2SO4. (b) 2NH3 + H2SO4   NH4NO3. (c) NH3 + HNO3   2NH4NO3 (a) Pb(NO3)2 + 2NH4OH  + Pb(OH)2   (b) Cr 2 (SO 4 ) 3 + 6NH 4OH 3(NH4)2SO4 + 2Cr(OH)3  3NH 4Cl + (c) FeCl 3 + 3NH 4OH  Fe(OH)3. (a)

15.

16.

2NO + O2   2NO2 2NO2 + H2O   HNO2 + HNO3 17. 18. 19. 20.

21.

22. 23. 24. 25. 26. 27.

Pt 8000 C

 4NO + 6H2O 4NH3 + SO2 

500 C 2NO + O2   2NO2. 28. (c) 4NO2 + 2H2O + O2  4HNO .  3 The quartz is acid resistant and packed in layers, 29. there by slowing down the movement of the gaseous NO2 entering from below and initiating better solvation of NO2 in water. A constant boiling mixture is one which boils 30. without change in composition. Hence on boiling, the above mixture evolves out the vapours of both acid and water in the same proportion as in the liquid. Thus dilute nitric acid cannot be concentrated beyond 68% by boiling. Lightening forms NO which is converted to NO2, HNO2 and HNO3 under atmospheric conditions.

(b) 14.

365

 3

Thus, NO is a source of nutrients to the plant. Thus, there are chances of better crops when lightning occurs. N2 + O2   2NO

CaOCl2 + 4NH3   3CaCl2 + 6H2O + 2N2. O2 . 4Mg + 2NO2   4MgO + N2. Hydrocarbons are oxidised to CO2 and H2O in the presence of an oxidation catalyst (Pt or Pd) and NO is reduced to N2 by reduction catalyst. Moist NH 3 can react with P 2O 5 forming phosphate and with conc. H2SO 4 forming sulphates. 6NH3 + P2O5 + 3H2O   2(NH4)3PO4 2NH3 + H2SO4   (NH4)2SO4 (A) : NO (B) : NO2 (C) : HNO2 (D) : HNO3 (E) : I2. (A) : NH4NO3 (B) : NH3 (C) : (NaOH3 + NaOH) (D) : N2O (E) : H2O Ca(HCO 3 ) 2 + 2NH 3   CaCO 3 + (NH4)2CO3. (A) : NH4NO2 (B) : NH4NO3 (C) : N2 (D) : AlN (E) : N2O There is formation of a protective oxide layer on the surface of the metal when it reacts with HNO3. Add KI solution. NO 2 would oxide I– to I2. NO 3 would give 'ring test'.  NH4NO3  NH 4 NO 2  N2O + 2H2O;    N2 + 2H2O. If NH4Cl or urea is added to a solution containing

NO 2 and NO 3 , NO 2 is decomposed as N2 and thus removed. (a) 2HNO3   H2O + 2NO + 3(O) (b)

2NO + O2   2NO2 4HNO3   4NO2 + O2 + 2H2O

IIT JEE WORKSHEET (1) (C) (4) (C) (6) (A) (9) (C) (11) (A) (14) (B)

KEY (2) (C) (5) (C) (7) (A) (10) (D) (12) (B) (15) (A)

(3) (B) (8) (C) (13) (B)

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366

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9th Class Chemistry

11. COMPOUNDS OF SULPHUR SOLUTIONS

FORMATIVE WORKSHEET HINTS/ANSWERS TO THE SELECTED QUESTONS 1. (A) 2. (A) 3. (C) 4. (D) 5. (A) 6. (D) 7. (C) 8. (A) 9. (C) 10. (D) 11. (C) 12. (D) 13. (A) 14. (D) 15. (C) 16. (C) 17. (C) 18. (A) 19. (A) 20.

i)

iii) 21.

43.

44.

It means a molecule of sulphurous acid in aqueous solution can furnish two hydrogen ions. V.D. of SO2 = Molecular wt.  2 = 64  2 = 32

38. (D) 42. (C)

39. (D)

40.(C)

(A)

(D) 16.

 444 C S  O2   SO2

A

 B

 444 C 2S 3O2   2SO3

A

i)

C

SO3  H2O  H2SO4

a) 2H2S+SO2  2H2O+3S b) 2Mg+SO2  2MgO+S iii) a) 2NaOH+SO2  Na 2SO3+ H2 O

C

b) SO2 + H2O  H2SO3 (B) 23. (A) 24. (D) 26. (C) 27. (D) 28. (B) (A) 30. (B)

D

45. 48. 51. 54. 56.

25. (D) 57.

S8  8O2  8SO2

ii)

 2SO2  O2   2SO3

iii)

SO3  H2SO4  H2S2O7

A

C 

 D

B

B

E 

 C

58. 59. 60.

 D

F 

E

 B

 F

47. (D) 50. (B) 53. (B)

220 5 44

(A) Number of moles



(B) i)

(D) 46. (B) (D) 49. (C) (D) 52. (D) (D) 55. (B) (D) The gas liberated is CO2

 Number of moles =

(C)

16.

 D

H 2SO 4  Na 2SO3   Na 2SO4  SO2  2H 2O

4FeS2  11O2  2Fe2O3  8SO2 32.

36. (C)

  x1  x 2  :  x 3  x 4   2 : 2  1:1

b) 2HNO2+SO2  2NO2+H2SO4 ii)

29. 31.

35. (C)

 x1  x 2  x 3  x 4  1

a) Cl2 + 2H2O+SO2  H2SO4 + 2HCl

22.

34. (D)

Na 2S  H2SO4  Na 2SO4  H2S

2H2S+SO2(solution)  2H2O + 3S 3Mg + SO2  2MgO+MgS

ii)

(C) (D) 41. (C) 33. 37.

wt 1  Number of moles  GMW GMW

As GMW of CO2 is least, it posses maximum number of moles. (D) (C) (B)

9th Class Chemistry

368

(iii) KHSO3+HCl(dil)  KCl + H2O + SO2

CONCEPTIVE WORKSHEET

(iv) S+O2  SO2(main reaction)

KEY 1

2

3

4

5

6

7

8

9

10

C

B

A

D

D

D

B

B

C

C

3.

11 12 13 14 15 16 17 18 19 20 A

C

C

A

B

D

B

C

B

B

21 22 23 24 25 26 27 28 29 30 A

C

C

B

B

D

D

A

B

A

31 32 33 34 35 36 37 38 39 40 4.

C

C

C

B

C

B

D

C

C

C

41 42 43 45 46 47 48 49 50 51 C

B

C

A

C

D

B

C

D

A

5.

CaO+H2O  Ca(OH)2

52 53 54 55 56 57 58 59 60 61 B

B

B

C

B

C

C

A

D

B

Ca(OH)2+SO2  CaSO3+H2 O 6.

62 63 64 65 66 67 68 69 70 71 D

A

D

B

A

C

C

D

C

7.

72 A

2.

burning (i) 4FeS2+11O2    2Fe2O3 + 8SO2 heat (ii) Cu+2H2SO4   CuSO4 + 2H2O + SO2

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Sulphur on burning in air forms sulphur dioxide and traces of sulphur trioxide. Sulphur trioxide dissolves in the water vapour present in air to form sulphuric acid mist which appears in the form of fumes. 2S + 3O2  2SO3 SO 3 + H2O

SUMMATIVE WORKSHEET

1.

Sulphur dioxide is called sulphurous anhydride because it can be obtained by the loss of a water molecule from a molecule of sulphurous acid. H2 SO3  H2O+SO2 

B

HINTS/ANSWERS TO THE SELECTED QUESTONS (B)

2S+3O2  2SO2(minor reaction) Both bleach animal and vegetable colouring matter when moist. Differences : i) Bleaching in case of chlorine is due to oxidation, whereas in case of sulphur dioxide is due to reduction. ii) Bleaching done by chlorine is permanent in nature, whereas that of sulphur dioxide in temporary in nature. iii) Bleaching done by chlorine damages delicate fibres like wool and silk, but bleaching by sulphur dioxide does not cause damage. The sulphur dioxide brings about bleaching by reduction, i.e., it removes oxygen atoms. The lost oxygen atoms are regained from atmosphere, over the passage of time, and hence the bleached clothes tend to regain colour. This is because SO2 reacts with calcium oxide.

(water

vapour)

 H2SO4(sulphuric acid mist)

8.

(a) It is because sulphur trioxide dissolves in water producing a large amount of heat which changes sulphuric acid to sulphuric acid mist. The latter is very difficult to liquefy.

Compounds of Sulphur Solutions

9.

10.

369

(b) Vanadium pentoxide does not get poisoned as easily as compared to platinised asbestos, if the mixture of sulphur dioxide and oxygen is slightly impure. (c) The reaction between sulphur dioxide and oxygen is exothermic in nature. Thus, once the reaction starts, it proceeds with its own heat. a) It is because the catalyst gets poisoned in the presence of impurities like As2O3, H2S and dust particles. Thus, oxidation of SO2 gas does not take place. b) i) 2NH4Cl + Ca(OH)2  CaCl2+2H2O + 2NH3 ii) CaOCl2 + 2HCl  CaCl2+H2O + Cl2 17. When sulphuric acid is diluted with water it ionizes almost completely into hydrogen ions (H+) and sulphate ions.  H+ + H2SO4 

HSO4 Hydrogen sulphate ion

 H+ + SO24 H2S O4 

Sulphate ion

11.

Since presence of H+ ions imparts acidic character, therefore solution of sulphuric acid in water behaves as an acid. i) Dilute sulphuric acid is a strong acid. It is proved by the following properties of sulphuric acid : 1) Dilute sulphuric acid reacts with bases and basic oxides forming sulphates (salts) and water. 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) Sod. sulpate CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(l) Cupric oxide Copper(II) sulphate 2) It reacts with carbonates to produces CO2 Na2CO3(s) + H2SO4(aq)  Na2SO4(aq) + H2O(l)+ CO2(g) Sod. carbonate Sod. sulphate MgCO3(s) + H2SO4(aq)  MgSO4(aq) + H2O (l)+ CO2(g) Mag. carbonate Mag. sulphate ii) Sulphuric readily absorbs moisture from atmospheric air. Therefore, it is hygroscopic in nature. That is why the bottle of sulphuric acid should always be keep stoppered.  HCOOH + H2SO4   CO+H2O(H2SO4 ) (Conc.)

12. 17.

18.

 C12H22O11 + H2SO4   12C + 11H2O(H2SO4) (C) 13. (D) 14. (C) (C)

15.

(A)

i)

2NaHCO3  H2SO4  Na 2SO4  2CO2  2H2O  X  CO2

ii)

2NaHSO3  H2SO4  Na 2SO4  2SO2  2H2O  Y  SO2

iii)

Fe  H2SO4  FeSO4  H2  Z  H2 Y