A Textbook of Physical Chemistry for JEE Main and Advanced [2] 931214703X, 9789312147030

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• H2, 0 +0---- > 02 and N +N---- > N2 (ii) Compound When two or more elements combine in a fixed ratio of mass then compound is formed. e.g. 2H +0---- > H20, C+20---- > C02 and N +3 H---- > NH? Properties of a compound are different from its constituent elements, e.g. H2 and 02 are gases but Hp is a liquid.

Chemical Classification of Matter

Properties of matter

Mixtures

The properties of matter are as follows

Mixtures contain two or more components in any ratio. Mixture can be classified as (i) Homogeneous mixture The mixture having components in a uniform composition and which are miscible into one each other is called homogeneous mixture. They are also called solutions.

(i) Mass and weight Mass represents the amount of substance present in a system, while weight represents the force that gravity exerts on that system.

[ Weight (Wj = mass (m) x gravity (g)

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

(ii) Volume It is the quantity of three-dimensional space enclosed by a matter. Volume of different bodies are calculated differently, e.g. For a cuboidal body, ^Volume (V) = length (Z)xbreadth (fr)xheight (A)] For a spherical body.

Volume = — nrJ 3

(iii) Density Density is defined as mass per unit volume. It gives us an idea that how denser or rarer a matter is packed. Density (d) =

mass (w) vo!ume(Vj y

(iv) Temperature Temperature of a body is the intensity of heat associated with it.

PHYSICAL CHEMISTRY

3

1.2 Measurement of physical quantities Comparison of a physical quantity with a standard, is called measurement. Every measurement compares a physical quantity to be measured with some fixed standard known as the unit of measurement, e.g. length of this room is 5.3 m, by this we can interpart that it is 5.3 times unit of measurement, which is one metre in this case. To express the result of a measurement, two pieces of information are required, i.e. (i) the magnitude or the numerical value which is 5.3 in the above example and (ii) the unit, which is metre in this case.

Unit

Energy and its form Energy is defined as the capacity to do work or to transfer heat. Energy can be in the form of (i) mechanical energy (ii) light energy (iii) electrical energy (iv) heat energy Kinetic energy (Energy by virtue of motion) and potential energy (energy by virtue of position) are two principle forms of energy. A big stone located at the top of mountain has potential energy (because of its height). When it rolls down the mountain side, potential energy is converted into kinetic energy.

In thermal plants, thermal energy due to combustion of fuel, is converted into electrical energy. Thus, one form of energy can be converted into other form.

Example 1.1 (Relating intermolecular forces and physical properties) Arrange the following substances in the order in which you expect their boiling points to increase, CCI4, Cl2, CINO and N2. (Given, CCI4, Cl2 and N2 are non-polar and CINO is polar.)

Strategy Boiling point depends on the molar masses and types of intermolecular forces. For non-polar species, substances of higher molar masses have higher boiling points.

Solution CCI4, Cl2 and N2 are non-polar substances. Thus, they have weak intermolecular forces. Thus, boiling point, N2 < Cl2 < CCI4. CINO being polar has stronger intermolecular forces and thus, has higher boiling point. Molar mass of CCI4(154) overcomes intermolecular forces of CINO. Thus, boiling point, N2 m. 4 2195x 1Q.m

Solution By unit factor,

1 km (from which unit)

1 km

3in = 3xUF 2.54 cm „„ 3 in = 3 in x ---------- = 7.62 cm 1 in

INTEXT EXERCISE

1.1 (See Solutions on

Page No. 25)

(b) 42.195 km = 42.195 x UF = 42.195 kmx 0.6215 mile = 1 km

mi|e

Subjective Questions 1. Arrange the following in the expected order of the increasing boiling point.

Ne, He, Cl2, (CH3)2 CO, O2 and O3 2. Melting point of solid A is 500 K and that of solid B is 600 K. Which has stronger packing of atoms? 3. Which would you expect to have the higher boiling point, the hydrocarbon fuel butane (C4H10) or the organic solvent acetone (CH3)2CO?

4. A red blood cell has a diameter of 7.5 pm (micrometer). What is the dimension in

(a) meters

(b) nanometers and

(c) picometers

5. How many kilogram are contained in 1 pg? 6. Density of water is 1 gem"3 at 298 K. Express this value in SI unit. 7. Express the following quantities using a SI prefix and a base unit. (e.g. 1.9x10 -6 m = 1.9 pm)

(iii) 7.82 x 10"3g

(i) 1.47 x 10"9m

(ii) 6.57 x 10"12 s

(v) 0.000123 s

(vi) 0.000000126 m

8. What SI prefixes correspond to the following multiples? (i)103 (ii) IO"6 (iii) 109

(iv) 9.7x103m

(iv) 10

-12

(v) 10"2

9. What do the following abbreviations stand for? (i) dL

(ii) cL

(iii) dm 2 for the following plot. 10. Derive value of the area in m

(iv) pm

60 ft 40 ft

11. What is the volume of 50 kg of water at a place where density is 0.98 g cm"3?

(v) nL

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

7

PHYSICAL CHEMISTRY

Multiple Choice Questions 1. One of the substances is out of order in the following list based on increasing boiling point. N2, O3, F2, Ar and Cl2 (a)N2 (b)F2 (c)Ar (d) O3

2. Which of the following substances would you expect to have the highest boiling point?

(a) C3H8

(c)CH3CN

(b)CO2

(d) All are equal

3. AHvap is called molar heat of vaporisation required to convert one molar mass of liquid into vapour, Liquid Vapour Match the liquid in Column I with their AH,Fvap values in Column II and select the answer from codes given below. Column II

Column I

i. ii.

ch4

iii.

c6h6

iv.

ch3no2

P-

34.0 kJ mol-1

q.

0.92 kJ mol-1

31.00 kJ mol-1

8.16 kJ mol'1

Codes ii iii iv ii iii iv ii iii iv r r r (c) q s (b) q P s s (a) p q P 4. Physical states of a matter in three vessels I, II and III respectively are

ii r

iii P

iv s

O

O O

o

(d) q

o

o O o o

O

o

o 0 o o

o o o o

o o o o

o o o o

ii

I i

II

(a) (b)

Liquid

Gas

Solid

Liquid

Solid

Gas

(c) (d)

Gas

Liquid

Solid

Gas

Solid

Liquid

III

5. Consider the following physical change in a substance. In this case, i n Solid Liquid Mn+ + ne-

Based on graphical analysis of a graph between E (along y-axis) and log[Mn+] (along X-axis), E° and n

can be derived from (a) slope of the line (c) intercept on Y-axis and slope of the line

(b) slope of the line and intercept on Y-axis (d) intercept on X-axis and slope of the line

10

CHAPTER 1 : MATTER : PROPERTIES ANO MEASUREMENTS

PHYSICAL CHEMISTRY

2. In graphical study, slope of the line between X and Y coordinates is zero if graph is

45 °

(b)

(a) O

,60°

(c)

(d) None of these

o

o

3. Which has least value in terms of joule? (b) 1 erg

(a) 1 cal

(d) 1 kcal

(c) 1 L atm

4. A measured temperature on Fahrenheit scale is 200° F. What will this reading be on Celsius scale? (NCERT Exemplar) (a) 40°C (b) 94°C (c) 93.3°C (d) 30°C

5. Average velocity =

I

(a)

(0

of O3 gas at 10° C in SI unit is derived from nM

tex 0,0821xW

(b)

48 k

8x8314x283

(d)

48 it

18x8314x10

\

48 7t

18 x8314 x283

V

48 7t

6. Variation of concentration (dx) (in mol L ’) with time (dt) (in seconds) is given by where, k is rate constant. If n = 2, then k is expressed in (a) L mor’s"1 (b) mol L"’s"’ (c) L2 mol'

n

(d) s'

S

7. x° is the same in °F and °C. Thus, x° is

(b)-273°

(a) 180°

(c)-40°

(d) -32°

8. The density of silver is 10.5 g cm-3. The value in SI unit would be (a)1.05 x104 kgm"3 (b) 10.5 x1044 kg m"3 (c) 1.05 x 103 kg m"3 (b)10.5x10 kgm

(d) 10.5 x102kg m‘

9. Which is equal to 100? (a) 1 MeV/1 GeV (b) 1 Tm/1 Gm

(d) 1 pldlnL

10. Which has the highest value? (a)100Tm (b) 1000 Gm

(c) 1 dm/ 1mm

(c)104Mm

(d) 106 nm

11. Increasing order of prefixes: micro, nano, pico and femto is (a) micro < nano < pico < femto (b) nano < micro < pico < femto (c) pico < nano < micro < femto (d) femto < pico < nano < micro 12. Multiplication by103 will convert

(a) pico into nano (c) femto into pico

(b) nano into micro (d) All of these

13. 1Ts in terms of the SI base unit is equal to (a)1x103s (b)1x109s

(c)1xl0l2s

(d)1x10~’2s

14. Name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantities = 103,10"2,10 ’and 10 3 respectively are

(a)k, c,d and m

(b)m,d, candk

(c)d,k, candm

(d) k,m,d and c

RT 15. — can be used to derive value of EMF. Using values of SI units, EMF is derived from (at 298 K) 8.314 J mol"’ K"’x 298 K (a)

(c)

(b) 8-314x 107 erg mol

i~F~

96500 C mol"’ 0.0821 L atm mol"1K"’ 96500 C mol"’

,2 cal mol ’K" (d) ------------------96500 C mol

K" x 298 K

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

1.4 Uncertainty in measurements

In addition and subtraction, if exponential terms are different, they are made equal and then required operation is performed. In the following (1.7 x 106 + 2.412 x107)

Experimental data as well as theoretical calculations in chemistry can be handled or processed in many ways. While dealing with calculations the meaningful way to handle is the number conveniently and present the data realistically with certainty to the possible extent. These ideas are discussed below:

Different exponential

(^Converted to same exponential)

Scientific notation

(1.7 x 106 + 24.12 x 106) = (1.7 + 24.12) x 106 = 25.82 x 106

When results or constants have very large number of digits (even spreading in decimal with zeros), then they are represented in scientific notation (exponential notation),

In the following, same rule is followed. (2.412 x107-1.7 xlO6) =

e.g. speed of light is 300000000 ms"1. Mass of one atom of oxygen (O) is found to be 0.000000000000000000027 kg etc. There can be error in writing such values and also appears absurd. In scientification notation, they can be expressed as N can vary between 1 to 9

n is an exponent having positive or negative . (integer) values

A/x10"

11

PHYSICAL CHEMISTRY

( Different)

Converted to same exponential)

(24.12 x 106- 1.7 x 106) = (24.12- 1.7) x 106 = 22.42 xlO6 = 2.242 x 107

or

N is a number with a single non-zero digit. If values are very very small ( > 1) as 24605, then in terms of N xlO" 24605

t

Move four places s.

to the left

Here, n =4 and N = 2.4605 Thus, it is 2.4605 xlO4 Thus, in scientific notations 15073 = 1.5073x10,4 0.0024 = 2.4 X10'3 While multiplying the two or more values given in scientific notations, exponentials are added algebraically Thus, (5.0 x IO6) x (1.2 x 107) = (5.0 x 1.2) x (106 + 7) = 6.0 x 1013 Same rule is followed for division.

Thus,

5.-2 x-12_ = = (5.2 (5.2 +1.3) +1.3) x x (1 (1008 “ 4) = 4.0 x 104 1.3 xlO4

Precision and accuracy For an experiment or observation, results should be precise and accurate. Precision refers to the closeness of various measurements for the same quantity.

Accuracy is the agreement of a particular value (measured) to the correct values (true value) of the result.

Illustration I Measurement of length True By student A value

3.00 m 2.99 m

By student B

3.01 m 2.95 m

By student C

2.96 m 3.05 m

2.95 m

Neither precise nor Precise and accurate Precise but not accurate accurate

(i) Values reported by student (A) have reproducibility and are very close to true value. Thus they are precise and accurate. (ii) Values reported by student (B) have reproducibility but are far away from true value. Thus, they are precise but not accurate. (iii) Values reported by student (C) do not have reproducibility and also are far away from true value. Thus, they are neither precise nor accurate.

12

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

A

Such numbers are written in scientific notation in a better way. 200 = 2.0 x 102 has two SF = 2.00 x 102 has three SF

B

= Nxl0"

Illustration II

Note All digits of N of scientific notation are significant. 8.256 x 103 has four SF

C Shooter or Archer

Shot by A (x) Shot by B (o) Shot by C (□) Shot to be hit (»)

Result Accurate and precise Precise but not accurate Neither precise nor accurate

Significant Figures Significant figures are meaningful digits known with certainty. The number of significant figures in a measurement is the number of figures that are known with certainity plus one that is uncertain begining with the first non-zero digit, e.g. 9.00 has three significant figures.

Rule V If name of objects is also written with digits, then digit has infinite number of SF. 12 balls or 5 eggs have infinite significant figure (SF) as these are exact numbers (without uncertainty) and can be represented by writing infinite number of zeros (after placing a decimal). Thus, 12 =12.00000 5 = 5.0000000

Significant figures in mathematical operation 1. In addition or subtraction of two or more values, total digits in the final result cannot be more to the right of the decimal point than either of the original numbers.

If we represent volume by 11.2 ml then

Un­ certain

Uncertain) and uncertainty in the last digit is ±1

Following rules are taken into consideration, while deciding the number of significant numbers.

Thus, result should be

Rule I All non-zero digits are significant. Thus, 102 cm has three SF 0.92 g has two SF Rule II Zeros preceding the first non-zero digit are not significant. Zeroes of such types indicate the position of decimal point.

Thus,

0.003 has one SF 0.0052 has two SF

Rule m Zeros between two non-zero digits are significant. Thus,

One digit after decimal

34.1 Similarly,

34.206 -11.1 23.106

Three digits after decimal One digit after decimal (minimum) Three digits after decimal _in the result

Thus, result should be

200.4 has four SF 105.01 has five SF

Rule IV Zeros at the end or right of a number are significant provided they are on the right side of the decimal point.

Two digits after decimal Three digits after decimal One digit after decimal Three digits after decimal in the result

10.11 + 8.015 + 16.0 34.125

One digit after decimal

23.1

2. In multiplication and division, the final result should have SF equal to the number of minimum SF.

1.5

002100 Has four SF / \ Non-significant Significant

+

QSF = 2>

2.22 SF = 3

=

3.33 SF = 3

(minimum) Zeros at the end are not significant if the digit is without decimal.

200 has one SF 200.0 has four SF

Result should be in two SF = 3.3

Thus, 3.33

1.5

Three SF) ------ = 2.22 Two SF

Three SF

Result should be in two significant figure = 2.2

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

13

Solution

Rounding off

--------------------- 3 (minimum)

If the result of a calculation contains the non-significant figures then it is covered off to nearest significant figures, e.g. we should express 15.433 as 15.5 and 14755 as 1.48 x!04.

(i)

3.24 x 0.08666-— 4 5.0061 — 5

= 0.05608725355

If the first digit you remove is less than 5, round down by dropping it and all the next digit.

= 0.0561

To express the following in three SF.

= 5.61 x 10“2

3 After three SF

(Minimum) |5

It is less than 5

JL2

5.663507 (ii)

Drop it and all the next digits to express it in three SF.

0.58 x 324.65 = 188.297 = 1.9 x 102 I--------- 2SF

|3

Thus it is 5.66 in three SF. (iii)

To express the following in two SF. 5.663507

t

. |4

|3

(943 x 0.00345)+101

= 3.25335 + 101 This digit is 6 or more than 6

= 104.25335 1—11 After 3SF, next digit is 2 < 5. Thus, omitted.

Round it by adding 1 to it.

3SF

= 104 — (Minimum) 12 | Four

Thus, 5.7 in two SF. If the digit to be removed is 5 then the preceding number is not changed, if it is an even number but it is increased by 1 number if it is an odd number. 5.665

next number is 8>5 thus 0 rounded off to 1

(iv)

| Five

24 + 6327-1.5678 t_________________ = 65.67-1.5678

It is to be dropped.

One digit is to be kept after decimal.

I = 64.1022

t

Preceding number is even before 5._

One digit after decimal.

Two SF

= 64.1

Thus, it is 5.66 in three SF. 5.655

It is to be dropped.

jl -"■"Preceding number is"-" odd before 5, it is increased „ by 1 number.

Example 1.14 (SF in algebraic operation) Perform the following calculation and round off the answers

to the correct number of significant figures:

(i)

2.568 x 5.8

(ii) 5.41 - 0.398

4.168

Thus, it is 5.66 in three SF.

Example 1.13 (SF in algebraic operation)

Solution 4

(i)

number of SF.

324 x 0.08666

(iv) 2.4 + 6327 -15678

J.

2

I

,(7 > 5) thus, 5 round off to 6

2.568x52 = 3.573512476

4.168

(ii) 0.58 x 324.65

= 3.6

|4

5.006

(iii) 943x0.00345 + 101

2

I

Express the result of following calculations in required

(i)

(iv) 4.11 - 58.16 x (3.38-3.01)

(iii) 3.38 - 3.01

(ii) 5.41 - 0.398 = 5.012 = 5.01 in 3 SF

Strategy Follow rules of algebraic operation and then round off the result in the required significant figure (SF).

significant figure (SF) are indicated as 3 (minimum).

each has two digits after decimal (iii)

3.38 -3.01

= 0.37—

Two digits after decimal

14

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

Example 1.17 (Significant numbers)

4 |3 3 3 4.18-58,16 x (3.38-3.05) = 4.18-58.16x0.37

(iv)

= 4.18-21.5192

r

This is round off to 4

= -17.3392

= -17.34

Example 1.15 (Scientific notation) Express the following numbers in scientific notation (A/ x 10n) and write the SF in the derived result.

By considering rules of writing number of SF, we can write SF of each value. Solution (i)

0.036653 m - Five SF. (ii) 7.2100 x 10"3g -Five SF.

(iv) 308000 (iii) 72.10 km-Four SF.

I

Strategy

Decimal inserted or shifted after digits in the left, then in scientific notation it is N x 1 CT1-

Solution

i

/

(ii)

843.4 LI (—I

Decimal shifted left to two digits. Thus, 8.434 x 102 has four SF.

0.00421 Decimal shifted right to three digits. I—I 4.21 x 10-3 has three SF.

(iii) 1.54 or 1.54x10°

(iv) 308000 t____ (v) 30860 t____

Has twoSF

3.08000 x 105

Has sixSF

3.0860 x 104

Has fiveSF

Example 1.16 (Scientific notation) Convert the following numbers given in scientific notation to usual form. (ii) 3.275 x102 (i) 6.39x10

(iii) 4.7610 x10":

(iv) 1.0213 x103

Solution

(i)

(iv) Rs 1400 -Infinite SF. Exact number

Significant

0.00125 Decimal shifted after n digit in the right then in I—i scientific notation it is N x 10-n-

(*)

II x

l- Non-significant Significant

(Zeros before a digit are not significant.)

(ii) 0.00421

(v) 30860

10205

(iv) ?1400

(iii) 72.10 km

Strategy Two digits after decimal

(i) 843.4 (iii) 1.54

How many significant figures does each of the following measurements have? (ii) 72100x10"3g (i) 0.036653 m

State the number of significant figures in each of the following numbers. (I) 2.653 x104 (ii) 0.00368 (iii) 653 (v) 400

(iv) 653.10

(viii) k

(vii) 400.0

Solution (i) 2.653 * 104 (ii) 0.00368 (iii)

(vi) 400.

653

Four SF

Three SF Three SF

(iv) 653.0 (v) 400 No decimal point

Four SF One SF

(vi)

Three SF

400.

Decimal point

(vii) 400.0 22 (viii) n = —

Four SF

Infinite number of SF

Example 1.19 (Rounding off)

6.39x1 O’4

(-4 indicates that, move four digits left of the decimal point) 00006.39 = 0.000639

t|

(ii) 3.275 x 102

Example 1.18 (Significant numbers)

(2 indicates that, move two digits right of the decimal point)

3.275 = 327.5 (iii)

4.7610 x 10-2 = 0.047610 V

(M

1.0213x103= 1021.3

Round off each of the following quantities to the number of significant figures indicated.

(i) 3.774499 L (4) (ii) 225.0974 K (3) (iii) 55.265 kg (4)

(iv) 55.275 kg (4)

Strategy Depending the number of significant figure required, preceding digit is round off, retained or removed.-

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

Solution

F"

This digit after four SF is less than 5, hence all the digits are removed.

(i) 3.774499 I I Four SF

Thus, 3.774499 is round off to 3.774 (four SF)

I

All the digits after three SF are removed as 0 < 5.

225.0974 K

(ii)

Three SF

15

Next digit is 8 > 5

(i) 5.607892 Four SF

Thus, 5.608 has four SF. Next digit is 2 < 5 I (ii) 32.392800 Four Thus, 32.39 Four SF

(iii) 0.007137

(iii) 55.265 Four SF

Next digit is 5 and before 5 there is even number thus no change.

Thus, 55.26 has four SF.

(iv) 55.275 Four SF

Next digit is 5 and before 5 there is odd number thus odd number is increased by 1.

= 7.137 x10-3 This, also has Four SF.

(iv) 1.78986 x 103 h t 8 after 9 can't be omitted Four Thus. 1.790 x 103 odd-j j-

(v)

Thus, 55.28 has four SF.

6.2255

•After four SF, next digit is 5 and before 5, odd number (5) exists.

Four SF.

Example 1.20 (Rounding off) Express the following numbers to four SF by rounding off.

(i) 5.607892 (iv) 1.78986 x 103

(ii) (ii) 32.392800 32.392800 (v) 62255

(iii) 0.007137 (vi) 6.2265

Thus, it is rounded off by adding 1 to it. Thus, 6.226 has four SF. (vi) Even —j j-

6.2265

After four SF, next digit is 5 and before 5, even number (6) exists. Thus, no change.

Four SF. Thus, 6.226 has four SF.

Solution

INTEXT EXERCISE

1.3 (See Solutions on Page No. 27)

Subjective Questions 1. How many significant figures are there in each of the following measurements? (a) 73.000 g (b) 0.0503 g (c) 0.360 cm (d) 2.001s

2. Express the following numbers to scientific notation. (a) 4.38 (b) 4380 (c) 0.000483

(d) 0.00000483

3. Convert the following number given in scientific notation to usual form (a)7.025x103 (b) 8.94x1 O’4

4. The circumference of the earth at the equator is 40.000 km. This value is precise to two significant figures. Write this in scientific notation to express correctly the number of significant number. 5. One sphere has a radius of 5.10 cm and the other 5 cm. What is difference in volume (in cm3) between the two spheres? 4 3 volume of the sphere of radius, r = — itr Give, the answer to the two significant figures 3

Multiple Choice Questions 1. Which of the following best describes the degree of uncertainty in the measurement of 16.30 g? (a) The uncertainty cannot be determined without additional information (b) The quantity is exact (c) + 0.10 g (d) ± 0.01 g 2. If the density of a solution is 3.12 g mL“ \ the mass of 1.5 mL solution in significant figures is (NCERT Exemplar) (b) 4680 x10'3g (d) 46.80 g (a) 4.7 g (c) 4.680 g

16

CHAPTER 1 : MATTER ! PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

3. Two students performed the same experiment separately and each one of them recorded two reading of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements.

Observation_________

Student

A B

(a) (b) (c) (d)

Ij

(iij

3.01 3.05

2.99 2.95

Results of both the students are neither accurate nor precise Results of student A are both precise and accurate Results of student B are neither precise nor accurate Results of student B are both precise and accurate

(NCERT Exemplar)

4. How many significant figures should be present in the answer of the following calculation? 2.5 X 1.25x3.5 (NCERT Exemplar) Z01 (0 4 (b) 3 . 1 5. The decimal equivalent of— upto four significant figures may be 0.6 ’ (a) 1.666 (b) 1.6666 (c) 1.667

(a) 2

(d) 1

(d) 01.666

6. In which of the following numbers, all the zeros are not significant? (a) 0.002 (b) 0.0020 (c) 0.00200

(d) 0.0002000

7. Which has maximum number of significant figures? (c)2.1x106 (a) 0.000010 (b) 0.216

(d) 0.021

8. If

1.40 x 2.67 x 1.93

= P, then P is 2300x5340x1.93 (b) 0.0025 x10-2 (a) 0.2500

9. If 1 + 2(17 -15.1) = P, then P is (a) 4.8 (b) 4.80

(c) 0.25

(d) 0.250

(0 5

(d) 5.0

10. 20.665 is rounded off to four significant figures as (a) 20.67 (b) 20.66 (c) 20.6650

(d) 22.665

The Compendium 1. Matter is anything that has mass and occupies space. 2. Solids, liquids and gases are three states of matter; iniermolecular force of attraction being in order: solids > liquids > gases. 3. Energy is the capacity to do work and can be in various forms which are interconvertible. 4. English system and metric system are two different systems of measurement. 5. SI has seven base units from, where all other units are derived. 6. One unit can be converted to other by Unit2 = Unit! x CF to be convene where, C.F. (Conversion factor) = in which converted

CF - --6?s- if minutes are to be converted into seconds. 1 min 7

°C _°F-32 : Kelvin = (°C+ 273.15) 100 ' 180

8. By unit analysis, unit of final result can be derived. 9. By graphical analysis, results can be derived from given graph without doing any tedious calculations. 10. In scientific notation every number is written as N x 10'irt such that N is a number with a single non-zero digit (N = 1 to 9) to the left of the decimal point and n is an integer, e.g. 2 xlO4. 11. Precision refers to the closeness of various measurement for the same quantity.

12. Accuracy is the agreement of a particular value (measured) to the correct value (true value) of the result. If correct value is (10.44 m) of a measurement, 10.32 m and 10.30 m are measured values of Mr X and 10.42 m and 10.45 m are measured values of Mr Y, then measurements Mr X are precise but not accurate, while measurements of Mr Y are accurate and precise. 13. All non-zero digits are significant. 14. Zeroes preceding the first non-zero digits are not significant. 15. Zeroes between two non-zero digits are significant. 16. Zeroes at the end or right of a number are significant provided they are on the right side of the decimal point. 17. In addition or subtraction of two or more values, total digits in the final result cannot be move to the right of the decimal point than either of the original number. 18. In multiplication and division, the final result should have the SF equal to the number of minimum SF. 19. In rounding off a digit • it is retained if next digit is less than 5. • it is increased by 1 if next digit is more than 5. • 1.726 is round off to 1.73 in three SF. • 1.724 is round off to 1.72 in three SF. • it is retained if it is even number before 5. • it is increased by 1 if it is odd number before 5. • 1.725 is round off to 1.72 in three SF. • 1.735 is round off to 1.74 in three SF.

Solved Examples

1 g/cm

3_ 1x10~3kg 1x10"6m3

= 103 kg/m3

103 kg m~3 (to be converted)

Unit factor

1 g cm-3 (from which unit)

Solution 1.025 g / cm3 = 1.025 g cm-3 x UF 1.025 g cm- 3 x 103 kg m'r3 1 g cm-3 = 1.025x 103 kg m-3

Sr Example 1.1 >4 laboratory beaker has a volume of 250 mL. What is the volume in (a) cubic centimetre? (b) litre? (c) cubic metre? (d) cubic decimetre?

(a) metre (m)?

Strategy

Strategy

Unit factor (UF) =

1 cm3 (to be converted)

1 mL (from which unit)

Unit factor

1 x 10~2m (to be converted)

1 pm (from which unit)

(b) 1 pm = 1 x 10"12 m = 1 x 10-3 x1 x10"9 m

(b) 1L = 1000 mL 1L ••• (UF) = 1000 mL

10~3 L

= 1 x 10-3 x 1 nm

1 mL

= 1x10~3 nm

3

it .4 . 1x 10-3 nm (to be converted) /. Unit factor =---------1 pm (from which unit)

(c) 1 mL = 1cm3 = | — I m3 = 10-6 ^3 m J00. 10~6m 3 (UF) = 1mL

Solution (a) 95.8 pm = 95.8 x unit factor 1 x 10~12 m = 95.8 pmx 1 pm

1 (d) 1 cm = — dm

10

1 mL = 1cm3 = f—1 k10j

= 95.8x 10’12 = 9.58x 10’

dm3 = 10’3 dm3

m

(b) 9.58 pm = 95.8 x unit factor 1 x 10"3 nm = 95.8 pmx = 9.58x10 2 nm 1 pm

10~3 dm3 1 mL

Solution

— Example 1.4 Calculate upto proper number of significant

(a) 250 mL = 250 mLx UF = 250 mLx

1 mL

= 250 cm3

Strategy

(c) 250 mL = 250 mLx

1Q-6 mi3: - = 0.00025 m3 = 2.5 x 10-4 m 1 mL

(d) 250mL=250 mLx

10~3 drri — =0.250 dm3 1 mL

)3

figures,

(4.50 x 102m)+ (3.00 x 106 mm)

10'3 L (b) 250 mL = 250 mLx = 0.250 L 1 mL

■>

Note 1 L= 1 dm3 — Example 1.2 Density of sea water is 1.025 g/cm3af 15°C. What is its density in kg/m3?

Data are in scientific notation. For algebraic operations, units and exponential power should be identical. 3.00 x106 3.00 x 106 mm = ------------- cm 100 3.00 x106 ------------- m 1000

Solution (4.50 x 102m) + (3.00 x 106mm) = (4.50 x 102m) + (3.00 x 103m)

Strategy 1 g = 1x 10’3kg

and

(b) nanometre (nm)?

(a) 1 pm = 1x10-12 m

(a) 1 mL= 1 cm3

(UF) =

— Example 1.3 The distance between an O-atom and a H-atom in a water molecule is 95.8 pm. What is the distance in

1 cm3= 1 x 10-6 m3

= (0.45 x 103m)+ (3.00 x 103m) = 3.45 x 103 m

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

-• Example 1.5 The density of a metal is 9.2 g cm"3.

Convert this value into corresponding SI unit. Strategy

Density in SI unit is kg m"3

ig , kg 1000 g kg 1 1000 . 1 cm 1 1 cm=------------- =— m 100 100 cm m’ 1g =

PHYSICAL CHEMISTRY

19

3- Example 1.7 Caesium atoms are the largest naturally occurring atoms. The radius of a caesium atom is 2.62 A How many caesium atoms should have to be laid side by side to give a row of caesium atoms 1.00 inch long? Assume that the atoms are spherical.

Strategy

Convert inch into cm. If n atoms are required for this length, then

n x 2r = required length.

Then,

1 cm-1 = 100 m-1 1cm-3 = 106 m"3

Solution 9.2g cm"3 =(— kglx (106 m"3) = 92 x 103 kg m' 11000 ) 1 gem

-3

1000 kg m -3

Solution 1.00 inch =2.54 cm

Radius (r)=2.62 A =2.62 x10"® cm.

nx2 x2.62 xIO^ =2.54 2.54 n =----------------- - = 4.85 x107 atoms 2x2.62x10'8

B Example 1.6 Perform the indicated operations and round off the answer to the proper number of significant figures. Also mention proper units. (i) 423.1cm + 0.256cm- 116cm (/'/) 6.057x 103m-9.35 m (Hi) 8.54 x 105 tickets/22 day. Strategy

Example 1.8 What is the specific gravity of a liquid, if 325 mL of the liquid has the same mass as 396 mL of water at a temperature at which 1 mL water has mass of 1 g?

Strategy ... ... , ., density of liquid Specific gravity of a liquid =------- -------- — density of water Solution

By using rules of algebraic operations, final result is round off to required number of SF.

Density of water =

Solution

Mass of 396 mL water = 396 x 1 = 396 g

(i) 423.1 cm + 0.256 cm -116 cm

no decimal = 307.356

Final result is = 307 cm ] without decimal (ii) 6.057 x 103m-9.35 m = (6057 - 9.35) m

= 6048 m ] Final result is without decimal Three SF Xy8.54 H 105 (iii) = 3.8818 x104 ^22 = 3.9 x 104 tickets day-1 Two SF

1 cm3

Density of liquid = — = 1.22 g cm"3 325 1 22 Thus, specific gravity = = 1.22

3 Example 1.9 Carry out the following conversions (a) 5pm = ....cm = nm (b) 8.5 cm3 =. m3-....mm3 (c) 65.2 mg =

= 6047.65 m

= 1gcm-3

volume

g=

pg (d) 5 cal = L atm =...... J fej 1 amu (= 1.6605 x 10"27 kg)=

J....MeV

Strategy By base units and derived units, one unit can be converted to other unit.

Solution (a) 1 pm = 1 x 10’ 12m= 1 x 10"12 x 102 cm = 1x 10"1ocm=10-3 nm

20

PHYSICAL CHEMISTRY

a

5 pm= 5x 10“1°cm

1 pm = 1x 10"12m=1x 10" 9 x 10"3 = 10"3nm a

5 pm = 5x 10“3nm

(b) 1 cm3 = 10" 6m3

1 cm3= 103mm3

8.5 cm3 = 8.5 x 10" 6m3 = 8.5 x 103mm3 (c) 1mg = 1x 10" 3g a 65.2 mg = 65.2 x W^g 1 pg = 1 x 10" 12g = 1 x 10"9mg 1 mg = 1 x 109 pg a

65.2 mg = 65.2 x 109pg

(d) By unit of ft 2 cal = 0.0821 L atm 5cal = 5x 0.0821 L atm =0.4105 L atm 1 cal =8314 J 5 cal = 8.314 x 5 = 41.57 J (e) Mass related to energy by E = mc2 = (1.6605x 10"27)(3.0x 108)2 = 1.6605x 10"27x9.0x 1016kg m2s"2

= 1.49445 x 10"1°J 1 amu = 1.4944x10"1°J 1 eV= 1.6022 x 10~19J

1 amu =

1.4944X 10"10 J 1.6022X 10"19JeV’1

= 932.7175 x 109eV = .933 MeV

CHAPTER 1 : MATTER : PROPERTIES ANO MEASUREMENTS

— Example 1.10 One cubic millimetre of oil is spread on the surface of water so that the oil film has an area of 1 sq m. What is thickness of the film formed? Strategy area x thickness = Volume

Area inm2 = (100)2 cm2 = (100)2 (10)2mm2 Solution

Volume of oil = 1 mm3 Surface area of water = 1 m2 on which oil is spread = (1 x 1000)2mm2 Let thickness be = x mm x mmx (1000)2mm2 = 1 mm3 x = 1 x 10~6mm= 1 x 10"9nm= 1 nm

3* Example 1.11 How much area in square metre, will one litre of paint cover if it is brushed out to a uniform thickness of 100|im?

Strategy 1L of paint = 1 dm3 = 1 x 10“ 3m i 3 1gm =1 x 10“®m area x thickness = Volume Solution

1L paint = 10" 3m3

Thickness = 100 pm = 100 x 10" 6m Area= Am2 •. A x thickness = Volume A x 100 x 10~6m= 1 x 10"3m3

A = 10 m2

Chapter Exercises Type I: MCQs with Only One Correct Option 1. An average adult has 5.2 L of blood. Volume of blood in m3 is (a)5.2x103m3 (b)5.2x10’3m 3 (c) 5.2 x 10-6 m3

2. 10.5 g cm-3 in SI unit will be (a) 10.5 x 10'3 kg m’3 (c) 10.5 kg m-3

(d) 5.2x10’2 m3

(b) 10.5x10’1 kg m-3 (d) 10.5x 103 kg m-3

3. Solder is an alloy of tin and lead that is used in electronic circuits. It has melting point of 225°C. It is............. °F. (a) 498 °F (b) - 437°F (c) 437°F (d) 400°F 4. x°C differ by 11° in Kelvin and Fahrenheit scale. Thus, x°C is (a) 300 (b) 273 (c) 132 (d) None of these

5. 0.445 and 0.435 are round off to two SF as (a) 0.44, 0.43 (b) 0.45, 0.44 (c) 0.44, 0.44 (d) 0.45, 0.45 6. For each of the following pairs, determine the quantity in which each value of pair have equal quantity.

l

9. At what temperature, will a Fahrenheit and Celsius thermometers will have same value but differ in sign? (a) 12.3° (b)16.6° (c) 11.43° (d)-11.43° 10. Least count of an analytical balance is 0.0002 g. Which of the following is the correct recorded value? (a) 9.3171g (b) 9.3170 g (c) 9.3175 g (d) 9.3173 g 11. Given, heat absorbed = mass x specific heat x rise in temperature If 75.0 g of water (specific heat = 1 cal degree-1g-1) at 30°C absorbs 900 cal of heat, then final temperature is (a)12°F (b)42°F (c)107.6°F (d) 94.2°F

12. Which choice best describes the degree of uncertainty in the measurement of 4.0032 g? (a) The measurement is between 4.0031 and 4.0033 (b) The measurement is between 4.002 and 4.004 (c) The quantity is exact (d) The uncertainty cannot be determined with given information 13. The largest volume of the following group is that of (a) 445 g of water at 4°C (d = 1g cm-3)

(b) 600 g of chloroform at 20°C (d = 15 g cm-3) (c) 155 cm3 of steel (d) 0.50 Lof milk

14. Of the following substance, the greatest density is that of (a) 1000 g of water at 4°C (b) 100.0 cm3 of chloroform weighing 198.9 g (c) 10.0 cm3 piece of wood weighing 7.72 g (d) an ethyl alcohol-water mixture of density 0.83 g cm-3 15. SI unit and metric unit of volume are (a) m3, L (b) m3, m

3

(d)dm3, m3

(c)L, m3

16. On the Reamur scale water freezes at 0’R and boils at 80°R. Hence, °R = — °C (b)^= °F-32 (a) — 80 180 80 100 (c)°R=0.8°C=^ (°F-32) (d) All of these

II

(a)

24.0 mg

24.0 eg

(b)

0.25 m

(c)

250 cm 0.8 nm

(d)

10L

10 m3

8A

7. The mercury thermometer used in homes can be accurate to ± 0.1° F and those of clinic can be accurate to ± 0.1 °C. If body temperature is 38.9° C then percentage error is higher in (a) home thermometer (b) clinic thermometer (c) equal in both (d) can’t be predicted 8. At what temperature will a Fahrenheit thermometer give a reading that is twice on the Celsius thermometer? (a) 53.3° (b)-123° (c) 80.0° (d) 160.0°

Directions Answers of Q. Nos. 17 to 20 have been based on the following codes. (a) If Statements I and II both are correct and Statement II is the correct explanation of Statement I (b) If Statements I and II both are correct but Statement II is not the correct explanation of Statement I (c) If Statement I is correct but Statement II is incorrect (d) If Statement II is correct but Statement I is incorrect 17. Statement I If U =

U=

3RT M

and at 300 KforO2 gas

3x8.314x300x10'I3 - ms 32

Statement II R, T and M have been taken in SI units.

22

CHAPTER 1 : MATTER ! PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

18. Statement I Values of R can be used to convert energy values from SI to non SI units.

Statement II R has dimension of work mol

K .

19. Statement I Significant figures for 0.200 is 3, whereas for 200 it is 1. Statement II Zeros at the end or right of a number are significant provided, they are not at the right side of a decimal point. (NCERT Exemplar)

20. Statement I 7.965 and 7.975 are round off to 7.97 to three significant figures. Statement II If digit is greater than 5 it is round off to next digit.

Type II: MCQs with One or More than One Correct Option

Select the correct statement(s). (a) Both I and II (b) Both II and III (c) Both I and III (d) I, II and III

27. Ratio of two units is 103 in the following. (a) pm/nm

(d) hm/dm

28. The vapour pressure at different temperature is given by the equation, . B log p = a - where, p is in mm of Hg. This equation in SI unit can be written as (a)logp = A-- 133.3 (b) logp = A-® - 21249 (d)logp = A-® + 1.1249

(c)logp = A-| + 133.3

29. In which of the following calculations, final result have SI units? (a) Average velocity of O2 gas at 300 K

21. Which of the following numbers are round off to same value in four SF? (a) 6.7745 (b) 6.7735 (c) 6.7736 (d) 6.7734

[2RT

2 x 0.0821 x 300

N M

32

(b) Pressure of 1 mole O2 gas at 300 K in 1 L flask

22. Number of SF is more than 1 in (a) 100 (b) 100.0 (c) 0.0001

_RT _ 8.314 x 300 x 1000

(d)2.01

23. Which of the following algebraic operations will have result in one SF? 1.24x 1,76 (a) 1.0+ 3.75-201 (b) 0.01 1.1 x 1.2 (c)3.25 + 3.00- 1.1 (d) 1.0 24. Select the correct statement about two sets of readings of students X and Y, when correct value is 12.7 mL.

X

(c) Gm/Mm

(b) nm/pm

Y

V 1 (c) KE of 1 g ball moving with velocity of light

= ~mv2 = - x 1 x 9 x 1013 2 2 (d) Acceleration of a ball moving with 1cm s-1 attained in 10 s = velocity = 1x10.3 second

30. Select derived SI units (a) molar volume-> m3mor1 (b) pressure -> kg m" 1s"2

I

II

I

II

12.4

12.3

12.3

12.3

(a) Reading of both are precise but not accurate (b) Reading of both are precise and accurate (c) Reading of X are precise but not accurate (d) Reading of Y are precise and accurate

25. Select the correct statement about metre (unit of length). The metre is the length of the path travelled by light.

(c) force -> g cm s“ 2 (d) power-> J s-1

Type III: Matrix Matching Type Questions 31. Match the given values (number) in Column I with number of significant figures in Column II and select the answer from the codes given below.

Column II

Column I

(a) In air during a time interval of 1/299792458 of a second (b) In vacuum during a time interval of 1/299792458 of a second

I.

100

P-

One

ii.

0.001

q-

Two

(c) In water during a time interval of 1/100 of a second

iii.

100.00

r.

Three

iv.

100.0

s.

Four

v.

0.0020

t.

Five

(d) In vacuum during a time interval of 1/100 of a second

26. Consider the following statements. I. Kilogram is equal to the mass of the international prototype of the kilogram. - •

II. Second is the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperline levels of the ground state of the Cs-133 atom.

III. National Metrology Institute (NMI) predicts climate conditions in each country.

Code (i) (a) P (b) r (c) r

(d) p

(ii) (iii) (iv) (v) t s q p t s q p r t s s r t s q

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

32. Match the given values (number) in Column I with the values (number) in column II as a result of rounding off to three SF and select answer from the codes given below. Column I

Column II

i.

7.9952

P-

7.99

q-

7.98

ii.

7.9857

iii.

7.9866

iv.

7.9759

s.

8.00

v.

8.0152

t.

8.01

vi.

7.9654

u.

8.02

Column II

Column I

0.1

p. deci/centi

ii.

1.0

q. micro x mega

iii.

10.0

r.

iv.

100.0

s. hecto/kilo

v.

1000.0

t.

deca/hecto

peta/tera

u. deca/deci

Codes (i) (ii) (a) r s (b) s r (c) r,s q (d) p tt

(iii) pP u pP u

(iv) qq tt uu ss

23

41. Second order rate constant (k) is related to initial concentration a and concentration (a - x) at time t by 1 1

2

equation (k) = -

t a-x

a

7.96

Code (i) (ii) (iii) (iv) (V) (Vi) t r (a) p q p q (b) s q p q u r (c) P q q p t r (d) s p q p u r 33. Match the values (number) in Column I with the prefix used in SI units in Column II and select the answer from the codes given below. i.

PHYSICAL CHEMISTRY

(V) t q t q

Type IV: Single Integer Answer (0-9) Type Questions 34. Boiling point of helium is -452°F. What will be the

temperature in Kelvin scale? 35. Number of significant figures in the answer of the following will be x. What is value of x?

(29.2-20.2) (1.09x 105) 1.69 36. At what SN unit of energy is highest? 1.1 L atm ; 2.1 erg ; 3.1 J 4.1 kcal; 5.1 kJ ; 6.1 cal

37. How many calorie are there in 20.92 J? 38. Symbol G is for giga and symbol h for hecto. Ratio of G : h is 10*. What is value of x?

39. What is value of 46.4°F in Celsius scale?

40. A liquid has freezing point at -12.8°C. What is freezing point in °F?

(a-x)' A

,45°

0.25

0

timet

What is value of (a - x) after t = 0.257 Assume all units imaginary. 42. Final result of the following will be expressed in x SF. What is value of x?

12.11+ 18.0+ 1.012- 1.02 43. One year is equal to 3.1536 x 107s. How many significant figure are there in this number expressed in seconds?

Type V: Linked Comprehension Based Questions Passage 1 (Q. Nos. 44-45) On the Reamur scale, water freezes 0°R and boils at 80°R Answer the following questions.

44. Select the correct alternate. ,,/R. °c

°F-32

(b) — = 100 180

80 100

(c)2R=^

C (d)^=^

180 80

100 80

45. If a liquid boils at 40°R, then it boils at (a) 50°C

’ (b) 323 K

(c) 122°F

(d) All of these are correct

Passage 2 (Q. Nos. 46-47) Zeroes at the end or right of a number are significant provided they are on the right side of a decimal point, terminal zeroes are not significant in same condition.

Answer the following questions.

46. Terminal zeroes are not significant if (a) there is no decimal point (b) there are digits less than 5 (c) there are digits larger than 5 (d) terminal zeroes are always significant

47. Select the correct statement(s). (a) 200 has one significant figure (b) 200 balls has infinite significant figures (c) 100° has three significant figures (d) All of the above are correct statements

24

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

Passage 3 (Q. Nos. 48-49) By rounding off a number, required number of significant figures may be attained.

Arrange the following questions.

51. Out of the given observations which has minimum error (a) at SN (1) of student A (b) at SN (2) of student A (c) at SN (1) of student B (d) at SN (2) of student B

48. 11.99 is round off to three digits as (a) 11.9

(b) 12.0

(c)11.1

(d) 11.99

Type VI: JEE Main & JEE Advanced Archive 52. The most abundant elements by mass in the body of a healthy human adult are oxygen (61.4%); carbon (22.9%), hydrogen (10.0%); and nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

49. 7.757 and 7.857 are round off to two SF as (a) 7.7, 7.8

(b) 7.8, 7.9

(c) 7.8, 7.8

(d) 7.7, 7.9

•

Passage 4 (Q. Nos. 50-51)

(a) 15 kg

In chemistry practical, two students A and B performed titration and burette readings were recorded as

Student B

Student A

Observation

Initial

Final

Initial

Final

1..

5.2 mL

10.2 mL

1.0 mL

6.4 mL

2.

15.2 mL

20.7 mL

6.4 mL

11.8 mL

Volume as recorded by teacher (which was correct as 5.6 mL) Answer the following questions.

50. Select the correct statement(s). (a) Observation by student A was accurate and precise (b) Observation by student B was accurate and precise (c) Observation by student A was accurate but not precise (d) Observation by student B was precise but not accurate

(b) 37.5 kg

(c) 7.5 kg

(d) 10 kg [JEE Main 2017]

53. If the value of Avogadro’s number is 6.023 x 1023 mol -1 and the value of Boltzmann constant is 1.380 x 10" 23 J K"1 then the number of significant digits in

the calculated value of the universal gas constant is [JEE Adv.2014 (Integer Type)] 54. A student performs a titration with different burettes and finds titre value of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average value is [IIT JEE 2010 (Integer Type)] 55. Silver (atomic weight = 108 g mol-1) has density of 10.5 g cm-3. The number of silver atoms on a surface area of 10-12 m2 can be expressed in scientific notation as

y x 10x. The value of x is Mole =

mass 1 mole = 6.022 x1023atoms mol atomic weight’ [IIT JEE 2010 (Integer Type)]

Hints and solutions Intext Exercise 1.1 Solutions to Subjective Questions

1. He °F= -x 42 + 32 °F= 107.6°F \5 J

12. (a) The measurement 4.0032 is between 4.0031 and 4.0033 13. (d) (a) 445 g of water = 445 mL of water (d = 1 g cm-3) 600 g (b) 600 g chloroform = = 400 mL 1.5 g cm-3 (c) 155 cm3 = 155mL (d) 0.50 L= 500 mL

14- (b) (a) d(H/) at 4° C) = 1 g cm-3 (b) d = -maSS = = 1.989gcm"3 volume 100

/

(C) d = ~ = 0.772 g cm-3 (d) d = 0.83 g cm3 Thus, option (b) is correct 15. (a) SI unit of volume = m3 Metric unit of volume = L (dm3) 16. (d) On the Reamur scale, water freezes at 0°R and boils at 80°R. Thus, 1^ = 1^ 80 100 Also j£=°F-32 _ °R jo °F-32 100 180 80 100 180 4 “/or nn) °R= 0.8 °C = -(°F- 32)

Thus, option (d) is correct. 17. (a) All data have been taken in SI unit. Thus, U (root mean square velocity) is in ms-1. 18. (a) R = 0.0821 L atm mor1K"1 = 8.314 x 107 erg mol"1K"1 = 8.314 J mor’K-1 =2 cal mor’K-1

R has dimension of work (or energy). Thus, one unit can be converted to other. 19. (c) 0.200J— Zeroes at the end of or right of a number are significant provided that they are on the right side of the decimal point. 200 ]— Zeroes at the end of number are not significant.

20. (d) 7.965

Three SF

CHAPTER 1 : MATTER : PROPERTIES AND MEASUREMENTS

21. (a, b,c) (a)

5 for is to be removed; digit before is 4 (even) number, thus 4 remains unchanged

6.7745

t

Four SF

Thus, 6.774

(b)

5 is to be removed; digit before it is 3 (odd) number, thus 3 increased to 4.

6.7735

V

Four SF

(c)

Thus, 6.774

6 is to be removed r~ (6 >5), hence 3 6.7736

is round off to 4

Four SF

Thus, 6.774

22. (b, d) (b)4 (a) 1 (c)1 23. (a, c) (a) 1 as only 1 digit after decimal. (b) 3, minimum 3 SF is each value. (c) 1 as only 1 digit after decimal. (d) 2, minimum 2 SF

24. (a, c) In both case, readings are far away (± 0.4) from correct value. Thus, are not accurate. Readings of X as well as that of Y are precise as close to one each other. 25. (b) By definition of metre.

26. (a) By definition of kilogram and second. 27. (b, c, d) (a) P22=lZ^=ix1O-3 nm 10-9m (H nm 10"9m (b) — = — = 1x103 pm 1O~12 (c) Gm_1°9m. = 1x103 Mm io6m (d) hm=WjJm 1x1o3 dm io-1m 28. (b) By conversion factor 1 mmHg = — atm 760 1 atm = 101325 Nm-2 (SI unit) 1

760

otm

101325 ..

-2

760 = 133.3Nm-2 /

p\

logp(inmmHg) =l A-yj

log (133.3p) (in SI) = A-log 133.3 + log p = I A - — I

® before 5 is even hence no change thus it is round off to 7.96.

2.1248 + log p= (a-®)

7.975

Three SF

7 before 5 is odd hence increases by 1 thus it is round off to 7.98.

(d)3

log p= A - y-2.1248

CHAPTER 1 : MATTER : PROPERTIES ANO MEASUREMENTS

29. (b, c, d)

(a)u =

[ORT

... nRT (b)p=—

PHYSICAL CHEMISTRY

deca / hecto =

R is in L atm and M in grams. Thus, incorrect 1x8.314x300 0.001 m3

8.314x300x1000

1

(Since, 1 L= 1 x 10"3 m-3)

10 102

= 0.1

102 hecto / kilo = — = 0.1 103 1015 peta I tera = —— = 1000 1012

Thus, correct (c) KE = - mv2 = - x 0.001 kg x (3 x 108)2 m2s’2 2 2 = lx9x1013 kg m22 s -2 o-2 Thus correct.

(d) Acceleration

Thus correct

velocity

—L ms -i pg

second

deca/deci = — = 100

0.1

Thus, (i) - (r, s), (ii) - (q). (iii) - (p), (iv) - (u), (v) - (t) °F-32 34. (4) 180 100

5

10 s

= - 269°C = - 269°C + 273= 4K 35. (3)

30. (a, b.d) In (c) mass is in gram Others are derived SI units

36. (4) 1. R = 0.0821 L atm = x (say)

31. (a) (i) 100-1 SF (Terminal zeroes are not significant if there is no decimal point). Thus, (i) - (p) (ii) 0.001 - 1 SF (Zeroes preceding to the first non-zero digit are not significant). Thus, (ii)-(p)

(iii) 100.00 - 5 SF (Zeroes at the end or right of a number are significant provided they are on the right side of decimal). Thus, (iii) - (t)

(iv) 100.0 - 4 SF (As above) Thus, (iv) - (s) (v) 0.0020-2 SF Thus, (v) - (q) 32. (b) 9 before 5 round off to zero but that is less (') 7.9952 than 7.99.

r

.-.1 L atm =—^—= 12.18 x 0.0821 2. R = 8.314 x 107ergs = x x ... 1 erg =------ ------ - = 1.2 x 10~8x 8.314 x107

3. R = 8.314J = x 1J = 0.12x

4. R = 0.002 kcal = x 1 kcal = 500x 5. R = 8.314 x10’3kJ=x 1kJ= 1.20 x 102x

6. R = 2 cal = x

.-.

37. (5)

Thus, 8.00 Then(i)-(s)

(ii)

r

7.9857

8 before 5 (even), no change.

(iii) 7.9866

— (6 > 5) Thus, 8 round off to 9.

Thus, 7.99 Then (iii)-(p)

r

(iv) 7.9759

Odd number before 5 thus 7 is round off to 8.

1 cal = - = 0.5x 2

8.314 J = 2 cal 2 20.92 J = x 20.92 8.314

38. (7)

Thus, 7.98 Then (ii)-(q)

r

39. (8)

(V)

8.0152

-1 (odd number) before 5 thus, 1 is round off to 2.

102

h

x =7 °F-32_ °C 100

180

46l4°F = 8°C

40. (9)

u,

41. (2)

w

Thus, 7.98 Then (iv)-(q)

r

5

°C =-(°F - 32)=-(-452 - 32) 9 9

= 1x10’3ms"2

t\(a-x) a) 1 1 (a - x)

1

1

1

a

= kt + -

(a-x)

a

Thus, 8.02 Then (v) - (u)

(vi) 7.9654

6 (even number) before 5 6 remains 6 (no change).

Thus, 7.96 Then (vi) - (r) . . deci 0.1 wr) 33. (c)^L = — = 10 centi 0.01

micro x mega = 10-6 x 106 = 1

(a - x)~’ = kt + a

y = mx + c OA = - = c = intercept = 0.25 a 1 .

a =---- = 4 0.25

m = tan45° = 1 = k

29

30

CHAPTER 1 : MATTER ! PROPERTIES AND MEASUREMENTS

PHYSICAL CHEMISTRY

.-. (a - x)-1 = kt + - = 1x0.25+ 0.25= 0.50 a (a-x) = 2 42. (3) 43. (5)

Solutions for Q. Nos. 44-45 If water freezes at 0° R and boils at 80° R in Reamur scale, then °R °C °F- 32 80 100 180

Thus, (a)

If boiling point = 40° R Then,

°C 100 100 x 40 = 50°C °C = 80 50° C = 50 + 273 = 323 K 40° 80

°F = — (°C) + 32 = — x 50 + 32 = 122° F 100 100 Thus, (d) Hence (44) - (a),

(45) - (d)

Solutions forQ. Nos.46-47 100 10.2 = 1 ox + 110 00 ~22 100 X = 80 1°B=80% => ”B= 20%

CHAPTER 2 : MOLES AND STOICHIOMETRY

PHYSICAL CHEMISTRY

Example 2.3 (Atomic mass of isotope)

Solution

Calculate the average atomic mass of hydrogen using the following data.

Mass number = Neutron + Proton

Isotopes of hydrogen ]H

(%) natural abundance

Molar mass

99.985

1

2H

0.015

2

(NCERT Exemplar)

Solution 7 _ Al Xi + A2X2 _ 1 x 99.985 + 2 x 0.055 _ , X, + X2 99.985+ 0.015

33

Protons in element = 65- 35 = 30 Atomic number = 30 Electrons in element = 30 Electrons in cation = 28 Thus, two electrons are lost in the formation of cation, M —> M2+ + 2e" Thus, cation is ^M2+.

K

Example 2.4 (Ionic mass/atomic mass)

Dalton atomic theory (1808) Main points of Dalton atomic theory are given below

A divalent cation M2+ is isoelectronic of CO2 and has (Z + 2) neutrons (Z is the atomic number of the divalent cation). What is the ionic mass of the divalent cation?

(i) All elements are composed of minute particles called atoms, which do not undergo subdivision during chemical reactions and cannot be created or destroyed by any chemical process. (Taylor)

Strategy In an ionic species,

(ii) Atoms of the same element are similar to one another in all respects and are equal in weights.

Z = number of protons = electrons in neutral species Ionic mass = Protons + Neutrons. Isoelectronic means containing same number of electrons (as in CO2). Solution M2+ is isoelectronic of CO2 Electrons in CO2 = 6 + 16 = 22 Electrons in M2+ = 22 Protons in M2+ = 22 + 2 = 24 ' Thus, atomic number of M2+ = 24 (Z) Neutrons in M2+ = Z+ 2 = 26

Thus, ionic mass of M2+ = P + N = 24 + 26 = 50

Example 2.5 (Ionic mass/atomic mass) Ionic mass of X2- is 16. It has 8 neutrons. Determine the number

of its (a) protons (b) electrons and (c) atomic number. Solution X2-has ionic mass =16

Number of neutrons = 8

Thus, protons in X2- = 16 - 8 = 8

Electron in X (atom) = 8 Electrons in

X2- = 8 + 2 = 10

Atomic number (Z) = Protons = 8

Example 2.6 (Ionic mass/atomic mass) One isotope of a metallic element has mass number 65 and 35 neutrons in the nucleus. The cation derived from the isotope has 28 electrons. Write the symbol for this cation.

(iii) Atoms of different elements have different masses (called atomic masses) and different properties. (vi) Chemical combination between two or more elements takes place by interaction of their atoms in simple numerical proportions (as 1 : 1; 1 : 2; 1 : 3; etc.) Ratio 2HC1 1 : 1 H2 +C1, 2NH, N2+3H2 1 :3 C + O2--- > CO, 1: 1 (v) Combining masses of the elements represent the combining masses of the atoms. Dalton atomic, theory can explain various laws of chemical combination.

Limitations and defects of Dalton's atomic theory (i) It can explain the laws of chemical combination by weight, but it fails to explain Gay-Lussac's law of gaseous volume.

(ii) Distinction between the ultimate particles of an element and the smallest particle of a compound, both of which are made of atoms, could not be made.

(iii) It assumes that atoms of same element are like and have same mass. But this is not true for isotopes (two or more atoms with same atomic number but different atomic mass are called isotopes). (iv) ‘Jz/JC.’JC.have atoms of different masses. have atoms of different masses.

34

CHAPTER 2 : MOLES AND STOICHIOMETRY

PHYSICAL CHEMISTRY

2.2 Mole

Example 2.7 [Conversion of grams into moles and vice-versa)

The mole (abbreviated as mole) is the base SI unit for a chemical species. One mole is the amount of a substance that contains as many particles or entities as there arc atoms in

Gypsum is a hydrate calcium sulphate. A 1.0 g sample

12

exactly 12 g of C isotope. Number of particles or entities in 1 mole of a substance is known as Avogadro constant or Avagadro number denoted by NA or No. One mole of a substance always contains the same number of particles, no matter what the substance may be.

contains 0.791 gCaSO4. How many moles of CaSO4 are

there in the sample? Assuming that rest of the sample is water, how many moles

of water are there in the sample. Show that the result is

consistent with the formula CaSO4 • 2H2O? Strategy

12

Moles =

Based on mass of one atom of C determined by mass spectrometer value of NA was derived.

To show the consistency with the formula, Moles of CaSO4 _ 1 CaSO4-2H2O = Moles of H2O 2

Mass of one atom of C = 1.992648 X10 23 g 12

Thus, 12 g of C =

12 1.992648 X1O“23

Mass of the species (in g) Molar mass (in g mol-1)

atoms

Solution = 6.022 1 3 67 x 1023 atoms mol

Molar mass ofCaSO4 = 40+ 32 + 64= 136gmol

CaSO4 = 0.791 g

Thus,

H2O= (1- 0.791)= 0.209 g

1 mole of H-atom = 6.022 x 1023 atoms 1 mole of H2O molecules = 6.022 x 1023 molecules

„ , . , mass in grams Mole of a substance =---------- - -----molar mass Mass of one molecule of a species =

Moles of CaSO4 = — = 5.816 x 10’3 4 136 Moles of H2O=

18 Ratio (CaSO4 : H2O)= 1 : 2

11.611 x 10"3

molar mass

Example 2.8 (Conversion of grams into molecules) How many molecules of CO2 are there in 88 g CO2?

itaKCnaMMMWzMt a>**
.

| Molecules of A |

Calculations are reversed if operations are reversed.

Example 2.9 [Conversion of grams into molecules and vice-versa] In a bank, there are as many coins as number of molecules in 1.6 jig of CH4. How many coins are there in the bank?

Solution 1.6jtgofCH4 = 1.6x10"6gCH4,

Mass of one mole of a substance in grams is called its molar mass.

Each water molecules has two H-atoms and one O-atom, thus 1 mole of H2O = NA waler molecules = Na oxygen atoms = 2Na hydrogen atoms

1.6x10~6

mol CH4 16 1.6x10'I-6 — x 6.02 x 10 Molecules of CH4 = 16 = 6.02 x 1016 coins

Moles of CH4 =

molecules

CHAPTER 2 : MOLES AND STOICHIOMETRY

Example 2.10 [Conversion of grams into atoms) How many g-atom and number of atoms are there in 70 g nitrogen?

Solution As 14g N has 1g atom 70 g has 70/14 = 5 g atom Further, 1 g-atom has NA or 6.02 x1023 atoms

PHYSICAL CHEMISTRY

Thus,

35

1 amu = —=1.66056 xlO"24 g Ho

A flow chart is given below to calculate the atomic mass unit of mass of one element of an atom. [Grams of Ap— Given

______ /Divide by ' L molar mass | Moles of T]

5 g-atom has 5 x 6.02 x 1023 atoms

Multiply

= 30.1 x1023 atoms

J>y No^

| Atoms

of a|

| Atomic mass

Example 2.11 (Conversion of grams into males

of Ap— Given

---- (Divide by No

and vice-versa] Analysis of chlorophyll shows that it contains 2.88% magnesium. How many atoms of magnesium does 1.20 g of chlorophyll contain?

Atomic mass _ Mass of one unit of A atom of A

An algorithm to find the mass of one atom Solution

1.20 g of chlorophyll has = 1.20 x ~~ g Mg, = 1.20X — x——2- mole Mg 100 24 g mol' = 1.44 x 10-3 mole Mg, = 1. 44x10"3 x 6.02x 1023 Mg atoms = 8.6688 x 1023 Mg atoms

To convert atomic mass into atomic mass unit, divide it by No Mass of an atom of hydrogen atom = 1.6736 x I0",-24 24g Thus, in term of amu, mass of hydrogen atom 1.6736 x 10~24 ” 1.66056x IO-24

Example 2.12 (Conversion of grams into moles

and vice-versa) A solution containing 10.0 mmol ofCaCI2-2H2O is diluted to 1L. Calculate the number of grams of CaCI2 • 2H2O per mL of final solution.

= 1.0080 amu In present notation "amu" has been replaced by 'u', (unified mass). Example 2.13 (Atomic mass unit) Mass of one atom of the element A is 3.9854 x 10"23g. How

Solution

1 L (= 1000 mL) has CaCI2-2H2O = 10 mmol = 10 x 10"3 mol = 10 x 10"3 x molar mass ofCaCI2 -2H2O = 10x10"3 x 147 gCaCI2-2H2O

Thus, 1 mL has CaCI2 =

10x10~3 x147 gmL"1 1000

= 1.47 xW"3 gmL”1

Note: g mL"1 = density

many atoms are contained in 1 g of the element A? Solution 89854 x 10-23g is the mass of = 1 atm

Thus, 1 g is the mass of =-------- ------ atoms 3.9854 x 10"23 = 2.5092 x 1022 atoms

Example 2.14 (Atomic mass unit]

Mass of one atom of an element X is 2.32481 x 10-23g.

Express this in amu.

Atomic mass unit One atomic mass unit (amu) is defined as "exactly one-twelfth the mass of an atom of and is equal to 1.66056 x 10“24g"

1 amu = ~ (mass of one C-atom)

Solution

Mass of one atom =

Atomic mass No

Atomic mass = Mass of one atom x No = 2.32481 x 802 xIO23 = 14g mol"1

1 12

12/

1 No

14

Mass in amu =— = 14 amu. N0

i

36

CHAPTER 2 : MOLES AND STOICHIOMETRY

PHYSICAL CHEMISTRY

Example 2.15 (Atomic mass unit)

Example 2.17 (Volume vs mole)

Mass of one atom of X is 6.642 x 10-23g. What is its atomic mass? How many moles of X are contained in 40 kg of it?

How many moles are there in 40 L of CO2 at STP?

Strategy Mass of one atom of an element =

Atomic mass

No Mass (in gram) Moles = Atomic mass (in g mol-1)

Solution Relating moles to volume, 22.4 L of CO2 = 1 mol at STP

40 L of CO? =

Thus,

Solution

Example 2.18 (Volume vs mole)

Atomic mass of X = Mass of one atom x No

Calculate the volume of 3.4 g NH3 gas at STP, assume its ideal behaviour.

= 6.642x 10-23g x 6.02 x 1023mol“1= 40.0g mol

Moles of X present in 40 kg

= 40x 1000 g 40.0 g mol 1

pmo|

Example 2.16 (Atomic mass unit) If x mole of 12C = 0.5 A/o atoms of12 C +1.2 g of 12C - 0.1 No atoms of12 C, derive the value of x. Solution

Strategy

Gram-----> Mole----- > Volume

Solution Moles of NH3 = Mass (in g) = — = 0.2 mol Molar mass 17 1 mol of a gas at STP = 22.4 L 0.2 mol of a gas at STP = 22.4 x 0.2 = 4.48 L

Number of atoms and grams of atoms are to be converted into moles. => 1.2 g 12C=1^ mol = 0.1 mol 12 0.1 A/012C= 0.1 mol => x= 0.5+0.1-0.1 = 0.5 mol 0.1/V C=0.1mol

0.5 A/012C = 0.5 mol I

mol = 1.786 mol

22.4

Example 2.19 (Volume vs mole) 16 g of an ideal gas SOX occupies 5.6 L at STP. What is the molar mass of the gas? What is the value of x?

Volume vs mole

Solution

Under STP conditions, when temperature is 273.15 K (0°C) and pressure is 1 atm, volume of one mole of an ideal gas is 22.4 L. „ nRT 1x0.0821 L atm mort-1 x 273.15 K V =-----=-------------------------------------------- = 22.4 L p 1.0 atm

5.6 L of SOX at STP of the gas = 16 g

1 mole of the gas = Molar mass of the gas = 22.4 L at STP

Thus, under STP conditions, 1 mole of an ideal gas = 22.4L To convert mass of a gas in mole Divide by .. . Multiply by Gram ---------------- Mole ■> Volume Molar mass 22.4 L

INTEXT EXERCISE

2.1 (See Solutions on Page No. 77)

22.4 L of the gas at STP =

x 22.4 = 64 g mol-1

Molar mass of the gas = 32 + 16x

32 + 16x = 64

x=2 Thus, gas isSO2.

Subjective Questions 1. How many moles of sucrose (C12H22On) are contained in 1.75 g?

2. How many grams are there in 0.02 mole of NaHCO3 the main ingredient in Alka-Seltzer tablet?

3. Convert the following masses to moles and to number of molecules (atoms as indicated) (a) 1.0 kg of H2O (b) 64.0 g of sulphur as sulphur (S) atoms and as S8 molecules (c) 4B2 x1O22 atoms of 37CI (d) 4.60 kg of CH3CH2OH (e) 500 g CaCO3 stone containing 20% CaCO3 4. Which sample in each of the following pairs contains the greater number of moles of atoms? (a) 25 g of 12C or 40 g of 16O (b) Wg of 16Oor Wg of 32S

(c) x atoms of H or x atoms of He

5. Which gas occupies the greater volume among 1 g H2 or 1 g He at STP? 6. Equal masses of SO2 and O2 are placed in a flask at STP (a) which has larger number of molecules? (b) which occupies larger volume? 7. What is the molecular mass of cholesterol if 0.5731 mole weighs 221.6 g?

CHAPTER 2 : MOLES AND STOICHIOMETRY

PHYSICAL CHEMISTRY

37

8. What mass of SO2 would contain the same mass of oxygen as it contained in 29.7 g of As2O3? (As = 75; O = 16 and S = 32) 9. The carat is the unit of mass used by jewellers. One carat is exactly 200 mg. How many carbon atoms are present in a 24 carat diamond?

10. Match the following. (i) 88 g CO2 (ii)

(a) 0.25 mole (b) 2 moles

6.022 X10 23 molecules of H2O

(iii) 5.6LO2atSTP

(c)

(iv) 96gO2

(d) 6.022 x 102J molecules

(v)

1 mole of any gas

1 mole

(e> 3 moles

11. The rest mass of an electron is 9.11 x 10

-31

(NCERT Exemplar)

kg. What is the molar mass of the electron ?

Multiple Choice Questions 1. One mole of oxygen gas at STP is equal to (a) 6.022 x 1023 molecules of oxygen

(NCERT Exemplar) (b) 12X944 x1023 atoms of oxygen

(c) 32 g of oxygen

(d) All of these

2. Which of the following pairs have the same number of atoms? (NCERT Exemplar) I. 16gofO2and4gofH2 II. 16 g of O2 and 44 g of CO2 III. 28gofN2and 32gofO2 IV. 12 g of C and 23 g of Na (a) Both I and II (b) II, III and IV (c) Both III and IV (d) I, III and IV (NCERT Exemplar)

3. 16 g of oxygen has same number of molecules as in (a) 16 g of CO (b)28gofN2

(c) 14 gofN2

(d) 2 g of He

4. Number of moles in 1.8 g H2O is equal to the number of moles in I. 18 g glucose Select the correct choice. (a) 1,11 and III

II. 6 g urea

III. 34.2 g sucrose

(b) Both 1 and II

(c) Both II and III

(d) Both I and III

5. One requires 0.01 mole of Na2CO3. Mass of Na2CO3-10 H2O to be taken is (a) 1.06 g (b) 2.86 g (c) 1.80 g (d) 3.02 g 6. If avogadro's number would have been 1 x 1023 mol-1, then mass of one atom of oxygen would be

(d)16x 6.02 amu

(a) 16 amu

(c)16 x1023 amu

(a) oxygen atoms in 11 g of CO2 (c) nitrogen atoms in 40 g of N2O4

(b) hydrogen atoms in 4 g of CH4 (d) sulphur atoms in 79 g of Na^Oj

(b----- amu 6D2 7. Number of atoms in 12g (22Mg) is equal to

8. Select the correct statement. (a) At STP volume occupied by 1 mole liquid water is 22.4 L (b) 1 mole of every substance at STP has 22.4 L as volume — (c) Volume occupied by 1 g of H2 gas at STP is equal to volume occupied by 2 g He at STP (d) CH4 is a real gas if 1 g of it at STP has volume 1.4 L 9. Following is the graphical presentation of volumes occupied by different gases at STP Y 11.2

Volume/l| 5.6

0

H2

He

nh3 ch4

T~ T~Mass/g3

Which is/are not placed at correct position? (a) H2, He (b) He, NH3

4

5

(c)NHj,CH4

(d) CH4, H2

38

CHAPTER 2 : MOLES ANO STOICHIOMETRY

PHYSICAL CHEMISTRY

10. Volume of an ideal gas at NTP is 1.12 x 10-7 cm3. The number of molecules in it is

(a) 3.01x1012

(b) 3.01x1 O'8

(c) 3.01 x1024

(d) 3.01x10

20

11. 1021 molecules are removed from 200 mg of CO2. Thus, number of moles of CO2 left is

(a) 2.88 x 10’3

(c) 288 x10’3

(b) 28.8 x 10'3

(d)28.8x103

12. A spherical ball of radius 7 cm contains 56% iron. If density of the metal (of the ball) is 1.4 g cm’3, number of moles of Fe present is (a) 10 (b) 15 (c) 20 (d) 28 13. Which of the following may contain one proton and one neutron? (a) H2 (b) 4He (c) 7D

(d) ?T

14. Bare proton is (a) H+

(d) He+

(c) T+

(b) D+

15. On heating KCIO3, 0.672 L of O2 gas at STP is formed KCIO3 taken is (a) 0.02 mole (b) 0.02 molecules (c) 0.03 mole

(d) 0.03 No molecules

16. Number of atoms in one gram each of isotope is placed in the increasing order (a) Mg24 < Mg25 < Mg26 (b)Mg26 < Mg25 25 < Mg24 (c) Mg26 < Mg25 < Mg24 (d) given data is insufficient

2.3 Laws of chemical combination Law of mass conservation According to this law "Mass is neither created nor destroyed in chemical reactions" i.e. total mass of reactants before chemical reaction = total mass of products after chemical reaction. When hydrogen gas bums and combines with oxygen to yield water (H2O), the mass of water formed is equal to the mass of hydrogen and oxygen consumed. This is in accordance with the law of mass conservation. 2H2(^)+O2(5)---- > 2H2O (/) Mass of reactants 4 g

36 g Mass of product

32 g

Example 2.21 (Law of mass conservation) A sample of MgCO3 is 50% pure. If on heating 4.0g of MgO and 4.4 g of CO2 are formed, calculate the amount of MgCO3 taken. (Mg = 24, C = 12, andO = 16)

Strategy MgO and CO2 are obtained only from pure MgCO3. From mass of MgO and CO2, mass of pure MgCO3 is known. Thus, amount of impure sample (50%) is derived.

Solution

MgCO3—> MgO+ CO2 4.0g 4.4 g

Mass of (MgO + CO2) = 8.4 g Thus, mass of pure MgCO3 = 8.4 g

Mass of impure sample = 8.4 x 19? = 16.8 g

CaCOj(s)-----> CaO(s) + CO2 (g) Mass of reactants 100 g 56 g 44 g Mass of product

Example 2.20 (Law of mass conservation)

Law of multiple proportions

In the decomposition of impure KCIO3, 4.9 g of it gave 1.49 g

According to law of multiple proportions "If two elements combine in different ways to form different substances, then the masses of one element that combine with a fixed mass of other element, are in the ratio of small whole numbers.

of KCI(s) and 0.03 mole of O2 gas. What is the per cent purity

of KCIO3? (K= 39, Cl = 35.5 and 0=6) Solution 2KCIO3(s) ---- > 2KCI(S) 4.9 g (impure) 1.49g

+ 3 O2(g) 0.003 mol = 0.003 x 32 = 0.96 g

By law of conservation of mass, mass of products = mass of reactants = 1.49+ 0.96= 2.45 g Thus, pure reactant = 2.45g Impure reactants = 4.90 (taken) 2 45

Thus, per cent purity = —x 100= 50%

Nitrogen and oxygen can combine either in a 7 : 8 mass ratio to make a substance denoted by NO or in a 7 : 16 mass ratio to make a substance denoted by NO2.

NO Mass ratio

14: 16

of N and O 7:8

Thus,

NO2

14 : 32

7 : 16

N:O mass ratio in NO _ (7gN/8gO) N : O mass ratio in NO2 (7gN/16gO)

CHAPTER 2 : MOLES AND STOICHIOMETRY

Also 1g of hydrogen can combine with 8 g of oxygen to yield H2O or with 16g of oxygen to yield H2O2. Thus, hydrogen reacts in a multiple of 2. H : O mass ratio in H2O (lg H/8gQ) H : O mass ratio in H2O2 (lgH/16gO)

H(2g) 16gO2

32 g O2 h2o2

h2o

Example 2.22 (Law of multiple proportion) Methane and ethane, both are constituents of natural gas. A sample of methane contains 11.4 g of carbon and 3.80 g of hydrogen, whereas a sample of ethane contains 4.47 g of carbon and 1.118 g of hydrogen. Show that the two substances obey the law of multiple proportions.

PHYSICAL CHEMISTRY

39

Law of reciprocal proportions (Richter-1792) “The masses of two or more different elements which separately combine with a definite mass of another element are either the same as or arc simple multiples of masses of these different elements when they combine amongst themselves." A combines with C to form AC B combines with C to form BC A combines with B to form AB Then ratio of A and B by mass in AB is same as ratio of A and B in AC and BC.

(a) C+ 2H2--- > CH4 (b) O2 +2H2---> 2H2O (O2H4)

(c) C + O2--- > CO2

Strategy

First, find the C : H mass ratio in each compound. If the law of multiple proportion is true, then (C : H) ratio in methane to ethane should be in multiple ratio. Solution

O2

C

S-No^

Carbon

Hydrogen

Methane

11.4g

3.80g

4.47g

Ethane

Ratio

3.80 4 47 ——=4.00 1.118

1.118g

3 Thus, ratio of (C : H) in methane to ethane is = -

4

Thus, law of multiple proportion is followed.

Example 2.23 (Law of multiple proportion] A metal (A4) forms two oxides A and B. On analysis it was found Mass O2 combined S. No.

Oxide A

1.45g

0.56 L al STP

Oxide B

1.125g

0.56 L at STP

Note 1 mole of O2 at STP = 22.4 L

If the law of multiple proportion is followed, then identify the

oxides. Solution

22.4 L of O2 at STP = 1 mol = 32 g of O2 0.56 L of O2 at STP = 0.80 g

4 g H2 combines with 12 g carbon in (a) 4 g H2 combines with 32 g oxygen in (b)

Thus, ratio of carbon and oxygen in (a) and (b) by mass that combines with a definite mass of H2 (in this case 4 g) = 12:32 =3:8 in (c) 12 g carbon combines with 32 g oxygen, thus ratio of carbon and oxygen (by mass) in CO2 is = 3:8 Compound

Elements

Mass ratio

CH4

C and H

,3: 1

h2o

Oand H

(8:1

CO,

C and 0

Law of definite proportions This slates "Different samples of a pure chemical substance always contain the same proportions of elements by mass". Ever)' samples of water (H2 0) contains 1 part hydrogen and 8 parts oxygen by mass. 0 H2

Metal M in oxide A = 1.45 - 0.80 = 0.650 g Metal M in oxide B = 1.125 - 0.80= 0.325 g

0.650 o . Thus, ratio of M in A and B = ------- =2.1 0.325 Thus, for fixed amount of oxygen, metal M in A and 8, is in the ratio of 2:1. Thus, A: M2O;

B:MO

2 :

16 :: 1:8

Every sample of carbon dioxide (CO2) contains 1 part carbon and 2.67 parts oxygen by mass

C O2 12 : 32 :: 1 : 2.67

40

CHAPTER 2 : MOLES AND STOICHIOMETRY

PHYSICAL CHEMISTRY

If the formula is to be derived then, than take molar ratio in which

„ . r , Mass Number of moles =----------------------------------Molar mass or atomic mass

Example 2.24 (Law of definite proportion) 1,4g of iron reacts with 0.8g of sulphur directly to form iron (II) sulphide. 5.6 g of iron was dissolved in dilute HCI and excess of potassium sulphide was added. If the law of definite composition is followed, how much iron (II) sulphide is formed in the second case? (Fe = 56 and S=32)

2.4 Gay-Lussac's law and Avogadro's law Based on experimental studies French chemist, Gay-Lussac concluded that the volumes of reactants and products in a large number of chemical reactions are related to each other by small integers, if volumes are measured at the same temperature and pressure. 2H2(5) + O2(^)---- > 2H2O (g) 2V IV IV 2V 100 mL 50 mL 100 mL

Strategy

Iron and sulphur in Iron (II) sulphide is formed in both cases are in definite molar composition.

If all the gases are measured at the same temperature and pressure (say 373 K and 1 bar pressure) then 100 mL of H2(^) combines with 50 mL of O2(^) to given exactly 100 mL of H2O (g).

Solution n , ,. .. .. Moles of Fe in case I Based on law of definite proportions =------------------------Moles of Fe in case II _ Moles of S in case I Moes of S in case II 71.4 W5.6> _ 0.8/32 15.6 J/156 J x/32

This is called Law of definite proportions by volume. This was explained by Avogadro's law. "The volume of a gas (at fixed temperature and pressure) is proportional to the number of moles (or molecules of gas present)".

i

V « n (moles) « nN0 (molecules)

x (sulphur in case II) = 3.2 g

Total mass of iron and sulphur = 5.6+ 3.2 (Based on law of conservation of mass) = 8.8 g

Example 2.25 (Law of definite proportion) (i) If H and 0 are in the ratio of 1 :16 by mass.

N2(