Chemistry

Class Exercise

Class Exercise - 1

Express the following numbers tothree significant figures.

(i) 6.022 × 1023 (ii) 5.356 g(iii) 0.0652 g (iv) 13.230

Solution

(i) 6.02 × 1023

(ii) 5.36 g(iii) 0.0652 g(iv)13.2

Class Exercise - 2

What is the sum of 2.368 g and1.02 g?

Solution

2.368 g

1.02 g

3.388

= 3.39 g

Class Exercise - 3

Express the result of the followingcalculation to the appropriate numberof significant figures816 × 0.02456 + 215.67

Solution

816 × 0.02456 = 20.0

Product rounded off to 3 significant figures becausethe least number of significant figure in thismultiplication is three.

67.235

67.215

0.20

Rounded off to 235.7

Class Exercise - 4

Solve the following calculations andexpress the results to appropriatenumber of significant figures.

(i) 1.6 × 103 + 2.4 × 102 – 2.16 × 102

(ii)23

20

6.02 10 5.00

4.0 10

Solution

(i) 1.6 × 103 + .24 × 103

3

3

3

1.6 10

.24 10

1.84 10

Rounded off to 1.8 × 103

3

3

3

1.8 10

.216 10

1.584 10

Class Exercise - 4

Rounded off to 1.6 × 103 or 16 × 102

(ii)23 23

20 20

6.02 10 5.00 30.10 10

4.0 10 4.0 10

= 7.525 × 103 (rounded off to 7.5 × 103)

Class Exercise - 5

Convert 10 feet 5 inches into SI unit.

10 feet 5 inches = 125 inches1 inch = 2.54 × 10-2 m

Solution

Rounded off to 317 × 10–2 m

125 inches = 2.54 × 10-2 × 125 m

= 317.5 × 10-2 m

Class Exercise - 6

A football was observed to travel at a speedof 100 miles per hour. Express the speedin SI units.

Solution

1 mile = 1.60 × 103 m100 miles per hour

3100 1.60 1060 60

= 4.4 × 10-4 × 105 m/s= 4.4 × 10 m/s= 44 m/s

Class Exercise - 7

What do the following abbreviationsstand for?

(i) O (ii) 2O (iii) O2 (iv) 3O2

Solution

(i) Oxygen atom(ii) 2 moles of oxygen atom(iii) Oxygen molecule(iv)3 moles of oxygen molecule

Class Exercise - 8

Among the substances given belowchoose the elements, mixtures andcompounds

(i) Air (ii) Sand(iii) Diamond(iv) Brass

Solution

(i) Air - Mixture(ii) Sand (SiO2) - Compound(iii) Diamond (Carbon) - Element(iv) Brass (Alloy of metal) - Mixture

Class Exercise - 9

Classify the following into elementsand compounds.

(i) H2O(ii) He(iii) Cl2(iv)CO(v) Co

Solution

Element: He, Cl2, Co

Compound: H2O and CO

Class Exercise – 10

Explain the significance of the symbol H.

Solution

(i) Symbol H represents hydrogen element(ii) Symbol H represents one atom of hydrogen atom(iii) Symbol H also represents one mole of atoms, that is,

6.023 × 1023 atoms of hydrogen.(iv)Symbol H represents one gm of hydrogen.

BASIC CONCEPTS OF CHEMISTRYSession - 2

Session Opener

Session Objectives

1. The law of conservation of mass

2. The law of definite proportions

3. The law of multiple proportions

4. The law of reciprocal proportions

5. Gay Lussac’s law of gaseous volumes

6. Dalton’s atomic theory

7. Modern atomic theory

Session Objectives

Law of conservation of mass

Total mass of the products remains equal to the total mass of the reactants.

H2 + Cl2 2 HCl

2g 71g 73g

Question

8.4 g of sodium bicarbonate on reaction with 20.0 g of acetic acid (CH3COOH) liberated 4.4 g of carbon dioxide gas into atmosphere. What is the mass of residue left?

Illustrative Problem

8.4 + 20.0 = m + 4.4 m = 24 g

It proves the the law of conservation of mass.

Solution:

A chemical compound always contains same elements combined together in same proportion of mass.

Law of definite proportions

Ice water H2O 1 : 8

River water H2O 1 : 8

Sea water H2O 1 : 8

Question

Two gaseous samples were analyzed.One contained 1.2g of carbon and3.2 g of oxygen. The other contained 27.3 % carbon and 72.7% oxygen. The experimental data are in accordancewith

(a) Law of conservation of mass(b) Law of definite proportions(c) Law of multiple proportions(d) Law of reciprocal proportions

Illustrative Problem

% of C in the 1st sample

%3.271002.32.1

2.1

Which is same as in the second sample.Hence law of definite proportion is obeyed.

Solution

the mass of one of the elements which combines with fixed mass of the others, bear a simple whole number ratio to one another.

Law of multiple proportions

8:16:24:32:40 1:2:3:4:5=

14:8 14:16 14:24 14:32 14:40

Two elements combine two or more compounds

Ratio of oxygen combining with 14 parts of nitrogen

The ratio of the weights of two elementsA and B which combine separately witha fixed weight of the third element ‘C’ iseither the same or some simple multipleof the ratio of the weights in which A andB combine directly with each other.

Statement of law of reciprocalProportions

)W:W(kW:W

W:WBA

CB

CA

k may be 1

Law of reciprocal proportions

SO2

OH

S

H2S

H2O

H2S 2 : 32, 1 : 16

SO2 32 : 32, 1 : 1

16:11

1:

16

1SO:SH 22

8:1,16:2OH2

2

1

1

8

16

1OH:SO:SH 222

Gases reacts in volume which bear a simple ratio to one another and to the volume of the products under similar conditions of temperature and pressure.

Gay Lussac’s law of gaseous volumes

Gay Lussac’s Law of gaseous volumes

N2 O2 2NO

2H2O2 2H2O

1 volumenitogen

gas

1 volumeoxygen

gas

2 volumenitrogen

oxide+

2 volumehydrogen

gas

1 volumeoxygen

gas

2 volumesteam

+

1 : 1 : 2

2 : 1 : 1

Ask yourself

Why Gay–Lussac’s law is not applicable to solids and liquids ?

because they have negligible volumes as compared to gases.

Do you know

The laws of chemical combinations are basedon quantitative results of chemical reactions.

Questions

Carbon is found to form two oxides,which contains 42.8% and 27.27% ofcarbon respectively. Find out which of the laws of chemical combination is proved correct by this data?

Illustrative Problem

In the first oxide, Carbon :Oxygen = 0.428 : 0.572

= 0.748 : 1

In the second oxide,

Carbon :Oxygen = 0.2727 : 0.7273

= 0.374 : 1

= 2:1

= 0.748 : 0.374

Solution:

Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Find out which of the laws of chemical combination is proved correct by this data?

Illustrative Problem

N

H

O

NH3

N2O3

H2O

15.63

)85.36(K

90.88

10.11

65.17

35.82

Hence k=1 which proves law of reciprocal proportion.

Solution:

Ask your self

The balanced chemical reaction is an expression of

Law of multiple proportion Law of

conservation of mass Law of

constant proportion

None of the above

Illustrative example

Zinc sulphate crystals contain 22.6% of zinc and 43.6% of water. How muchZinc should be used to produce 13.7 gm of zinc sulphate crystals and how muchwater they will contain?

Solution:

100 gm of ZnSO4 will have 43.9 gm water and 22.6 gm zinc

4Zn required to produce 100gm of ZnSO crystals 22.6 gm

(Hint: Law of constant composition)

Solution

43.913.7gm of crystal contain water 100

13.76.0143gm

100gm of crystal contain water 43.9gm

4

22.6Zn required to produce 13.7gm of ZnSO crystals 13.7

100

3.0962gm

1. Matter is made up of indivisible particles called atoms.

2. Atoms of the same element are similar with respect to shape, size and mass.

3. Atoms combine in simple whole number ratio to form molecule.

4. An atom can neither be created nor destroyed.

Dalton’s Atomic Theory

5. Atoms of two elements may combine in different ratios to form more than one compound.

SO2 1:2

SO3 1:3

S+O2

Limitations of Dalton’s Atomic Theory

It could not explain:

Why atoms of different elementshave different masses, sizes, etc. Nature of binding force

between atoms and molecules.

Limitations of Dalton’s Atomic Theory

1. Can not explain causes of chemical combination

2. Can not explain law of combining volume

3. It does not give idea about structure of atom

4. It does not distinguish between the ultimate particleof an element and a compound

5. It does not give idea about isotopes and isobars.

Modern Atomic Theory

1. Atom is divisible

2. Same atom may have differentatomic masses like 1H, 2H and 3H.

3. Different atoms may have same atomic mass like 40Ca and 40Ar.

4. Atom is the smallest particle that takes part in a chemical reaction.

5. The mass of an atom can be changed into energy.

Class Test

Class Exercise - 1

Percentage of copper and oxygen insamples of CuO obtained by differentmethods were found to be the same.This proves the law of

(a) constant proportions (b) reciprocal proportions(c) multiple proportions (d) none of these

Solution

Definition

Class Exercise - 2

A balanced chemical equation is inaccordance with

(a) Boyle’s law(b) Avogadro’s law(c) Gay Lussac’s law(d) law of conservation of mass

Solution

Definition

Class Exercise - 3

Different samples of water were foundto contain hydrogen and oxygen in theapproximate ratio of 1 : 8. This shows the law of

(a) multiple proportion(b) constant proportion(c) reciprocal proportion(d) none of these

Solution

Definition

Class Exercise - 4

Law of multiple proportion is illustratedby the following pair of compounds.

(a) HCl and HNO3

(b) KOH and KCl(c) N2O and NO(d) H2S and SO2

Solution

Definition

Class Exercise - 5

The oxides of nitrogen contain63.65%, 46.69% and 30.46%nitrogen respectively. These dataproves the law of

(a) definite proportions (b) multiple proportions(c) reciprocal proportions (d) conservation of mass

Solution

Definition

Class Exercise - 6

12 g carbon combine with 64 g sulphurto form CS2. 12 g of carbon alsocombine with 32 g oxygen to form CO2.10 g sulphur combine with 10 g oxygento form SO2. These data proves the

(a) law of multiple proportions(b) law of definite proportions(c) law of reciprocal proportions(d) Cray Lussac’s Law of gaseous volume

Solution

Ratio of the weights of S and O combining with fixed weight of C is 64 : 32 = 2 : 1. Ratio of weights of S and O combining directly = 10 : 10 = 1 : 1. The two ratios are simple multiple of each other. This proves the law of reciprocal proportions.

C

S O

CS2 CO2

SO2

12 g

10 g 10 g

64 g 32 g

Class Exercise - 7

Nitrogen forms five stable oxides withoxygen of the formula, N2O, NO,N2O3, N2O4, N2O5. The formation ofthese oxides explain the

(a) law of definite proportions(b) law of partial pressure(c) law of multiple proportion(d) law of reciprocal proportions

The mass of oxygen which combine with the fixed massof nitrogen (= 14 g) is N2O, NO, N2O3, N2O4, N2O5 are 8,16, 24, 32, 40 g respectively. They are in the ratio of 1 : 2: 3 : 4 : 5. This proves the law of multiple proportions.

Solution

Class Exercise - 8

Two metallic oxides contains 27.6%and 30.0% oxygen respectively.If the formula of the first oxide isM3O4, then that of the second will be

(a) MO (b) MO2 (c) M2O5 (d) M2O3

Solution

72.43M

27.6 416

72.4 16 3

M 27.6 4M = 56

70 16 x56 30 y

x 2y 3

Hence oxide is M2O3.

Class Exercise - 8

Two metallic oxides contains 27.6%and 30.0% oxygen respectively.If the formula of the first oxide isM3O4, then that of the second will be

(a) MO (b) MO2 (c) M2O5 (d) M2O3

Solution:

Class Exercise - 9One litre of nitrogen combines withthree litre of hydrogen to form twolitre of ammonia under the sameconditions of temperature and pressure. This explain the

(a) law of constant composition(b) law of multiple proportion(c) law of reciprocal proportions(d) Gay Lussac’s law of gaseous volumes

Solution

The ratio of volumes of N2, H2 and NH3 is 1 : 3 : 2, whichis a simple ratio. This proves Gay Lussac’s law of gaseousvolumes.

Class Exercise - 102.16 g of copper metal when treatedwith nitric acid followed by ignitionof the nitrate gave 2.70 g ofcopper oxide. In anotherexperiment 1.15 g of copperoxide upon reduction with hydrogen gave0.92 g of copper. Show that the abovedata illustrate the law of definite proportions.

Solution

% of Cu in copper oxide in 1st case 2.16

100 80%2.70

% of oxygen = 20% % of Cu in copper oxide in 2nd case

0.92

100 80%1.15

% of oxygen = 20%

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