NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

NCERT Solutions for Class 9 Science Chapter 3 (in-text Exercise Questions)

All the in-text exercise questions with detailed solutions are given below:

Exercise-3.1 (Page: 27)

Question.1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

Answer:

Given the reaction

sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

5.3g 6g 8.2g 2.2g 0.9g

Now,

Total Mass on the Left Hand Side = 5.3g + 6g = 11.3g

Total Mass on the Right Hand Side = 8.2g + 2.2g + 0.9g = 11.3g

As Mass on the LHS is equal to the RHS, the observation is in agreement with the law of conservation of mass.

Question.2 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?

Answer:

Given:

Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.

Now,

For every 1 g of Hydrogen, 8 g of oxygen is needed for the reaction to take place.

Therefore, for 3 g of Hydrogen, the mass of oxygen needed = 8 $\times$ 3 = 24 g.

Hence, 24 g of Oxygen is needed to complete a reaction with 3 g of Hydrogen.

Question.3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton’s atomic theory is the result of the law of conservation of mass is

"Atoms can neither be created nor destroyed".

Question.4 Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer:

The postulate of Dalton’s atomic theory can explain the law of definite proportions is

"The relative number and kinds of atoms are equal in given compounds."

Exercise-3.2 (Page: 30)

Question 1. Define the atomic mass unit.

Answer:

An atomic mass unit is a unit of mass used to express the weight subatomic particles, where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.

Question 2. Why is it not possible to see an atom with naked eyes?

Answer:

We can't see the atom with the naked eye because they are minuscule in nature. they are measured in the nanometres. Also, except for noble gases, all-atom do not exist independently. they exist in the form of any compound.

Exercise 3.3 and 3.4 (Page: 34)

Question 1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Answer:

The Formula of the given compounds are :

(i) sodium oxide :

$\Rightarrow N{a}_{2}O$

(ii) aluminium chloride:

$\Rightarrow AlC{l}_3$

(iii) sodium sulphide:

$\Rightarrow N{a}_{2}S$

(iv) magnesium hydroxide:

$\Rightarrow Mg(OH{)}_2$

Question 2. Write down the names of compounds represented by the following formulae:

(i) $A{l}_2(S{O}_4{)}_3$

(ii) $CaC{l}_2$

(iii) $K_{2}S{O}_4$

(iv) $KN{O}_3$

(v) $CaC{O}_3$

Answer:

The Names of the following compounds are :

(i) $A{l}_2(S{O}_4{)}_3$

=Aluminium Sulphate

(ii) $CaC{l}_2$

=Calcium Chloride

(iii) $K_{2}S{O}_4$

=Potessium Sulphate

(iv) $KN{O}_3$

= Potassium Nitrate

(v) $CaC{O}_3$

= Calcium Carbonate.

Question 3. What is meant by the term chemical formula?

Answer:

The chemical formula of a compound is a symbolic representation of its composition. For example, The chemical formula for common salt is NaCl as it is made up of Sodium (Na) and Chlorine (Cl).

Q 4.(i) How many atoms are present in a

• $H_{2}S$ molecule

Answer:

$H_{2}S$ molecule has 2 atoms of Hydrogen and 1 atom of Sulphur.

and hence,

A total of 3 atoms are present in $H_{2}S$ molecule.

Question 4.(ii) How many atoms are present in a

• $P{O}_4^{3-}$ ion?

Answer:

$P{O}_4^{3-}$ ion has one atom of Phosphorus and 4 atoms of Oxygen.

And Hence,

A total of 5 atoms are present in $P{O}_4^{3-}$ ion.

Exercise-3.5.1-3.5.2 (Page: 35)

Question 1. Calculate the molecular masses of $H_2,O_2,C{l}_2,C{O}_2,C{H}_4,C_2{H}_6,C_2{H}_4,N{H}_3,C{H}_{3}OH$

Answer:

The molecular mass of $H_2$ :

= 2 $\times$Atomic mass of Hydrogen

= 2 $\times$1u

=2u.

The molecular mass of $O_2$ :

= 2 $\times$ Atomic mass of Oxygen

= 2 $\times$16u

= 32u.

The molecular mass of $C{l}_2$ :

= 2 $\times$ Atomic mass of Chlorine

= 2 $\times$ 35.5uu

= 71u.

The molecular mass of $C{O}_2$ :

= Atomic mass of Carbon +2 $\times$ Atomic mass of Oxygen

= 12u + 2 $\times$16u

= 44u.

The molecular mass of $C{H}_4$ :

= Atomic mass of Carbon +4 $\times$Atomic mass of Hydrogen

= 12u + 4 $\times$1u

= 16u.

The molecular mass of $C_2{H}_6$ :

= 2 $\times$ Atomic mass of Carbon + 6 $\times$ Atomic mass of Hydrogen

= 2$\times$12u+ 6 $\times$1u

=24u+ 6u

= 30u.

The molecular mass of $C_2{H}_4$ :

= 2 $\times$ Atomic mass of Carbon + 64$\times$Atomic mass of Hydrogen

= 2$\times$12u+ 4$\times$1u

=24u+ 4u

= 28u.

The molecular mass of $N{H}_3$ :

= Atomic mass of Nitrogen + 3 $\times$Atomic mass of Hydrogen

= 14u+ 3 $\times$ 1u

= 17u.

The molecular mass of $C{H}_{3}OH$ :

= Atomic mass of Carbon +4$\times$Atomic mass of Hydrogen + Atomic mass of Oxygen

= 12u+ 4 $\times$ 1u + 16u

= 32u.

Question 2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃ , given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u

Answer:

Given,

the atomic mass of Zn = 65 u,

the atomic mass of Na = 23 u,

the atomic mass of K = 39 u,

the atomic mass of C = 12 u,

and the atomic mass of O = 16 u

Now,

Formula unit mass of ZnO = Atomic mass of Zinc + Atomic mass of O

= 65 u + 16 u

= 81 u.

Formula unit mass of Na₂O = 2 $\times$ Atomic mass of Na+ Atomic mass of O

= 2 $\times$23 u + 16 u

= 62 u.

Formula unit mass of K₂CO₃ = 2$\times$ Atomic mass of K+ Atomic mass of C + 3 $\times$ Atomic mass of O

= 2 $\times$ 39 u + 12 u + 3 $\times$ 16 u

= 138 u.

Question 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer:

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

So, by the law of definite proportions,

When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 gram of carbon dioxide. The remaining 42.00 g of oxygen will remain unreacted. Law of constant proportion is Held.

Q 3. What are polyatomic ions? Give examples.

Answer:

Polyatomic ions are ions that contain more than one atom. These atoms can be of the same type or of a different type.

Some examples of polyatomic ions are NH₄+ , OH^- , SO₄^2- , and SO₃^2- .

Question 4. Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Answer:

The chemical formula of Given compounds are :

(a) Magnesium chloride

$=MgC{l}_2$

(b) Calcium oxide

$=CaO$

(c) Copper nitrate

$=Cu(N{O}_3{)}_2$

(d) Aluminium chloride

$=AlC{l}_3$

(e) Calcium carbonate

$=CaC{O}_3$

Question 5. Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Answer:

The names of the elements present in the given compounds are :

(a) Quick lime :

$CaO$ = Calcium and Oxygen.

(b) Hydrogen bromide:

$HBr$ = Hydrogen and Bromine.

(c) Baking powder:

$NaHC{O}_3$ = Sodium, Hydrogen, Carbon, and Oxygen.

(d) Potassium sulphate:

$K_{2}S{O}_4$ = Potassium, Sulphur, and oxygen.

Question 6.(a) Calculate the molar mass of the following substances.

• Ethylene , $C_2{H}_2$

Answer:

The molecular mass of ethyne, $C_2{H}_2$ = 2 $\times$Atomic Mass of C + 2 $\times$ Atomic Mass of H

= 2 $\times$ 12u + 2 $\times$ 1u

= 24u + 2u

= 26u

Question 6. (b) Calculate the molar mass of the following substances.

• Sulphur molecule , S₈

Answer:

The molecular mass of Sulphur molecule, = 8 $\times$ Atomic Mass of S

= 8 $\times$ 32u

= 256u

Question 6.(c) Calculate the molar mass of the following substances

• Baking soda

Answer:

The molecular formula of Baking soda is NaHCO₃

So the molecular mass will be some of the individual atoms involved in the formula.

Molecular Mass of baking soda = Molecular mass (Na + H + C + 3xO)

= 23 + 1 + 12 + 3x16

= 84 g/mol

Question 6.(d) Calculate the molar mass of the following substances.

• Hydrochloric acid, HCl

Answer:

The molecular mass of Hydrochloric acid, HCl = 1 $\times$ Atomic Mass of Cl + 1 $\times$ Atomic Mass of H

= 35.5 u + 1u

= 36.5u

Question 6. (e) Calculate the molar mass of the following substances.

• Nitric acid, HNO₃

Answer:

The molecular mass of Nitric acid, HNO₃ = 1 $\times$ Atomic Mass of N + 1 $\times$ Atomic Mass of H + 3 $\times$ Atomic Mass of O

= 1$\times$ 14u + 1 $\times$1u + 3 $\times$ 16u

= 14u + 1u + 48u

= 63u