3, Atoms and Molecules Class 9 Science Chapter 3 NCERT Solutions (CBSE)

CBSE Class 9 Science Chapter 3: Atoms and Molecules – Full NCERT Textbook Solutions

Chapter 3: Atoms and Molecules

Internal Questions:

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 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

Solution:Given, Sodium carbonate () = 5.3 g  , Acetic acid () = 6 g , Sodium acetate ( ) = 8.2 g , Carbon dioxide () = 2.2 g  , Water () = 0.9

We have, Sodium carbonate () + Acetic acid () → Sodium acetate () + Carbon dioxide () + Water ()

LHS : Sodium carbonate () + Acetic acid ()

= 5.3 g + 6 g =11.3 g

RHS : Sodium acetate () + Carbon dioxide () + Water ()

= 8.2 g + 2.2 g + 0.9 g = 11.3 g

The total mass of reactants (11.3 g) is equal to the total mass of products (11.3 g). Therefore, these observations are in agreement with the law of conservation of mass. The mass is conserved in the chemical reaction.
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: According to the given ratio of hydrogen and oxygen by mass = 1 : 8

Given, Mass of hydrogen gas = 3 g

Let  be the mass of oxygen gas .

A/Q,  

So , 

 

 

Therefore, 24 grams of oxygen gas would be required to react completely with 3 grams of hydrogen gas to form water.
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer: The postulate stating that atoms cannot be created or destroyed in a chemical reaction is based on the law of conservation of mass, as total mass remains constant during a reaction.

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

Answer: The postulate that the relative number and kinds of atoms are constant in a given compound explains the law of definite proportions, as a compound always contains the same elements in fixed proportions by mass.

Internal Questions:

1. Define the atomic mass unit.

Answer: The atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12 .
2. Why is it not possible to see an atom with naked eyes?

Answer: It is not possible to see an atom with naked eyes because atoms are extremely small in size. Their size is of the order of nanometres, which is much smaller than what our eyes can detect, so we cannot see them directly.

Internal Questions:

1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide

Answer:  (i) Sodium oxide:

(ii) Aluminium chloride:

(iii) Sodium sulfide:

(iv) Magnesium hydroxide:

2. Write down the names of compounds represented by the following formulae:
(i)        (ii)       (iii)      (iv)      (v)   

Answer: (i)  : Aluminum sulphate
(ii)   : Calcium chloride
(iii)  : Potassium sulphate
(iv)  : Potassium nitrate
(v)  : Calcium carbonate
3. What is meant by the term chemical formula?

Answer: A chemical formula is a symbolic representation of a compound that shows which elements are present and the number of atoms of each element in it.

4. How many atoms are present in a
(i)  molecule and  (ii)   ion ?

Answer: (i) The number of atoms of  is 3 (= 2 + 1) .
(ii) The number of atoms of  is 5 (= 1 + 4) .

Example 3.1 : (a) Calculate the relative molecular mass of water (  ).
(b) Calculate the molecular mass of  .

Solution: (a) Atomic mass of hydrogen = 1u

Atomic mass of oxygen = 16 u

The molecular mass of water (  ) = 2 × 1 +1 × 16 = 2 + 16 = 18 u

 (b) Atomic mass of hydrogen = 1u

Atomic mass of nitrogen = 14 u

Atomic mass of oxygen = 16 u

The molecular mass of  = 1 × 1 + 1 × 14 + 3 × 16 = 1 + 14 + 48 = 63 u

Example 3.2 : Calculate the formula unit mass of .

Solution: Atomic mass of calcium = 40 u

Atomic mass of chlorine = 35.5 u

The formula unit mass of  = 1 × 40 + 2 × 35.5 = 40 +70 = 110 u

Internal Questions:

1. Calculate the molecular masses of   ,  ,  , ,  ,  , , ,  ,  

Solutuion: The molecular masses of  = 2 × 1 = u

The molecular masses of  = 2 × 16 =32 u

The molecular masses of = 2 × 35.5 = 70.0 u

The molecular masses of  = 2 × 1 = 2 u

 The molecular masses of  = 1 × 12 + 2 × 16 = 12 + 32 = 44 u

The molecular masses of  = 1 × 12 + 4 × 1 = 12 + 4 = 16 u

The molecular masses of = 2 × 12 + 6 × 1 = 24 + 6 = 30 u

The molecular masses of = 2 × 12 + 4 × 1 = 24 + 4 = 28 u

The molecular masses of = 1 × 14 + 3 × 1 = 14 + 3 = 17 u

The molecular masses of  = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 12 + 3 +16 + 1 =32 u
2. Calculate the formula unit masses of  , , ,
[Given atomic masses of Zn = 65 u , Na = 23 u, K = 39 u, C = 12 u and O = 16 u. ]

Solution: The formula unit masses of = 1 × 65 + 1 × 16 = 81 u

The formula unit masses of  = 2 × 23 + 1 × 16 = 46 +16 =62 u

The formula unit masses of = 2 × 39 +1×12 +3×16 =78+12+48 = 138 u

Exercise :

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Solution:  Given, Mass of the compound = 0.24 g

Mass of boron in the compound = 0.096 g

Mass of oxygen in the compound = 0.144 g

Percentage composition of boron

Percentage composition of boron

Percentage composition of boron = 40%

Percentage composition of oxygen

Percentage composition of oxygen

Percentage composition of oxygen = 60%

So, the compound of oxygen and boron has a percentage composition by weight of 40% boron and 60% oxygen.

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?

Solution: In the first case:                                               
Carbon (3.0 g) + Oxygen (8.0 g) → Carbon dioxide (11.0 g)
Ratio of C : O = 3 : 8

In the second case:
Carbon = 3.00 g
Oxygen = 50.00 g (but carbon dioxide needs only 8.00 g oxygen for 3.00 g carbon)
So, oxygen is in excess. Only 8.00 g oxygen will react.

Mass of CO₂ formed = 3.00 g (C) + 8.00 g (O) = 11.00 g

According to the Law of Constant Proportions (or Law of Definite Proportions), a chemical compound always contains the same elements in a fixed mass ratio. Therefore, this reaction is governed by the Law of Constant Proportions (or Law of Definite Proportions).

3. What are polyatomic ions? Give examples.

Answer: Clusters of atoms that act as an ion are called polyatomic ions. Examples of polyatomic ions are Ammonium ion (NH₄⁺) , Nitrate ion ( NO₃⁻) , Sulphate ion (SO₄²⁻) and Carbonate ion: CO₃²⁻

4. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Answer: (a) Magnesium chloride: MgCl₂

(b) Calcium oxide: CaO

(c) Copper nitrate: Cu(NO₃)₂

(d) Aluminium chloride: AlCl₃

(e) Calcium carbonate: CaCO₃
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: (a) Quick lime – Calcium, Oxygen
(b) Hydrogen bromide – Hydrogen, Bromine
(c) Baking powder – Sodium, Hydrogen, Carbon, Oxygen
(d) Potassium sulphate – Potassium, Sulphur, Oxygen

6. Calculate the molar mass of the following substances.
(a) Ethyne,   
(b) Sulphur molecule,  
(c) Phosphorus molecule,   (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid,  

Answer: (a) Ethyne, C₂H₂:

Molar mass of C₂H₂ = 2 × 12 + 2 × 1 = 24 + 2 = 26 g

 (b) Sulphur molecule, S₈:

 Molar mass of S₈ = 8 × 32 = 256 g

 (c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31):

Molar mass of  = 4 × 31 = 124 g

 (d) Hydrochloric acid, HCl:

Molar mass of HCl = 1 × 1 + 1 × 35.5 = 1 + 35.5 = 36.5 g

 (e) Nitric acid, HNO₃ :

Molar mass of HNO₃ = 1 × 1 + 1 × 14 + 3 × 16 =1 + 14 + 48 = 63 g

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Posted 4 months ago