3 . Atoms and Molecules

Assam Class 9 Science Chapter 3 . Atoms and Molecules

Chapter 3 . Atoms and Molecules

Class 9 Science Chapter 3 . Atoms and Molecules Internal / Example Questions and Answers

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: We have,

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

Here, Sodium carbonate () = 5.3 g

Acetic acid () = 6 g

Sodium acetate () = 8.2 g

Carbon dioxide () = 2.2 g

Water () = 0.9

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 law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. The total mass of reactants should be equal to the total mass of products.

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?

Solution : 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?

Answers : The postulate of Dalton's atomic theory that is a result of the law of conservation of mass is:

‘‘Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. ’’

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

Answers : The postulate of Dalton's atomic theory that can explain the law of definite proportions is the postulate stating that compounds are formed by the combination of atoms in fixed ratios, leading to consistent proportions of elements in a compound.

Internal Questions :

1. Define the atomic mass unit.

Answers : 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?

Answers : It is not possible to see an atom with naked eyes because atoms are very small, they are smaller than anything that we can imagine or compare with.

Internal Questions :

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

Answers : (i) Sodium oxide:

(ii) Aluminium chloride:

(iii) Sodium sulphide:

(iv) Magnesium hydroxide:

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

Answers : (i) : Aluminum sulphate
(ii) : Calcium chloride
(iii) : Potassium sulphate
(iv) : Potassium nitrate
(v) : Calcium carbonate

3. What is meant by the term chemical formula?

Answers : The term chemical formula is a symbolic representation of the elements present in a compound and the ratio of their atoms. It provides information about the composition of a substance at the atomic level.

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

Answers : (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 , , , , , , , , ,

Solution : 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

Example 3.3 :
1. Calculate the number of moles for the following:
(i) 52 g of He (finding mole from mass)
(ii) number of He atoms (finding mole from number of particles).

Solution : (i) Here, M = 4 g , m = 52 g , n = ?

We know that, the number of moles

(ii) Here, , and

We have, the number of moles

Example 3.4 : Calculate the mass of the following:
(i) 0.5 mole of   gas (mass from mole of molecule)
(ii) 0.5 mole of N atoms (mass from mole of atom)
(iii)    number of N atoms (mass from number)
(iv)      number of   molecules (mass from number)

Solution :

Solution: (i) Here, n = 0.5 , M = 2 × 14 = 28 g , m = ?

We know that,

(ii) Here, n = 0.5 , M = 14 g , m = ?

We know that,

(iii) Here, , , M = 14 g and

We know that,

(iv) Here, ,, M = 2 × 14 g = 28 g and

We know that,

Example 3.5 Calculate the number of particles in each of the following:

(i) 46 g of Na atoms (number from mass)

(ii) 8 g molecules (number of molecules from mass)

(iii) 0.1 mole of carbon atoms (number from given moles)

Solution: (i) Here, M = 23 g , m = 46 g , and

We know that,

Therefore, the number of particles is .

(ii) Here, M = 2×16 = 32 g , m = 8 g , and

We know that,

Therefore, the number of particles is .

(iii) Here, , n = 0.1 and

We have,

Internal Questions :

1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon ?

Solution : Given, 1 mole of carbon atoms = 12 g

Avogadro's number (

Mass of 1 atom of carbon

So, one atom of carbon weighs grams.

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u) ?

Solution : For sodium (Na): Given, atomic mass of Na (M) = 23 g

Mass of Na in grams (m)= 100 grams

Avogadro's number ( , N = ?

We know that

Therefore, number of atoms of Na is atoms

For iron (Fe) : Given, atomic mass of Fe = 56 g

Mass of Fe in grams = 100 grams

Avogadro's number ( , N = ?

We know that

Therefore, number of moles of Fe is atoms .

Class 9 Science Chapter 3 . Atoms and Molecules Exercies Questions and Answers :

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 = 40%

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 : Given, Molar mass of carbon (C) = 12 g

Molar mass of oxygen () = 16 g

Molar mass of carbon dioxide () = 12 g (C) + 16 g() = 44 g

The chemical equation of the reaction is :

The ratio between carbon and carbon dioxide:

Moles of C mole

Moles of

Moles of mole

Since, 1:1 stoichiometric ratio between carbon and carbon dioxide, the moles of produced will be 0.25 mol.

According to the given information,

When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

When 11.0 g of carbon dioxide is formed when 3.00 g of carbon will be burnt in 50.00 g of oxygen consume 8 g of oxygen. So, the mass of oxygen = 50 g – 8 g = 42 g

The law of chemical combination that governs this answer is the law of definite proportions. This law states that in a chemical compound, the elements are always present in fixed and definite proportions by mass.

3. What are polyatomic ions? Give examples.

Answers : A group of atoms carrying a charge is known as a polyatomic ions .

For Example :

(i) Sodium Chloride: NaCl

Here, is positive charged ion and is negative charged ion .

(ii) Ammonium ion: NH₄⁺

It is a positively charged ion formed by combining one nitrogen atom with four hydrogen atoms.

(iii) Nitrate ion: NO₃⁻

It is a negatively charged ion formed by combining one nitrogen atom with three oxygen atoms.

(iv)Sulphate ion: SO₄²⁻

It is a negatively charged ion formed by combining one sulfur atom with four oxygen atoms.

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

Answers : (a) Magnesium chloride: MgCl₂

(b) Calcium oxide: CaO

(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.

Answers : (a) Quick lime :

The chemical formula for quicklime is CaO.

It contains the elements Calcium (Ca) and Oxygen (O).

(b) Hydrogen bromide:

The chemical formula for hydrogen bromide is HBr.

It contains the elements Hydrogen (H) and Bromine (Br).

The chemical formula for hydrogen bromide is NaHCO₃ .

It contains the elements Sodium (Na), Hydrogen (H), Carbon (C), and Oxygen (O).

(d) Potassium sulphate:

The chemical formula for potassium sulfate is K₂SO₄.

It contains the elements Potassium (K), Sulphur (S), and Oxygen (O).

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,

Solution: (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

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

7. What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite ( ) ?

Solution: (a) Here, mole , Molar mass g and Given mass,

We know that,

Therefore, the mass of nitrogen atom is 14 g .

(b) Here, mole , Molar mass g and Given mass,

We know that,

Therefore, mass of 4 moles of aluminium atoms = 108 g

Here, moles ,

Molar mass = 2 × 23 + 1 × 32 + 16 × 3 = 46 + 32 + 48 =126 g

We know that,

Therefore, mass of 10 moles of sodium sulphite = 1260 g

8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.

Solution: (a) 12 g of oxygen gas (O₂): Here, , m = 12 g

Molar mass of oxygen gas (M) = 2×16 g = 32 g

We know that,

Number of moles

Therefore, number of moles = 0.375 moles

(b) 20 g of water (H₂O): Here, , g

Molar mass of H₂O (M) = 18 g

We know that,

Number of moles

Therefore, number of moles = 1.11 moles

Molar mass of CO₂ (M) = 1 × 12 + 2 × 16

=12 +32 = 44 g

We know that,

Number of moles

Therefore, number of moles = 0.500 moles.

9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Solution: (a) Mass of 0.2 mole of oxygen atoms (O): Here, n = 0.2 mole ,

Molar mass of oxygen atoms (M) = 16 g , Given mass (m)= ?

We know that,

Therefore, the mass of oxygen atoms = 3.2 g .

(b) Mass of 0.5 mole of water molecules (H₂O): Here, mole

Molar mass of H₂O (M) = 18 , Given mass (m) = ?

We know that,

Therefore, the mass of water molecules = 9 g

10. Calculate the number of molecules of sulphur () present in 16 g of solid sulphur.

Solution: Given, Mass of solid sulphur (m) = 16 g

Molar mass of sulphur (M) = 8 × 32 g = 256 g

Avogadro's number ,

We know that,

So, there are molecules of S₈ present in 16 g of solid sulphur.

11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u ) .

Solution: Given, aluminium oxide =

Mass of aluminum oxide (m) = 0.051 g

Mass of aluminum (Al) = 27 g

Mass of aluminium oxide (M) = 2 × 27 + 3 × 16 = 54 + 48 = 102 g

Avogadro's number of ,

We know that,

So, there are aluminum ions present in 0.051 g of aluminum oxide.