Atoms and Molecules

Class 9 – NCERT Science Solutions

Intext Questions 1

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

As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.

Sodium carbonate + acetic acid ⟶ Sodium acetate + carbon dioxide + water

5.3 g + 6 g ⟶ 8.2 g + 2.2 g + 0.9 g

⇒ 11.3 g ⟶ 11.3 g

As per the above reaction, L.H.S. = R.H.S. = 11.3 g

Hence, the observations are in agreement with the law of conservation of mass.

Question 2

Hydrogen and oxygen combine in a 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 : water = 1 : 8

So, for every 1 g of hydrogen, 8 g of oxygen is required

∴ For 3 g of hydrogen, oxygen required is 8118​ x 3 = 24 g

Hence, 24 g of oxygen would be required for the complete reaction with 3 g of hydrogen gas.

Question 3

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

Answer

One of the postulates of Dalton’s atomic theory states that : ‘Atoms cannot be created nor be destroyed in a chemical reaction’. This law is the result of law of conservation of mass.

Question 4

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

Answer

The postulate of Dalton’s atomic theory that can explain the law of definite proportions is : ‘Relative number and kinds of atoms are equal in given compounds.’

Intext Questions 2

Question 1

Define the atomic mass unit.

Answer

Atomic mass unit is defined as 1⁄12 ^th the mass of an carbon atom C-12.

Question 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:

1. Atoms are very small in size, measured in nanometres.
2. Except for atoms of noble gases, they do not exist independently.

Intext Questions 3

Question 1

Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Answer

(i) Sodium oxide:

Step 1 — Write each symbol with its valency

Na1+O2−Na1+↗O2−

Step 2 — Interchange the valencies

Na21 ↘↙ O2⇒Na12 ↘↙ O1Na21 ↘↙ O2⇒12Na​ ↘↙ 1O​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Sodium oxide : Na2ONa2​O

(ii) Aluminium chloride

Step 1 — Write each symbol with its valency

Al3+Cl1−Al3+↗Cl1−

Step 2 — Interchange the valencies

Al23 ↘↙ Cl1⇒Al11 ↘↙ Cl3Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Aluminium chloride : AlCl3AlCl3​

(iii) sodium sulphide

Step 1 — Write each symbol with its valency

Na1+S2−Na1+↗S2−

Step 2 — Interchange the valencies

Na21 ↘↙ S2⇒Na12 ↘↙ S1Na21 ↘↙ S2⇒12Na​ ↘↙ 1S​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Sodium sulphide : Na2SNa2​S

(iv) Magnesium hydroxide

Step 1 — Write each symbol with its valency

Mg2+OH1−Mg2+↗OH1−

Step 2 — Interchange the valencies

Mg22 ↘↙ OH1⇒Mg11 ↘↙ OH2Mg22​ ↘↙ OH1⇒11Mg​ ↘↙ 2OH​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Magnesium hydroxide : Mg(OH)2Mg(OH)2​

Question 2

Write down the names of compounds represented by the following formulae:

(i) Al₂(SO₄)₃

(ii) CaCl₂

(iii) K₂SO₄

(iv) KNO₃

(v) CaCO₃

Answer

(i) Al₂(SO₄)₃ — Aluminium sulphate

(ii) CaCl₂ — Calcium chloride

(iii) K₂SO₄ — Potassium sulphate

(iv) KNO₃ — Potassium nitrate

(v) CaCO₃ — 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. It denotes in a compound, the number of atoms of each element present.

Question 4

How many atoms are present in a

(i) H₂S molecule and

(ii) PO₄^3- ion ?

Answer

(i) In one molecule of H₂S, 2 atoms of hydrogen and 1 atom of sulphur are present. Hence, 3 atoms in total are present.

(ii) In one ion [PO₄^3-], 1 atom of phosphorus and 4 atoms of oxygen are present. Hence, 5 atoms in total are present.

Intext Questions 4

Question 1

Calculate the molecular masses of

1. H₂
2. O₂
3. Cl₂
4. CO₂
5. CH₄
6. C₂H₆
7. C₂H₄
8. NH₃
9. CH₃OH.

Answer

1. The molecular mass of H₂ = 2 x atoms atomic mass of H = 2 x 1u = 2u
2. The molecular mass of O₂ = 2 x atoms atomic mass of O = 2 x 16u = 32u
3. The molecular mass of Cl₂ = 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u
4. The molecular mass of CO₂ = atomic mass of C + (2 x atomic mass of O) = 12 + (2 × 16)u = 12 + 32 = 44u
5. The molecular mass of CH₄ = atomic mass of C + (4 x atomic mass of H) = 12 + (4 x 1) = 16u
6. The molecular mass of C₂H₆ = (2 x atomic mass of C) + (6 x atomic mass of H) = (2 x 12) + (6 x 1) = 24 + 6 = 30u
7. The molecular mass of C₂H₄ = (2 x atomic mass of C) + (4 x atomic mass of H) = (2 x 12) + (4 x 1) = 24 + 4 = 28u
8. The molecular mass of NH₃ = atomic mass of N + (3 x atomic mass of H) = 14 + (3 x 1) = 17u
9. The molecular mass of CH₃OH = atomic mass of C + (3 x atomic mass of H) + atomic mass of O + atomic mass of H = [12 + (3 × 1) + 16 + 1] = (12 + 3 + 17) = 32u

Question 2

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

Answer

Given:

Atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u

The formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u

The formula unit mass of Na₂O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K₂CO₃ = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u

Exercises

Question 1

A 0.24 g sample of a 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.

Answer

Given,

Mass of the sample compound = 0.24 g,

mass of boron = 0.096 g,

mass of oxygen = 0.144 g

To calculate the percentage composition of the compound,

Percentage of boron :

= mass of boronmass of the compoundmass of the compoundmass of boron​ x 100

= 0.0960.240.240.096​ x 100

= 40%

Percentage of oxygen

= mass of oxygenmass of the compoundmass of the compoundmass of oxygen​ x 100

= 0.1440.240.240.144​ x 100

= 60%

Hence, percentage of boron = 40% and percentage of oxygen = 60% in the compound.

Question 2

When 3.0 g of carbon is burnt in 8.00 g of 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

Given,

C 3 g+O2 8 g⟶CO211 g 3 gC​+ 8 gO2​​⟶11 gCO2​​

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

Now,

C 3 g+O2 50 g⟶CO2? 3 gC​+ 50 gO2​​⟶?CO2​​

Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO₂ and 42 g [i.e., 50 g – 8 g] of oxygen will be left unused.

The above answer is governed by the law of constant proportions.

Question 3

What are polyatomic ions? Give examples.

Answer

Cluster of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them.

Example: Hydroxide [OH^–], Cyanide [CN^–]

Question 4

Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Answer

(a) Magnesium chloride

Step 1 — Write each symbol with its valency

Mg2+Cl1−Mg2+↗Cl1−

Step 2 — Interchange the valencies

Mg22 ↘↙ Cl1⇒Mg11 ↘↙ Cl2Mg22​ ↘↙ Cl1⇒11Mg​ ↘↙ 2Cl​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Magnesium chloride : MgCl2MgCl2​

(b) Calcium oxide

Step 1 — Write each symbol with its valency

Ca2+O2−Ca2+↗O2−

Step 2 — Interchange the valencies

Ca22 ↘↙ O2⇒Ca12 ↘↙ O2Ca22 ↘↙ O2⇒12Ca​ ↘↙ 2O​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Calcium oxide : CaOCaO

(c) Copper nitrate

Step 1 — Write each symbol with its valency

Cu2+NO31−Cu2+↗NO31−​

Step 2 — Interchange the valencies

Cu22 ↘↙ NO31⇒Cu11 ↘↙ NO32Cu22 ↘↙ NO3​1​⇒11Cu​ ↘↙ 2NO3​​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of copper nitrate : Cu(NO3)2Cu(NO3​)2​

(d) Aluminium chloride

Step 1 — Write each symbol with its valency

Al3+Cl1−Al3+↗Cl1−

Step 2 — Interchange the valencies

Al23 ↘↙ Cl1⇒Al11 ↘↙ Cl3Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Aluminium chloride : AlCl3AlCl3​

(e) Calcium carbonate –

Step 1 — Write each symbol with its valency

Ca2+CO31−Ca2+↗CO31−​

Step 2 — Interchange the valencies

Ca22 ↘↙ CO31⇒Ca11 ↘↙ CO32Ca22 ↘↙ CO3​1​⇒11Ca​ ↘↙ 2CO3​​

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of copper nitrate : Ca(CO3)2Ca(CO3​)2​

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

(a) Quick lime (CaO) — Calcium and oxygen

(b) Hydrogen bromide (HBr) — Hydrogen and bromine

(c) Baking powder (NaHCO₃) — Sodium, hydrogen, carbon and oxygen

(d) Potassium sulphate (K₂SO₄) — Potassium, sulphur and oxygen,

Question 6

Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂

(b) Sulphur molecule, S₈

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

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO₃

Answer

(a) Molar mass of Ethyne C₂H₂ = (2 x Mass of C) + (2 x Mass of H) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g

(b) Molar mass of Sulphur molecule S₈ = (8 x Mass of S) = 8 x 32 = 256 g

(c) Molar mass of Phosphorus molecule, P₄ = 4 x Mass of P = 4 x 31 = 124 g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl = 1 + 35.5 = 36.5 g

(e) Molar mass of Nitric acid, HNO₃ = Mass of H + Mass of N + (3 x Mass of O) = 1 + 14 + (3 × 16) = 15 + 48 = 63 g