Chapter 3 Atoms and Molecules

NIOS Class 10 Science Chapter 3 Question Answers

INTEXT QUESTIONS 3.1

1. Name the scientists who proposed the law of conservation of mass and law of constant proportions.

Ans: Lavoisier proposed the law of conservation of mass, and Proust proposed the law of constant proportions

2. 12 g of magnesium powder was ignited in a container having 20 g of pure oxygen. After the reaction was over, it was found that 12 g of oxygen was left unreacted. Show that it is according to law of constant proportions.

2Mg + O₂ ⎯→ 2MgO

Ans: In the container, 12g of Oxygen was left unreacted.

Therefore, the amount of unreacted Oxygen = (20 – 12)g = 08g.

Thus, 12g of magnesium reacted with 8g of oxygen in the ratio 12:8.

This is what we expected for MgO, i.e 24g of Mg reacted with 16g of Oxygen or 12g of Mg will react with 8g of Oxygen.

INTEXT QUESTIONS 3.2

1. Nitrogen forms three oxides: NO, NO₂ and N₂O₃. Show that it obeys law of multiple proportions.

Ans: Atomic mass of nitrogen is 14u and that of oxygen is 16u.

In NO, 14g of nitrogen reacted with 16g of oxygen

In NO₂, 14g of nitrogen reacted with 32g of oxygen

In N₂O₃, 28g of nitrogen reacted with 48g of oxygen

14g of nitrogen reacted with 24g of Oxygen.

Therefore, the amount of oxygen which reacts with 12g of nitrogen in the case of NO, NO₂ and N₂O₃ will be in the ratio of 16:32:24 or 2:4:3. This proves the law of multiple proportions.

2. Atomic number of silicon is 14. If there are three isotopes of silicon having 14, 15 and 16 neutrons in their nuclei, what would be the symbol of the isotope?

Ans: (ii) Atomic number of Si is 14

Mass number of silicon atoms having 14,15, and 16 neutrons will be 28, 29, and 30, respectively, and therefore symbols of isotopes of silicon will be

3. Calculate molecular mass of the compounds whose formulas are provided below:

C₂H₄, H₂O and CH₃OH

Ans:

Molecular mass of C₂H₄ = mass of two atoms of carbon + mass of 4 atoms of hydrogen

= 2 × 12u + 4 × 1u = 28u

Molecular mass of H₂O = mass of two atoms of hydrogen + mass of one atom of oxygen

= 2 × 1u + 1 × 16u = 18u

Molecular mass of CH₃OH = mass of one atom of carbon + mass of 4 atoms of hydrogen + mass of one atom of oxygen

= 1 × 12u + 4 × 1u +1 × 16u = 32u

INTEXT QUESTIONS 3.3

1. Work out a relationship between number of molecules and mole.

Ans: 1 mole of a substance contains 6.023 × 10²³ molecules of that substance

i.e 1 mole of a substance = 6.023 × 10²³ molecules of that substance.

2. What is molecular mass? In what way is it different from the molar mass?

Ans: Molecular mass is the sum of the atomic masses of all the atoms present in that molecule. Molecular mass is the mass of one molecule, whereas molar mass is the mass of 1 mol or 6.023 × 10²³ elementary entities (atoms, molecules, ions)

3. Consider the reaction

C (s) + O₂ (g) ⎯→ CO₂ (g)

18 g of carbon was burnt in oxygen. How many moles of CO2 are produced?

Ans:

12 g of carbon gives 1 mole of CO₂

18g of carbon will give 1.5 moles of CO₂

4. What is the molar mass of NaCl?

Ans: Molar mass of NaCl = ( 23.0 + 35.5 )g mol^-1= 58.5 g mol^-1

INTEXT QUESTIONS 3.4

1. Write the name of the expected compound formed between

(i) hydrogen and sulphur

(ii) nitrogen and hydrogen

(iii) magnesium and oxygen

Ans:

(i) H₂S
(ii) NH₃
(iii) MgO

2. Propose the formulas and names of the compounds formed between

(i) potassium and iodide ions

(ii) sodium and sulphate ions

(iii) aluminium and chloride ions

Ans:

(i) KI, Potassium iodide
(ii) Na₂SO₄, Sodium Sulphate
(iii) AlCl₃, Aluminium Chloride

3. Write the formula of the compounds formed between

(i) Hg²⁺ and Cl^–

(ii) Pb²⁺ and PO ^3–₄

(iii) Ba²⁺ and 2^–

Ans:

(i) HgCl₂
(ii) Pb₃(PO₄)₂
(iii) BaSO₄

TERMINAL EXERCISE

1. Describe the following:

(a) Law of conservation of mass

(b) Law of constant proportions

(c) Law of multiple proportions

Ans:

(a) Law of conservation of mass: In every chemical reaction, the total mass of all the reactants is equal to the mass of all the products.
(b) Law of constant proportions: In a given chemical compound, the proportions by mass of the elements that compose it are fixed, independent of the origin of the compound or its mode of preparation.
(c) Law of multiple proportions: When two elements form more than one compound, the masses of one element in these compounds for a fixed mass of the other element are in the ratio of small whole numbers.

2. What is the atomic theory proposed by John Dalton? What changes have taken place in the theory during the last two centuries?

Ans: The following statements comprise the atomic theory of matter:

1. Matter consists of indivisible atoms.
2. All the atoms of a given chemical element are identical in mass and in all other properties.
3. Different chemical elements have different kinds of atoms, and in particular, such atoms have different masses.
4. Atoms are indestructible and retain their identity in chemical reactions.
5. The formation of a compound from its elements occurs through the combination of atoms of unlike elements in a small whole number ratio.

Changes in this theory during the last two centuries.

All atoms of naturally occurring elements are not of the same mass.
An atom consists of several fundamental particles such as electrons, protons and neutrons.
The compound atoms which Dalton referred to are today known as a molecule.

3. Write the number of protons, neutrons, and electrons in each of the following isotopes :

²₁H

¹⁸₈O

¹⁹₉F

⁴⁰₂₀Ca

Ans:

²₁H: 1 Proton and 1 electron. Neutron = 2-1=1
¹⁸₈O: 8 protons and 8 electrons. Neutron = 18-8=10.
¹⁹₉F: 9 Protons and 9 electrons. Neutron = 19-9= 10
⁴⁰₂₀Ca: 20 Protons and 20 electrons. Neutrons = 40-20= 20.

4. Boron has two isotopes with masses 10.13 u and 11.01 u and abundance of 19.77% and 80.23% respectively. What is the average atomic mass of boron?

(Ans. 10.81 u)

Ans:

Abundance 19.77% = 0.1977

Abundance 80.23% = 0.8023

Avg atomic mass = (10.13 x 0.1977) + (11.01 x 0.8023)

= 2.002 + 8.834

= 10.836

Avg atomic mass of boron = 10.836

5. Give symbol for each of the following isotopes:

(a) Atomic number 19, mass number 40

(b) Atomic number 7, mass number 15

(c) Atomic number 18, mass number 40

(d) Atomic number 17, mass number 37

Ans:

(a) Element with atomic number 19 = Potassium (K) → ⁴⁰₁₉K

(b) Element with atomic number 7 = Nitrogen (N) → ¹⁵₇N

(d) Element with atomic number 17 = Chlorine (Cl) → ³⁷₁₇Cl

6. How does an element differ from a compound? Explain with suitable examples.

Ans: An element may be defined as a substance where all the atoms have the same atomic number. A compound is a type of matter containing the atoms of two or more elements in a small whole-number ratio. Eg., Carbon dioxide is a compound that contains one carbon atom and 2 oxygen atoms to form CO₂.

7. Charge of one electron is 1.6022×10^–19 coulomb. What is the total charge on 1 mol of electrons?

Ans:

Charge of one electron = 1.6022 × 10⁻¹⁹ C

Number of electrons in 1 mole = 6.022 × 10²³ (Avogadro’s number)

Total charge = (1.6022 × 10⁻¹⁹) × (6.022 × 10²³)

= 9.65 × 10⁴ C

8. How many molecules of O₂ are in 8.0 g of oxygen? If the O₂ molecules were completely split into O (Oxygen atoms), how many mole of atoms of oxygen would be obtained?

Ans:

Number of moles, if oxygen molecules are split into atoms = 0.25 x 2 = 0.50 mol.

9. Assume that human body is 80% water. Calculate the number of molecules of water that are present in the body of a person whose weight is 65 kg.

Ans:

Mass of water in the body

The human body is 80% water.

Weight of person = 65 kg

Mass of water = 0.80 × 65 = 52 kg

Mass of water=0.80×65=52 kg = 52000 g

Moles of water

Molar mass of water (H₂O) = 18 g/mol

Moles of water = 52000/18 ≈ 2888.9 mol

Number of molecules of water

Number of molecules = moles × Avogadro’s number

Avogadro’s number = 6.022 × 10²³

= 2888.9 × 6.022 × 10²³

= 2888.9×6.022×10²³≈ 1.74 × 10²⁷ molecules

≈1.74×10²⁷ molecules

10. Refer to atomic masses given in the Table (3.2) of this chapter. Calculate the molar masses of each of the following compounds :

HCl, NH₃, CH₄, CO and NaCl

Ans: Molecular mass is the sum of the atomic mass of all the atoms present in that molecule.

HCl:

Molecular mass of HCl = Mass of hydrogen + Mass of Chlorine

= 1.008 + 35.46

= 36.468u

NH₃:

Molecular mass of NH₃ = Mass of nitrogen + Mass of 3 atoms of hydrogen

= 14.01 + (3 x 1.008)

= 14.01 + 3.024

= 17.024u

CH₄:

Molecular mass of CH₄ = Mass of Carbon + mass of 4 atoms of hydrogen

= 12.01 + (4 x 1.008)

= 12.01 + 4.032

= 16.042u

NaCl:

Molecular mass of NaCl = Mass of sodium + Mass of chlorine

= 23 + 35.45

= 58.45u

11. Average atomic mass of carbon is 12.01 u. Find the number of moles of carbon in

(a) 2.0 g of carbon.

(b) 8.0 g of carbon.

Ans:

12. Classify the following molecules as di, tri, tetra, penta and hexa atomic molecules:

H₂, P₄, SF₄, SO₂, PCl₃, CH₃OH, PCl₅, HCl

Ans:

H₂ – Diatomic
P₄ – Tetratomic
SF₄ – Pentatomic
SO₂ – Triatomic
PCl₃– Tetratomic
CH₃OH – Hexatomic
PCl₅ – Hexatomic
HCl – Diatomic

13. What is the mass of

(a) 6.02×10²³ atoms of oxygen

(b) 6.02×10²³ molecules of P₄

(c) 3.01×10²³ molecules of O₂

Ans:

(a) 6.02 × 10²³ atoms of oxygen

6.02 × 10²³ atoms = 1 mole
Atomic mass of oxygen (O) = 16 g/mol

Mass = 16 g

(b) 6.02 × 10²³ molecules of P₄

6.02 × 10²³ molecules = 1 mole
Atomic mass of P = 31 g/mol (approx.)
Molar mass of P₄ = 4 × 31 = 124 g

Mass = 124 g

(c) 3.01 × 10²³ molecules of O₂

3.01 × 10²³ = 0.5 mole (half of Avogadro’s number)
Molar mass of O₂ = 32 g/mol

Mass = 0.5 × 32 = 16 g

14. How many atoms are present in:

(a) 0.1 mol of sulphur

(b) 18 g of water (H₂O)

(c) 0.44 g of carbon dioxide (CO₂)

Ans:

(a) 0.1 mol of sulphur

One mole of Sulphur contains 6.022 x 10²³ atoms

No. of atoms in 0.1 mol = 0.1 x 6.022 x 10²³

= 6.022 x 10²² atoms

(b) 18 g of water (H₂O)

Molar mass of H₂O is 18

So 18g of H₂O = 1 mol.

No. of molecules in H₂O = 3

No. of atoms = 3 x 6.022 x 10²³

= 1.8066 x 10²⁴ atoms

= 1.81 x 10²⁴ atoms

(c) 0.44 g of carbon dioxide (CO₂)

Molar mass of CO₂ = 44 g

Number of CO₂ molecules = 0.01 x 6.022 x 10²³ = 6.022 x 10²¹molecules

Since each CO₂ carries 3 atoms, the total number of atoms is,

= 3 x 6.022 x 10²¹

= 18.066 x 10²²atoms

15. Write various postulates of Dalton’s atomic theory.

Ans: The postulates of Dalton’s atomic theory are:

Matter consists of indivisible atoms.
All the atoms of a given chemical element are identical in mass and in all other properties.
Different chemical elements have different kinds of atoms, and in particular, such atoms have different masses.
Atoms are indestructible and retain their identity in chemical reactions.
The formation of a compound from its elements occurs through the combination of atoms of unlike elements in a small whole number ratio.

16. Convert into mole:

(a) 16 g of oxygen gas (O₂)

(b) 36 g of water (H₂O)

(c) 22 g of carbon dioxide (CO₂)

Ans:

(a) 16 g of oxygen gas (O₂)

Molar mass of Oxygen = 16

Since two atoms make one oxygen molecule, the molar mass is 32 g.

17. What does a chemical formula of a compound represent?

Ans: The formula of a compound indicates

(i) elements constituting the compound

(ii) the number of each constituent element.