CBSE Class IX ( 9th) Science | Chapter 3. Atoms And Molecules | Lesson Exercises |Chapter |CBSE Class 6th, 7th, 8th, 9th, 10th|Worksheet |Wikipedia|PPT|PDF|Video|Download|MCQs

CBSE Class IX ( 9th) Science | Chapter 3. Atoms And Molecules | Lesson Exercises

Question 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.

Answer :

Mass of boron as given = 0.096 g

Mass of oxygen as given = 0.144 g

Mass of sample as given = 0.24 g

∴ Percentage of boron by weight in the compound = (0.096 ÷ 0.24 ) x 100 = 40%

∴ Percentage of oxygen by weight in the compound = (0.144 ÷ 0.24) x 100 = 60%

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 willbe 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 : We know that in burning process carbon utilises oxygen to form Carbon Dioxide .

Carbon + Oxygen => Carbon Dioxide + Heat Energy

Here as given, in burning, 3.0 g of carbon utilises 8.00 g of oxygen to form 11.00 g of carbon dioxide.When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from the total of 50 gm of Oxygen to produce 11 gm of Carbon Dioxide. 42 gm of Oxygen will be left behind without any change.

The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction.

Here in terms of mass 3g of Carbon and 50 gm of Oxygen combines to give 11 gm of Carbon Dioxide and 42 gm of unused Oxygen keeping the total mass of combining

chemicals 53 gm (3+50 gm) by weight constant before and after the reaction.

2. Law of constant proportions which states that In a chemical substance the elements are always present in definite proportions by mass.Here respective of the excess quantity of Oxygen available for burning of same weight of Carbon, the amount of carbon dioxide will remain same.

Question 3. What are polyatomic ions? Give examples.

Answer : Polyatomic ions are clusters or group of atoms that always have a fixed charge attach to them. This charge may be positive or negative .

Ammonium NH₄⁺

Hydroxide OH^-

Nitrate NO₃^-

Carbonate CO₃^2-

Sulphate SO₄^2-

Question 4. Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(d) Aluminium chloride

(e) Calcium carbonate.

Answer :

(a) Magnesium chloride => MgCl₂

(b) Calcium oxide => CaO

(d) Aluminium chloride => AlCl₃

(e) Calcium carbonate. => CaCO₃

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

(a) Quick lime

(b) Hydrogen bromide

(d) Potassium sulphate.

Answer :

Compound	Chemical Formula	Elements
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.	CaO HBr NaHCO₃ K₂SO₄	Calcium, Oxygen Hydrogen, Bromine Sodium, Hydrogen, Carbon, Oxygen Potassium, Sulphur, Oxygen

Question 6. Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂

(b) Sulphur molecule, S₈

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO₃

Answer :

(a) Molar Mass of Ethyne, C₂H₂; = 2 × 12 + 2 × 1 = 26 g

(b) Molar Mass of Sulphur molecule, S₈ = 8 × 32 = 256 g

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

(e) Molar Mass of Nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 63 g

Question 7. What is the mass of.

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?

Answer :The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams. The atomic mass of an element gives us the mass of one atom of that element in atomic mass units (u). To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.
Molar mass of atoms is also known as gram atomic mass. For example, atomic mass of hydrogen=1u. So, gram atomic mass of
hydrogen = 1 g.

(a) The mass of 1 mole of Nitrogen atoms = 14 g.

(b) The mass of 4 moles of Aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of Sodium Sulphite (Na₂SO₃) =10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Question 8. Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

Answer: To get the mass of 1 mole of atom of that element, that is, molar mass, we have to take the same numerical value but change the units from 'u' to 'g'.

∴ 1 mole substance = Total Atomic mass of a substance (u) Or

= Total gram atomic mass of substance (g) Or

= Molar Mass

= 1 Mole of substance Or

= 6.022 × 10²³ in number of particles of that substance

Here required mole of substance = Given mass of substance in 'g' ÷ Molar Mass

(a) 12 g of oxygen gas = Given mass of oxygen in 'g' ÷ Molar Mass of Oxygen

= 12 ÷ 32

= 0.375 mole of oxygen

(b) 20 g of water = Given mass of water in 'g' ÷ Molar Mass of Water

= 20 ÷ 18

= 1.11 mole of water

= 22 ÷ 44

= 0.5 mole of carbon dioxide

Question 9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Answer : Mass of atoms (m) in given moles (n) = Molar Mass of substance (M)× Given Moles of substance (n)

(a) ∴ Mass of 0.2 mole of oxygen atoms = Molar Mass of oxygen × Given Moles = 16g × 0.2 = 3.2 g

(b) ∴ Mass of 0.5 mole of water molecules =Molar Mass of water × Given Moles = 18 g × 0.5 = 9 g

Question 10. Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Answer :

1 mole of Sulphur (S₈) = 8 × 32 g = 256 g
also as 1 mole = 6.022 × 10²³ in number of particles of that substance

So 256 g of sulphur contains = 6.022 × 10²³ molecules of sulphur
Therefor 16 g of sulphur will contains = 3.76 × 10²² molecules (approx)

Question 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)

Answer :

1 mole of Aluminium Oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 102 g

We know, 102 g of Aluminium Oxide (Al₂O₃) = 6.022 × 10²³ molecules of Aluminium Oxide (Al₂O₃)

We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number

Then, 0.051 g of Aluminium Oxide (Al₂O₃) contains

= (0.051 ÷ 102 ) × 6.022 × 10²³molecules of Aluminium Oxide (Al₂O₃)

= 3.011 × 10²⁰ molecules of Aluminium Oxide (Al₂O₃)

The number of Aluminium ions (Al₃⁺) present in one molecule of aluminium oxide is 2.

Therefore, the number of Aluminium ions (Al₃⁺) present in 3.011 × 10²⁰ molecules (0.051 g ) of Aluminium Oxide (Al₂O₃)

= 2 × 3.011 × 10²⁰

= 6.022 × 10²⁰