shobha rollno. 16

atoms and molecules

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.

Answer

Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)

Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100%

= 40%

And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%

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 combinations will govern your answer?

Answer

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.

Page No: 44

3. What are polyatomic ions? Give examples?

Answer

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, Nitrate (NO₃^-) , hydroxide ion (OH - ).

4. Write the chemical formulae of the following:

(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
► Calcium and oxygen

(b) Hydrogen bromide
► Hydrogen and bromine

(c) Baking powder
► Sodium, hydrogen, carbon, and oxygen

(d) Potassium sulphate
► Potassium, sulphur, and oxygen

6. Calculate the molar mass of the following substances:

(a) Ethyne, C₂H₂
► Molar mass of ethyne, C₂H₂ = 2 × 12 + 2 × 1 = 26 g

(b) Sulphur molecule, S₈
►Molar mass of sulphur molecule, S₈ = 8 × 32 = 256 g

(c) Phosphorus molecule, P₄ (atomic mass of phosphorus = 31)
► Molar mass of phosphorus molecule, P₄ = 4 × 31 = 124 g

(d) Hydrochloric acid, HCl
► Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO₃
► Molar mass of nitric acid, HNO₃ = 1 + 14 + 3 × 16 = 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 (Na₂SO₃)?

Answer

(a) The mass of 1 mole of nitrogen atoms is 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) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

8. Convert into mole.

(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

Answer

(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12 / 32 mole = 0.375 mole

(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole = 1.111 mole

(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22 / 44 mole = 0.5 mole

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer

(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

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

Answer

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 10²³molecules
Then, 16 g of solid sulphur contains = 6.022 × 10²³ / 256  = 16 molecules
= 3.76375 × 10²² molecules

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

mole of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al₂O₃= 6.022 × 10²³molecules of Al₂O₃
Then, 0.051 g of Al₂O₃contains = 6.022 × 10²³ / 102 × 0.051 molecules
= 3.011 × 10²⁰ molecules of 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 × 10²⁰molecules (0.051 g ) of aluminium oxide (Al₂O₃) = 2 × 3.011 × 10²⁰
= 6.022 × 10²⁰