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: Boron and oxygen compound — Boron + Oxygen
0.24 g — 0.096 g + 0.144 g
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 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: The reaction of burning of carbon in oxygen may be written as:
It shows that 12 g of carbon bums in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO2. Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 – 8 = 42 g of O2. The answer governs the law of constant proportion.
Question 3. What are poly atomic ions? Give examples.
Answer: The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g OH` S042-, C032-
Question 4. Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer: (a) Magnesium chloride
Symbol — Mg Cl
Change — (+2) (-1)
Formula — Mg Cl 2
(b) Calcium oxide
Symbol — Ca O
Charge —(+2 )(-2)
Formula —CaO
(c) Copper nitrate
Symbol — Cu NO
Change (+2 )(-1)
Formula …4 CU(N 0 3)2
(d) Aluminium chloride
Symbol — Al Cl
Change — (+3)( -1)
Formula — Al Cl 3
(d) Calcium carbonate
Symbol —Ca C O 3
Change —(+2)( -2)
Formula —Ca C0 3
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 —Calcium oxide
Elements —Calcium and oxygen
(b) Hydrogen bromide
Elements — Hydrogen and bromine
(c) Baking powder —Sodium hydrogen carbonate
Elements —Sodium, hydrogen, carbon and oxygen
(d) Potassium sulphate
Elements — Potassium, sulphur and oxygen
Question 6. Calculate the molar mass of the following substances.
(a) Ethyne : C 2 H 2
(b) Sulphur molecule : S8
(c) Phosphorus molecule : P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid : HCl
(e) Nitric acid : H N O 3
Answer: The molar mass of the following:
Unit is ‘g’ (gram)
(a) Ethyne : C 2 H2 = 2 x 12 + 2 x 1 = 24 + 2 = 26 g
(b) Sulphur molecule : S8 = 8 x 32 = 256 g
(c) Phosphorus molecule : P4=4 x 31 = 124g
(d) Hydrochloric acid : HCl = 1 x 1 + 1 x 35.5 = 1 + 35.5 = 36.5 g
(e) Nitric acid : H N 03 = 1 x 1 + 1 x 14 + 3 x 16 = 1 + 14 + 48 = 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)?
(c) 10 moles of sodium sulphite (Na 2S0 3).
Answer: (a) Mass of 1 mole of nitrogen atoms = 14 g
(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium atoms = 27 g
∴ Mass of 4 moles of aluminium atoms = 27 x 4 = 108 g
(c) 10 moles of sodium sulphite (Na 2 SO 3)
Mass of 1 mole of Na 2 SO 3 = 2 x 23 + 32 + 3 x 16 = 46 + 32 + 48 = 126 g
∴ Mass of 10 moles of Na 2 SO 3 = 126 x 10 = 1260 g
Question 8. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of Carbon dioxide.
Answer: (a) Given mass of oxygen gas = 12 g
Molar mass of oxygen gas (O2) = 32 g
Mole of oxygen gas 12\32 = 0.375 mole
(b) Given mass of water = 20 g
Molar mass of water (H 2 O) = (2 x 1) + 16 = 18 g
Mole of water = 20/18 = 1.12 mole
(c) Given mass of Carbon dioxide = 22 g
Molar mass of carbon dioxide (CO2) = (1 x 12) + (2 x 16)
= 12 + 32 = 44 g
∴ Mole of carbon dioxide = 22/44 = 0.5 mole
Question 9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer: (a) Mole of Oxygen atoms = 0.2 mole
Molar mass of oxygen atoms = 16 g
Mass of oxygen atoms = 16 x 0.2 = 3.2 g
(b) Mole of water molecule = 0.5 mole
Molar mass of water molecules = 2 x 1 + 16= 18 g .
Mass of H 2 O = 18 x 0.5 = 9 g
Question 10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer: Molar mass of S8 sulphur = 256 g = 6.022 x 1023 molecule
Given mass of sulphur = 16 g
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: Molar mass of aluminium oxide Al2 03
= (2 x 27) + (3 x 16)
= 54 + 48 = 102 g.
