Question 1:
A 0.24 g sample of the 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 = 0.096g (Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = (0.096/0.24)×100 % = 40%
Thus, percentage of oxygen by weight in the compound = (0.144/ 0.24) ×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 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 carbon burning in oxygen can be written as:
Carbon + Oxygen = Carbon dioxide
C + O2 → CO2
It is given that 3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide.
So, when 3g of carbon is burnt in 50g of oxygen, 3g of carbon will react with 8 g of oxygen.
42 g of oxygen will remain unreacted. In this case, also, only 11g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Question 3:
What are polyatomic ions? Give examples.
Answer:
A polyatomic ion is a group of atoms carrying a charge (positive or negative).
The ions contain more than one atom (same kind or may be of a different kind) and behave as a single unit.
For example, ammonium ion (NH4+), hydroxide ion (OH−), carbonate ion (CO32−), and Sulphate ion (SO42−).
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:
The chemical formulae of the following are as follows:
(a) Magnesium chloride →MgCl2
(b) Calcium oxide →CaO
(c) Copper nitrate →Cu (NO3)2
(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
(c) Baking powder (d) Potassium sulphate.
Answer:
- Quick lime CaO – Calcium, Oxygen
- Hydrogen bromide HBr – Hydrogen, Bromine
- Baking powder NaHCO3 – Sodium, hydrogen, carbon, oxygen
- Potassium sulphate K2SO4 – Potassium, Sulphur, 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 × mass of C) + (2 × Mass of H) = 24 + 2 = 28g
(b) Molar mass of sulphur molecule, S8 = (8 × Mass of S) = 8 ×32 = 256g
(c) Molar mass of phosphorus molecule, P4 = (4 × Mass of P) =4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = (1 + Mass of C) = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, HNO3 = (Mass of H) + (Mass of N) + (3 × Mass of O) = 1 + 14 + 48 = 63g
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 (Na2SO3)?
Answer:
Atomic mass of N atoms = 14 u
Mass of 1 mole of nitrogen atoms = Atomic mass of N atoms =14 g
4 moles of aluminium atoms
Atomic mass of aluminium = 27 u
Mass of 1 mole of Aluminium atoms = 27g
Mass of 4 moles of aluminium atoms = (27 x 4) = 108 g
10 moles of sodium sulphite (Na2SO3)
Mass of 1 mole of Na2SO3 = Molecular mass of Sodium Sulphite =10 x [(2 x 23) + (32) + (3 x 16)]g = (46 + 32 + 48)g = 126 g
Mass of 10 moles of Na2SO3 = (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) mass of oxygen gas = 12 g
Molar mass of oxygen = 32
32 g of oxygen gas = 1 mole
Therefore, number of moles = mass given/ molar mass = (12/32) mole = 0.375 mole
(b) Given Mass of water = 20 g
The molar mass of water = 18
18g of water = 1 mole
Therefore, a number of moles = mass given/ molar mass = (20/18) mole = 1.11 moles (approx.)
(c) Given the mass of carbon dioxide = 22 g
The molar mass of carbon dioxide = 44 g
44g of carbon dioxide = 1 mole
Therefore, number of moles = mass given / molar mass of water = (22/44) mole = 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) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = (0.2 × 16g) = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = (0.5 × 18g) = 9g
Question 10:
Calculate the number of sulphur molecules (S8) present in 16 g of solid sulphur.
Answer:
1 mole of solid sulphur (S8) = (8 × 32g) = 256g
i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16g of solid sulphur contains ((6.022×1023)/ (256)) × 16 molecules
= 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:
Molar mass of aluminium oxide (Al2O3) = (2 × 27) + (3 × 16)
= 54 + 48 = 102 g
=> 102g of Al2O3 will have 6.022 × 1023 molecules of Al2O3
Therefore, 0.051 g of Al2O3 contains
= ((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 × 1020 molecules (0.051g) of aluminium oxide = (2 × 3.011 × 1020)
= 6.022 × 1020