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Practice each question, then open the answer or worked solution.
Assertion (A): Combustion of 16 g of methane gives 18 g of water.
Reason (R): In the combustion of methane, water is one of the products.
(i) Both A and R are true but R is not the correct explanation of A.
(ii) A is true but R is false.
(iii) A is false but R is true.
(iv) Both A and R are false.
Option (iii) is correct.
16g of CH4 on complete combustion will give 36g of water.
Long Answer Type Question
A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The
gas is now transferred to another vessel at a constant temperature, where
the pressure becomes half of the original pressure. Calculate
(i) the volume of the new vessel.
(ii) a number of molecules of dioxygen.
(i) Moles of oxygen = 1.6/32 = 0.05mol
At STP, 1 mol of O2 = 22.4L
Then volume of O2 = 22.4 × 0.05 = 1.12L
V1 = 1.12L
V2 =?
P1 = 1atm
P2 = ½ = 0.5atm
According to Boyle’s law, p1V1 = p2V2
Substituting the values
V2 = 1 × 1.12/0.5 = 2.24L
(ii) No of molecules in 1.6g or 0.005mol = 6.022 × 1023 × 0.05 = 3.011 × 1022
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according
to the reaction given below:
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with
1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles
of CaCl2 formed in the reaction.
No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19
No: of moles of CaCO3 = Mass/Molar mass = 1000/100 = 10
1. When CaCO3 is completely consumed
1 mol of CaCO3 = 1 mol CaCl2
10 mol CaCO3 = 10mol CaCl2
2. When HCl is completely consumed.
2 mol HCl = 1 mol CaCl2
0.19mol HCl = ½ × 0.19mol CaCl2 = 0.095 mol CaCl2
HCl will be the limiting reagent and the number of moles of CaCl2 formed will be 0.095mol
Define the law of multiple proportions. Explain it with two examples. How
does this law point to the existence of atoms?
When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other bear a simple ratio to one another is the law of multiple proportions.
For example, carbon combines with oxygen to form two compounds they are carbon dioxide and carbon monoxide
The masses of oxygen which combine with a fixed mass of carbon in carbon dioxide and carbon monoxide are 32 and 16. Therefore oxygen bear: 32:16 ratio or 2:1
Example 2: Sulphur combines with oxygen to form sulphur trioxide and sulphur dioxide
The masses of oxygen which combine with a fixed mass of sulphur in SO3 and SO2 are 48 and 32. Therefore oxygen bear a ratio of 48:32 or 3:2
A box contains some identical red coloured balls, labelled as A, each weighing
2 grams. Another box contains identical blue coloured balls, labelled as B,
each weighing 5 grams. Consider the combinations AB, AB2, A2B and A2B3
and show that the law of multiple proportions is applicable.
| AB | ab2 | A,B | A2B3 | |
| Mass of A (in g) | 2 | 2 | 4 | 415 |
| Mass of B (in g) | 5 | 10 | 5 |
Masses of B which combine with a fixed mass of A are
10g, 20g, 5g, 15g
2 : 4 : 1 : 3
This is the simple whole-number ratio.
Important Topics of Chemistry Chapter 1 Some Basic Concepts of Chemistry
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