Question 1

Assertion (A): The pair water (H₂O) and hydrogen peroxide (H₂O₂) illustrates the law of multiple proportions.
Reason (R): For a fixed 2 g of hydrogen, the masses of oxygen in H₂O and H₂O₂ are 16 g and 32 g, which are in the simple ratio 1 : 2.
Choose the correct option.

Accepted Answer

Both A and R are true and R is the correct explanation of A. H₂O and H₂O₂ contain the same two elements (H and O). For a fixed 2 g of hydrogen the oxygen masses are 16 g and 32 g, a ratio of 1 : 2 — exactly the small whole-number ratio required by the law of multiple proportions. R correctly explains A.

Question 2

For a fixed mass of element A, the masses of element B that combine with it in three different compounds of A and B are found to be in the ratio 1 : 2 : 3. This whole-number pattern is a direct consequence of the:

Accepted Answer

law of multiple proportions. When the same two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a ratio of small whole numbers — this is the law of multiple proportions (Dalton, 1803). A simple ratio like 1 : 2 : 3 is exactly what the law predicts.

Question 3

A metal M forms three oxides containing 40.0%, 50.0% and 57.1% oxygen by mass. For a fixed mass of the metal, the ratio of the masses of oxygen in the three oxides is:

Accepted Answer

2 : 3 : 4. Oxygen per unit mass of metal = %O ÷ %M: for the oxides these are 40/60 = 0.667, 50/50 = 1.000 and 57.1/42.9 = 1.333. Dividing by the smallest (0.667) gives 1 : 1.5 : 2 = 2 : 3 : 4, a simple whole-number ratio (law of multiple proportions).

Question 4

2 g of hydrogen combines completely with 16 g of oxygen to form water. If instead 2 g of hydrogen is mixed with 32 g of oxygen in a closed vessel and ignited, the mass of water formed and the mass of oxygen left unreacted are respectively:

Accepted Answer

18 g of water and 16 g of oxygen. By the law of definite proportions hydrogen and oxygen combine only in the ratio 2 : 16 (i.e. 1 : 8). So 2 g of hydrogen uses just 16 g of oxygen, forming 2 + 16 = 18 g of water (conservation of mass). The extra oxygen, 32 − 16 = 16 g, stays unreacted.

Question 5

At the same temperature and pressure, equal volumes of oxygen (O₂, molar mass 32 g mol⁻¹) and an unknown gas G weigh 1.6 g and 2.2 g respectively. Using Avogadro’s law, the molar mass of G is:

Accepted Answer

44 g mol⁻¹. By Avogadro’s law equal volumes at the same T and P contain equal numbers of molecules (equal moles). Moles of O₂ = 1.6 ÷ 32 = 0.05 mol, so G is also 0.05 mol. Molar mass of G = 2.2 ÷ 0.05 = 44 g mol⁻¹.

Question 6

At constant temperature and pressure, 1 volume of a gaseous compound of nitrogen and hydrogen decomposes to give 0.5 volume of nitrogen and 1.5 volumes of hydrogen. Applying Gay-Lussac’s law and Avogadro’s law, the molecular formula of the compound is:

Accepted Answer

NH₃. Volumes are proportional to numbers of molecules. Doubling to remove the fraction: 2 volumes of compound → 1 volume N₂ + 3 volumes H₂. So 2 molecules of compound contain 2 nitrogen atoms and 6 hydrogen atoms, giving 1 N and 3 H per molecule — the formula NH₃.

Question 7

In copper(II) oxide, copper and oxygen are always present in the fixed mass ratio 63.5 : 16 (law of definite proportions). The mass of copper that combines with 32 g of oxygen to form pure copper(II) oxide is:

Accepted Answer

127 g. The Cu : O mass ratio is fixed at 63.5 : 16. For 32 g of oxygen (which is 2 × 16 g), the copper is 2 × 63.5 = 127 g. Using the ratio only once (taking 63.5 g) ignores that the oxygen has been doubled.

Laws of Chemical Combination — IISER Chemistry MCQs with Solutions

Questions with solutions