Question

Two oxides of nitrogen $A$ and $B$ gave the following results :

$4.4 g$ of ' $A$ ' gave $2.24$ litre nitrogen and $60 g$ of ' $B$ ' gave $22.4$ litre of nitrogen at NTP. Show these results illustrate the law of multiple proportions.

Answer 1

SOLUTION : In 'A', volume of nitrogen obtained $=2.24$ litre at N.T.P.

Wt. of nitrogen obtained $=\frac{28}{22.4} \times 2.24,2.8 g [\because 22.4$ litre of nitrogen at N.T.P.

wcight $=28 gm .]$

Wt. of ' $A$ ' $=4.4 g$

$\therefore$ Wcight of oxygen $=4.4-2.8=1.6 g$.

In ' $B$ ', volume of nitrogen obtained $=22.4$ litre at N.T.P.

$\therefore Wt$. of nitrogen obtained $=28 g ; Wt$. of ' $B$ '. $=60 g ; \therefore$ Wt. of oxygen $=60-28=32 g$.

Let us fix the weight of nitrogen as $2.8 g$ and find the different weights of oxygen which combine with $2.8 g$ of nitrogen in the two compounds.

In ' $A$ ', $2.8 g$ of nitrogen combine with oxygen $=1.6 g$

In ' $B$ ', $28 g$ of nitrogen combine with oxygen $=32 g$

$\therefore 2.8 g$ of nitrogen combine with oxygen $=3.2 g$.

The ratio of the weights of oxygen which combine with the fixed weight of nitrogen $(2.8 g )$ is

$1.6: 3.2$ or $1: 2$

which is a simple ratio and hence illustrates the law of multiple proportions.