Question

on analysis of0.11 GM of an oxide of Nitrogen 56 ml of nitrogen is obtained at same temperature and pressure and at same temperature and pressure by 0.15 gm of another oxide of Nitrogen gives 56 ml nitrogen verify that this that are verify law of multiple proportion

Answer 1

22400 ml of N2 = 28 gm so,

56 ml of N2 = 28 x 56/ 22400 = 0.07 gm

Here weight of oxide (A) =0.11 gm 

After reduction the weight = 0.07gm

Loss of weight = 0.11-0.07 = 0.04gm (It is the amount of oxygen)--1.

Similarly weight of oxide(B) =0.15 gm

Weight after reduction = 0.07 gm

Hence, in case (A)

0.07 gm of nitrogen combines with 0.04 gm of oxygen 

so, weight of nitrogen which would combine with 0.08 gm of oxygen should be equal to

 = 0.07 x 0.08/ 0.04 = 0.14 gm--.---2

Here,in both the cases 1 and 2 the weight of  nitrogen which combines with fixed amount of oxygen ( 0.04 gm) are 0.07 gm and 0.14 gm which are in the simple ratio of 2:1

Hence the oxides follow the law of multiple proportions.