Two oxides of a metal, M were heated separately in hydrogen. The water obtained in each case were carefully collected and weighed. It was observed that- (1) 0.725g of the first oxide gives 0.18g of water and (2) 2.86g of the second oxide gives 0.36g of water. show that these results are in acccordance with the law of multiple proportions.

Explanation

The amount of water obtained by the reduction of first oxide=0.18g.
Now, 18g water contains=16g of oxygen
`therefore`0.18g water contains=0.16g of oxygen.
`therefore`0.725g of the first oxide contains=0.16g of oxygen
So, the mass of metal in the first oxide`=(0.725-0.16)`
=565g
`therefore`The mass of oxygen which combines with 0.565g of metal M=0.16g
Again, 0.36g of water is obtained by reduction of 2.86g of second oxide.
Now, 0.36g of water contains`=(16xx0.36)/(18)=0.32`g oxygen.
So, the amount of metal in the second oxide
`=(2.86-0.32)=2.54g`
`therefore`The mass of oxygen which combines with 2.54g of metal, M=0.32g
`therefore`The mass of oxygen which combines with 0.565g of metal, `M=(0.32xx0.565)/(2.54)=0.071g`
thus, the ratio of the masses of oxygen which combine separately with a fixed mass (0.565g) of the given metal to form two different oxides is given by=0.16:0.071=2:1 (approx), which is a simple whole number ratio. so, the given data are in accordance with the law off multiple proportions.