One litre of mixture of `CO` and `CO_(2)` is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out % composition of mixture by volume. All measurements are made at same `P` and `T`

Explanation

(a). Let the volume of CO in the mixture be V.
The volume of `CO_(2)` in the mixture is `(1-V)`
`CO_(2)+Cto2CO`
`(1-V)LCO_(2)` forms `2(1-V)LCO`.
Total CO,
`V+2(1-V)=1.6`
or `V=0.4`
`CO=0.4L`
`CO_(2)=1-V=0.6L`
(b). 28 g of nitrogen combines with 72 g of metal. three atoms of metal. Three atoms of metal combine with two atoms of nitrogen. This means the valency of metal is 2.
" Eq of "`N=(28)/((14)/(3))=(1)/(6)`
" Eq of "metal `=(72)/((M)/(2))=(144)/(M)`
`therefore(1)/(6)=(144)/(M)impliesM=(144)/(6)=24g`