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

On passing through a charcoal only, `CO_(2)` reduces to `CO`.
`CO + C rarr` No reaction
volume `a`
`CO_(2) + C rarr 2 CO`
Volume before reaction `b` `0`
volume after reaction 0 `2b`
As given `a + b =1` and `a + 2 b = 1.4`
`:. b = 0.4 L`
`implies % of b = (0.4)/(1) xx 100 = 40%`
`:. a = 0.6 L`
`implies % of a = (0.6)/(1) xx 100 = 60%`