A mixture of `C_(3)H_(8)` and `CH_(4)` exerts a pressure of 320 mm Hg at temperature T K in a V litre flask. On complete combustion, gaseous mixture contains `CO_(2)` only and exerts a pressure of 448 mm Hg under identical conditions. Hence mole fraction of `C_(3)H_(8)` in the mixture is

Options

A. `0.2`
B. 0.8
C. 0.25
D. 0.75

Answer

A

Explanation

Let V L at T K and 320 mm Hg represent 1 mol Then V L at T K and 448 mm Hg represents
`= (448)/(320) mol = 1.4 mol`
`underset(x mol)(C_(3)H_(8) (g)) + 5O_(2) (g) rarr underset(3x mol)(3CO_(2) (g)) + 4H_(2) (l)`
`underset((1 - x)mol)(CH_(4) (g)) + 2O_(2) (g) rarr underset((1 - x) mol)(CO_(2) (g)) + 4H_(2) O (l)`
Let moles of `C_(3)H_(8) = x`
`:.` Moles of `CH_(4) = 1 - x`
Moles of `CO_(2)` produced `= 3x + 1-x = 1 + 2x`
`1 + 2x = 1.4`
`2x = 0.4`
`x = 0.2`
Mole fraction of `C_(3)H_(8) = (x)/(1) = 0.2`