A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm.
The moles fraction of `C_(3) H_(8)` in the mixture is

Options

A. 0.25
B. 0.75
C. 0.45
D. 0.55

Answer

A

Explanation

`underset(a "mol")(C_(3) H_(8)) + underset(5a "mol")(5O_(2)) rarr underset(3a "mol")(3CO_(2)) + underset(-)(4H_(2) O(l))`
`underset(b "mol")(C_(2) H_(4)) + underset(3b "mol")(3O_(2)) rarr underset(2b "mol")(2CO_(2)) + underset(-)(4H_(2) O (l)`
Initially,
`PV = nRT`
`4.93 xx V = (a + b) RT`
After combustion, pressure is due to the total moles of `CO_(2)`
`11.8 xx V = (3a + 2b) RT`
Divide equation (ii) by equation (i), we get
`(11.08)/(4.93) = 2.25 = (3a + 2b)/(a + b)`
`2.25 a + 2.25b = 3a + 2b`
`0.25b = 0.75a`
`b = 3a`
`chi_(C_(2)H_(8))` or `chi_(a) = (a)/(a + b) = (a)/(a + 3a) = (1)/(4) = 0.25`