Q 2741123923.     A solution is prepared by mixing `8.5 g` of `CH_2 Cl_2` and `11.95 g` of `CHCl_3`. If vapour pressure of `CH_2Cl_2` and `CHCl_3` at `298 K` are `415` and `200 mmHg` respectively, the mole fraction of `CH Cl_3` in vapour form is : (Molar mass of `Cl =35.5 g mol^(-1)`)

JEE 2017 Mains 9 April
A

`0.675`

B

`0.325`

C

`0.162`

D

`0.486`

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