A solution is prepared by mixing 8.5g of `CH_(2)Cl_(2)` and 11.95g of `CHCl_(3)`. If vapour pressure of `CH_(2)Cl_(2)` and `CHCl_(3)` at 298 K are 415 and 200 mm Hg respectively, the mole fraction of `CHCl_(3)` in vapour form is :
`("Molar mass of Cl"="35.5 g mol"^(-1))`

Options

A. 0.675
B. 0.162
C. 0.486
D. 0.325

Explanation

No. of moles of `CH_(3)Cl_(3)=11.95/119.5=0.1` mole
No. of moles o `CH_(2)Cl_(20=8.5/85=0.1` mole
More fraction of `CHCl_(3), x_(a)=0.1/(0.1+0.1)=0.5`
Mole fraction of `CH_(2)Cl_(2), x_(B)=1-0.5=0.5`
`P_("total")=P_(CHCl_(3))=+p_(CH_(2))Cl_(2)=x_(A)xxp_(CHCl_(3))^(@)+x-(B)xxp_(CH_(2)Cl_(2))^(@)`
`=0.5xx200+0.5xx415=307.5` mm Hg
As, `p_(CHCl_(3)=100mm, P_("total")=307.5` mm Hg.
`:.` Mole fraction of `CHCl_(3)` in vapour phase will be `(p_(CHCl_(3)))/(p_("total"))=100/37.5=0.325`