question_answer

A solution is prepared by mixing 8.5 g of \[C{{H}_{2}}C{{l}_{2}}\] and 11.95 g of \[CHC{{l}_{3}}\]. If vapour pressure of \[C{{H}_{2}}C{{l}_{2}}\] and \[CHC{{l}_{3}}\] at 298 K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of \[Cl=35.5\,g\,mo{{l}^{-1}}\]) [JEE Online 09-04-2017]

A)  0.162                   

B)  0.675

C)  0.325                   

D)  0.486