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. 0.162
B. 0.675
C. 0.325
D. 0.486

Answer

C

Explanation

Molar mass of `CHCl_3` = 119.5 g/mol.
Molar massof `CH_2Cl_2` = 85g/mol.
Moles of `CHCl_3 = 11.95/119.5 = 0.1` mol.
Moles of `CH_2Cl_2 = 8.5/85 = 0.1 mol`
Mole fractionof `CHCl_3 = 0.1/0.2 = 0.5` mol.
Mole fraction of `CH_2Cl_2= 0.1/0.2 = 0.5` mol.
(Given -
Vapour pressure of `CHCl_3= 200 m m Hg= 0.263 atm`.
Vapour pressure of `CH_2Cl_2 = 415 m m Hg= 0.546 atm`.)
(1atm = 760 mm Hg)
`:.P_(("above solution"))`
= Mole fraction of `CHCl_3xx ("Vapour pressure of " CHCl_3)`
+ Mole fraction of `CH_2Cl_2 xx`(vapour pressure of `CH_2Cl_2`)
`=0.5 xx 0.263 + 0.5 xx 0.546 = 0.4045`
Mole fractionof `CHCl_3` in vapour form
`=(0.1315)/(0.4045) = 0.325`