A solution is preparedb mixind `8.5` of g `CH_(2) Cl_(2) and 11.95` g of `CHCl_(3).` If vapour pressure of `CH_(2)Cl_(2) and CHCl_(2) and CHCl_(3)` at 298 K are 415 and 200 mmHg 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 os `CHCl_(3) = 119.5` g/mol
Molar mass of `CH_(2) Cl_(2) =85` g/mol.
molcs of `HCCl_(3 ) = (11.95)/( 119.5) =0.1` mol.
Moles of `CH_(2)Cl_(2)=(8.5)/(85)=0.1` mol
Molefraction of `ChCl_(3)=(0.1)/(0.2) =0.5` mol.
Mole fraction of `CH_(2) Cl_(2) =(0.1)/(0.2) = 0.5` mol.
Given-Vapour opressure of `CHCl_(3)=200 mm Hg = 0.263atm.` Vapour prcssure of `CH_(2) Cl_(2) =415 m m Hg = 0.546` atm.)
(1 atm = 760 mm Hg)
`therefore P _(("above solution"))`
= Mole fractin of `CHCl_(4)xx` (Vapour pressure of `CHCl_(3)`) + Mole fraction of `CH_(2) Cl_(2)xx ` (Vapour pressure of `CH_(2) Cl_(2)`)
`=0.5 xx0.263+ 0.5xx 546 =04045`
Mole fraction of `CHCl_(3)` in vapour from
`= (0.1315)/(0.4045) =0.325`