The vapour pressure of chloroform `(CHCl)_(3)` and dichlorocethene `(CH_(2)Cl_(2))` at `298 K` is `200 mm Hg ` and `415 mm Hg`, respectively. Calculate
a. The vapour pressure of the solution prepared by mixing `25.5 g` of `CHCl_(3)` and `40 g` of `CH_(2)_Cl(2)` at `298 K`.
b. Mole fractions of each components in vapour phase .

Options

A. 615.0 mm Hg
B. 347.9 mm Hg
C. 285.5 mm Hg
D. 173.9 mm Hg

Answer

B

Explanation

`{:(""CHCl_(3),""CH_(2)Cl_(2)),("(A)","(B)),(n_(A)=(25.5)/(119.5)=0.213,n_(B)=(40)/(85)=0.47):}`
`X_(A)=(n_(A))/(n_(A)+n_(B))=(0.213)/(0.684)=0.31`
`X_(B)=1-0.31=0.69`
`P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`=200xx(0.31)+41.5(0.69)`
`=62.28+28.63=90.38`
Note: This question will be bonus because V.P. of Dichloromethane cannot be less than chloroform. (it would be 415 mm of Hg)