the molar volume of liquid benzene (0.877 gm/ml) increases by a factor of 2750 as its vapourises at 293k and that of liquid toluene ( 0.867gm/lt) increases by a factor of 7720 at the same temp

a solution of benzene and toulene at 293k has a vapour pressure of 46 torr . calculate the mole fraction of benzene in the vapour above the solution?/

Given -
Density of benzene = 0.877 g ml^-1
Molecular mass of benzene =78
Temperature = 293 K

Case of benzene -

Part-I
Molar volume of benzene in liquid form
= 78/0.877 x (1/1000)
= 0.08894 L

Part-II
Molar volume of benzene in vapour phase
= 0.08894 x 2750
= 244.58 L

Case of Toluene

Molar volume of toluene in liquid form
= 92/0.867 x ( 1/1000)
= 0.106 L

Molar volume of toluene in vapour phase
= 0.106 x 7720
= 819.19 L

Formula
PV = nRT

Case of  Benzene

P₁ = nRT / V
= 1 x 0.082 x 293 / 244.58
= 0.098 atm
= 74.48 torr

Case of Toluene

P₂ = 0.029 atm
= 22.04 torr

Using Raoult’s law,

P_B = P₁ X_B = 74.48 X_B
P_T = P₂ X_T = 22.04 ( 1 – X_B)
P = P_B + P_T  = 74.48 X_B + 22.04 ( 1 – X_B) = 46

Solve the above equation-
X_B = 0.457

According to Dalton’s law-
P_B = P X_B ^’ 

Mole fraction of benzene in vapour form.

X_B = 74.48 x 0.457 / 46
= 0.74 Answer