Question

At ${88}^{0}C$ benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at ${88}^{0}C$ at 1 atm pressure?

• A. 0.416
• B. 0.588
• C. 0.688
• D. 0.74

Answer 1

The correct option is D 0.74

Given

Partial pressure of toluene 360 torr

Partial pressure of benzene 900 torr

Total vapour pressure 0f their mixture is 760 torr

Solution

$P_{total}=X_{benz}\ast P_{benz}^0+X_{tole}\ast P_{tole}^0$

X{benz}=1- X(toluene)

putting values

$760=X_{benzene}\ast 900+(1-X_{benzene})\ast 360$

$X_{benzene}$=0.74

The correct option is D