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, benzene -toluene from an ideal solution?

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

Answer 1

Answer: (D)

0.74

Given,

VP of benzene $=900torr$

VP pf toluence $=360torr$

To find mole fraction of benzene,

$x_{benzene}=\frac{P-P_{toluence}^{sat}}{P_{benzene}^{sat}-P_{toluene}^{sat}}$

[when solution boils at the same temperature, vapour pressure = saturation pressure]

$=\frac{760-360}{900-360}$ [1 atn = 760 torr]

$x_{Benzene}=0.74$

Hence, option D is correct