Question

The total vapour pressure of an ideal binary liquid mixture of benzene and toluene is 0.3 bar. The vapour pressure of pure benzene is 0.5 bar and that of toluene is 0.2 bar. The mole fraction of benzene in this mixture is _______.
(rounded off to two decimal places)

Answer 1

Correct Answer: 0.32 - 0.34

For an ideal solution, the total vapour pressure is given by Raoult’s Law, which states:

P_total = P_benzene + P_toluene

Where:

• P_benzene = P_{0 benzene} · X_benzene
• P_toluene = P_{0 toluene} · X_toluene

Here, P_{0 benzene} and P_{0 toluene} are the vapour pressures of pure benzene and toluene, and X_benzene and X_toluene are the mole fractions of benzene and toluene, respectively.

The mole fraction of benzene X_benzene is related to the total vapour pressure by the equation:

P_total = P_{0 benzene} · X_benzene + P_{0 toluene} · (1 − X_benzene)

Substitute the given values into the equation:

0.3 = 0.5 · X_benzene + 0.2 · (1 − X_benzene)

Simplify and solve for X_benzene:

0.3 = 0.5 · X_benzene + 0.2 − 0.2 · X_benzene

0.3 − 0.2 = 0.5 · X_benzene − 0.2 · X_benzene

0.1 = 0.3 · X_benzene

X_benzene = \( \frac{0.1}{0.3} = 0.34 \)

Thus, the mole fraction of benzene in the mixture is 0.34.