Question

A binary mixture of benzene and toluene under vapour-liquid equilibrium at 80 \( ^\circ {C} \) follows ideal Raoult’s law. At this condition, the saturation pressures of benzene and toluene are 101 kPa and 40 kPa, respectively. If the mole fraction of benzene in the liquid phase is 0.6, the corresponding mole fraction of benzene in the vapour phase will be ________. (Round off to two decimal places)

Hint:

For ideal solutions, Raoult's law relates the mole fraction in the vapor phase to the mole fraction in the liquid phase using the saturation pressures.

Answer 1

Correct Answer: 0.76

The ideal Raoult's law gives the relation between the mole fraction of a component in the liquid phase and its vapor pressure: \[ y_{{benzene}} = \frac{P_{{benzene}} x_{{benzene}}}{P_{{benzene}} x_{{benzene}} + P_{{toluene}} x_{{toluene}}} \] where:
\( P_{{benzene}} \) = 101 kPa (saturation pressure of benzene),
\( P_{{toluene}} \) = 40 kPa (saturation pressure of toluene),
\( x_{{benzene}} \) = 0.6 (mole fraction of benzene in the liquid phase),
\( x_{{toluene}} = 1 - x_{{benzene}} = 0.4 \) (mole fraction of toluene in the liquid phase).
Substituting these values into the equation: \[ y_{{benzene}} = \frac{101 \times 0.6}{101 \times 0.6 + 40 \times 0.4} \] Now, calculate the value: \[ y_{{benzene}} = \frac{60.6}{60.6 + 16} = \frac{60.6}{76.6} \approx 0.79. \] Thus, the mole fraction of benzene in the vapour phase is \( \boxed{0.79} \).