\text{kPa} $.

The mole fraction of toluene in the solution is $ x_{\text{Toluene}} = 0.4 $.

We need to determine the composition of the vapor in equilibrium with the solution, assuming it is an ideal solution.

Calculations:

Identify the mole fraction of benzene:

$ x_{\text{Benzene}} = 1 - x_{\text{Toluene}} = 1 - 0.4 = 0.6 $

Calculate the partial pressures using Raoult's Law:

Partial pressure of benzene:

$ p_{\text{Benzene}} = p_{\text{Benzene}}^{\circ} \times x_{\text{Benzene}} = 9.7 \, \text{kPa} \times 0.6 = 5.82 \, \text{kPa} $

Partial pressure of toluene:

$ p_{\text{Toluene}} = p_{\text{Toluene}}^{\circ} \times x_{\text{Toluene}} = 3.63 \, \text{kPa} \times 0.4 = 1.452 \, \text{kPa} $

Calculate the total pressure of the solution:

$ p_{\text{Total}} = p_{\text{Benzene}} + p_{\text{Toluene}} = 5.82 \, \text{kPa} + 1.452 \, \text{kPa} = 7.272 \, \text{kPa} $

Determine the mole fraction of toluene in the vapor phase using Dalton's Law:

$ Y_{\text{Toluene}} = \frac{p_{\text{Toluene}}}{p_{\text{Total}}} = \frac{1.452}{7.272} \approx 0.20 $

Conclusion:

The mole fraction of toluene in the vapor phase is approximately $ 0.20 $. Therefore, the composition of the vapor in equilibrium with the solution is $\approx 0.20$ for toluene.

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