A solution is prepared by mixing 50.0 g of benzene (C6H6) and 50.0 g of toluene (C7H8) at 25°C. The vapor pressures of pure benzene and pure toluene at this temperature are 95.1 mmHg and 28.4 mmHg, respectively. Assuming ideal behavior, what is the mole fraction of benzene in the vapor above the solution?

• A. 0.50
• B. 0.95
• C. 0.26
• D. 0.74

Answer: (D)

Verified step by step guidance

1. Calculate the number of moles of benzene and toluene separately using their given masses and molar masses. Use the formula: \(n = \frac{\text{mass}}{\text{molar mass}}\). The molar mass of benzene (C6H6) is approximately 78.11 g/mol, and that of toluene (C7H8) is approximately 92.14 g/mol.
2. Determine the mole fraction of benzene (\(x_{\text{benzene}}\)) and toluene (\(x_{\text{toluene}}\)) in the liquid solution using the formula: \(x_i = \frac{n_i}{n_{\text{benzene}} + n_{\text{toluene}}}\).
3. Calculate the partial vapor pressures of benzene and toluene above the solution using Raoult's Law: \(P_i = x_i \times P_i^\circ\), where \(P_i^\circ\) is the vapor pressure of the pure component at 25°C.
4. Find the total vapor pressure above the solution by summing the partial pressures: \(P_{\text{total}} = P_{\text{benzene}} + P_{\text{toluene}}\).
5. Calculate the mole fraction of benzene in the vapor phase (\(y_{\text{benzene}}\)) using Dalton's Law: \(y_{\text{benzene}} = \frac{P_{\text{benzene}}}{P_{\text{total}}}\). This gives the composition of benzene in the vapor above the solution.