Question

At \( T \)(K), the vapour pressure of pure benzene and toluene are \( 75 \) and \( 22 \) mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture, the mole fraction of toluene in vapour phase is:

• A. \( 0.406 \)
• B. \( 0.594 \)
• C. \( 0.539 \)
• D. \( 0.461 \)

Hint:

For vapour phase mole fraction calculations, use Raoult’s Law: \[ Y_{\text{component}} = \frac{P_{\text{component}}}{P_{\text{total}}} \]

Answer 1

The Correct Option is A

Step 1: Determine the Number of Moles The molecular masses of benzene and toluene are: \[ M_{\text{benzene}} = 78 \text{ g/mol}, M_{\text{toluene}} = 92 \text{ g/mol} \] \[ n_{\text{benzene}} = \frac{23.4}{78} = 0.3 \text{ moles} \] \[ n_{\text{toluene}} = \frac{64.4}{92} = 0.7 \text{ moles} \] Step 2: Compute Mole Fractions in Liquid Phase \[ X_{\text{benzene}} = \frac{0.3}{0.3 + 0.7} = 0.3 \] \[ X_{\text{toluene}} = \frac{0.7}{1} = 0.7 \] Step 3: Calculate Partial Pressures (Raoult’s Law) Using \( P_i = X_i \cdot P_{\text{pure}} \): \[ P_{\text{benzene}} = 0.3 \times 75 = 22.5 \text{ mm Hg} \] \[ P_{\text{toluene}} = 0.7 \times 22 = 15.4 \text{ mm Hg} \] Total pressure: \[ P_{\text{total}} = 22.5 + 15.4 = 37.9 \text{ mm Hg} \] Step 4: Compute Mole Fraction in Vapour Phase \[ Y_{\text{toluene}} = \frac{P_{\text{toluene}}}{P_{\text{total}}} = \frac{15.4}{37.9} \approx 0.406 \] Conclusion Thus, the correct answer is: \[ 0.406 \]