Question

At $T$ (K), 1 mol of benzene is mixed with 1 mol of toluene. The mole fraction of benzene and toluene in its vapour state is respectively
($P^0_{\text{Benzene}} = 160$ torr, $P^0_{\text{Toluene}} = 60$ torr)

• A. 0.5 & 0.5
• B. 0.4 & 0.6
• C. 0.62 & 0.38
• D. 0.73 & 0.27

Hint:

Vapour phase mole fractions depend on partial vapour pressures, not mole ratios in the liquid.

Answer 1

The Correct Option is D

Using Raoult’s law, partial pressures: \[ P_{\text{benz}} = x_{\text{benz}} \cdot P^0_{\text{benz}}, \quad P_{\text{tol}} = x_{\text{tol}} \cdot P^0_{\text{tol}} \] Mole fractions in solution: $x_{\text{benz}} = x_{\text{tol}} = 0.5$ \[ P_{\text{total}} = 0.5 \cdot 160 + 0.5 \cdot 60 = 80 + 30 = 110 \] Vapour phase mole fractions: \[ y_{\text{benz}} = \frac{80}{110} = 0.73, \quad y_{\text{tol}} = \frac{30}{110} = 0.27 \]

Answer 2

At T (K), 1 mol of benzene is mixed with 1 mol of toluene. The mole fraction of benzene and toluene in its vapour state is respectively:

Given:
- 1 mol benzene and 1 mol toluene → equimolar mixture
- Vapour pressure of pure benzene \( P^0_{\text{benzene}} = 160 \) torr
- Vapour pressure of pure toluene \( P^0_{\text{toluene}} = 60 \) torr

Step 1: Calculate mole fraction in liquid phase:
Since the mixture is equimolar:
Mole fraction of benzene in liquid phase \( x_{\text{benzene}} = \frac{1}{1+1} = 0.5 \)
Mole fraction of toluene in liquid phase \( x_{\text{toluene}} = 0.5 \)

Step 2: Use Raoult’s Law to calculate partial vapour pressures:
\[ P_{\text{benzene}} = x_{\text{benzene}} \cdot P^0_{\text{benzene}} = 0.5 \cdot 160 = 80 \ \text{torr} \]
\[ P_{\text{toluene}} = x_{\text{toluene}} \cdot P^0_{\text{toluene}} = 0.5 \cdot 60 = 30 \ \text{torr} \]

Step 3: Calculate total pressure:
\[ P_{\text{total}} = P_{\text{benzene}} + P_{\text{toluene}} = 80 + 30 = 110 \ \text{torr} \]

Step 4: Find mole fractions in vapour phase:
Mole fraction of benzene in vapour phase:
\[ y_{\text{benzene}} = \frac{P_{\text{benzene}}}{P_{\text{total}}} = \frac{80}{110} \approx 0.727 \approx 0.73 \]
Mole fraction of toluene in vapour phase:
\[ y_{\text{toluene}} = \frac{P_{\text{toluene}}}{P_{\text{total}}} = \frac{30}{110} \approx 0.273 \approx 0.27 \]

Final Answer:
\[ \boxed{0.73 \ \text{and} \ 0.27} \]