The mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene `(P_(B)^(@)=120 "torr")` and toluene `(P_(T)^(@)=80 "torr")` having `2.0 mol` of each, is

Options

A. 0.5
B. 0.25
C. 0.6
D. 0.4

Explanation

To find the mole fraction of toluene in the vapor phase that is in equilibrium with a solution of benzene and toluene, we can follow these steps:

### Step 1: Determine the mole fractions in the liquid phase
Given:
- Moles of benzene (B) = 2.0 mol
- Moles of toluene (T) = 2.0 mol

Total moles in the solution:
\[ n_{total} = n_B + n_T = 2 + 2 = 4 \text{ mol} \]

Mole fraction of benzene (\(X_B\)):
\[ X_B = \frac{n_B}{n_{total}} = \frac{2}{4} = 0.5 \]

Mole fraction of toluene (\(X_T\)):
\[ X_T = 1 - X_B = 1 - 0.5 = 0.5 \]

### Step 2: Calculate the partial pressures
Using Raoult's Law, the partial pressures of benzene and toluene in the vapor phase can be calculated as follows:

Given:
- Vapor pressure of pure benzene (\(P_B^0\)) = 120 torr
- Vapor pressure of pure toluene (\(P_T^0\)) = 80 torr

Partial pressure of benzene (\(P_B\)):
\[ P_B = P_B^0 \cdot X_B = 120 \, \text{torr} \cdot 0.5 = 60 \, \text{torr} \]

Partial pressure of toluene (\(P_T\)):
\[ P_T = P_T^0 \cdot X_T = 80 \, \text{torr} \cdot 0.5 = 40 \, \text{torr} \]

### Step 3: Calculate the total pressure
Total pressure (\(P_{total}\)) is the sum of the partial pressures:
\[ P_{total} = P_B + P_T = 60 \, \text{torr} + 40 \, \text{torr} = 100 \, \text{torr} \]

### Step 4: Calculate the mole fraction of toluene in the vapor phase
The mole fraction of toluene in the vapor phase (\(y_T\)) is given by:
\[ y_T = \frac{P_T}{P_{total}} = \frac{40 \, \text{torr}}{100 \, \text{torr}} = 0.4 \]

### Final Answer
The mole fraction of toluene in the vapor phase is \(0.4\).

---