The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises at `20^(@)C` and that of liquid toluene(density `0.867 g mL)` increases by a factor of `7720`at `20^(@)C`. A solution of benzene and toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

Explanation

To find the mole fraction of benzene in the vapor above the solution, we will follow these steps:

### Step 1: Calculate the molar volume of benzene and toluene
Given:
- Density of benzene, \( \rho_B = 0.877 \, \text{g/mL} \)
- Increase factor for benzene, \( f_B = 2750 \)
- Density of toluene, \( \rho_T = 0.867 \, \text{g/mL} \)
- Increase factor for toluene, \( f_T = 7720 \)

**Molar volume of benzene:**
\[ V_B = \frac{M_B}{\rho_B} \times f_B \]
Where \( M_B = 71 \, \text{g/mol} \) (molar mass of benzene).

Calculating:
\[ V_B = \frac{71 \, \text{g/mol}}{0.877 \, \text{g/mL}} \times 2750 = 244.58 \, \text{L} \]

**Molar volume of toluene:**
\[ V_T = \frac{M_T}{\rho_T} \times f_T \]
Where \( M_T = 92 \, \text{g/mol} \) (molar mass of toluene).

Calculating:
\[ V_T = \frac{92 \, \text{g/mol}}{0.867 \, \text{g/mL}} \times 7720 = 819.19 \, \text{L} \]

### Step 2: Calculate the vapor pressures of pure benzene and toluene
Using the ideal gas law \( PV = nRT \):
\[ P_0B = \frac{nRT}{V_B} \]
\[ P_0T = \frac{nRT}{V_T} \]

Assuming 1 mole:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 293 \, \text{K} \)

Calculating \( P_0B \):
\[ P_0B = \frac{1 \times 0.0821 \times 293}{244.58} = 0.098 \, \text{atm} \]

Calculating \( P_0T \):
\[ P_0T = \frac{1 \times 0.0821 \times 293}{819.19} = 0.029 \, \text{atm} \]

### Step 3: Use Raoult's Law to find the mole fraction of benzene in the vapor
Given total vapor pressure \( P_t = 46 \, \text{torr} = \frac{46}{760} \, \text{atm} = 0.060 \, \text{atm} \).

Using Raoult's Law:
\[ P_t = P_0B \cdot X_B + P_0T \cdot X_T \]
Where \( X_B + X_T = 1 \) (mole fractions).

Substituting \( X_T = 1 - X_B \):
\[ 0.060 = 0.098 \cdot X_B + 0.029 \cdot (1 - X_B) \]

Expanding:
\[ 0.060 = 0.098X_B + 0.029 - 0.029X_B \]
\[ 0.060 = (0.098 - 0.029)X_B + 0.029 \]
\[ 0.060 - 0.029 = 0.069X_B \]
\[ 0.031 = 0.069X_B \]
\[ X_B = \frac{0.031}{0.069} \approx 0.449 \]

### Step 4: Calculate the mole fraction of benzene in the vapor
Using the formula for the mole fraction of benzene in vapor:
\[ Y_B = \frac{P_0B \cdot X_B}{P_t} \]
Substituting the values:
\[ Y_B = \frac{0.098 \cdot 0.449}{0.060} \]
Calculating:
\[ Y_B = \frac{0.04396}{0.060} \approx 0.733 \]

### Final Answer
The mole fraction of benzene in the vapor above the solution is approximately **0.733**.