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 solve the problem, we will follow these steps:

### Step 1: Calculate the volume of benzene and toluene when vaporized
1. **Calculate the molar volume of benzene**:
- Density of benzene = 0.877 g/mL
- Molar mass of benzene (C6H6) = 78 g/mol
- Molar volume of benzene when vaporized = (Molar mass / Density) * Increase factor
- Molar volume of benzene = (78 g/mol / 0.877 g/mL) * 2750 mL = 244.583 L

2. **Calculate the molar volume of toluene**:
- Density of toluene = 0.867 g/mL
- Molar mass of toluene (C7H8) = 92 g/mol
- Molar volume of toluene when vaporized = (Molar mass / Density) * Increase factor
- Molar volume of toluene = (92 g/mol / 0.867 g/mL) * 7720 mL = 819.192 L

### Step 2: Calculate the pure vapor pressures of benzene and toluene
1. **Using the ideal gas law (PV = nRT)**:
- For benzene:
- P_benzene = (nRT) / V
- P_benzene = (1 mol * 0.0821 L·atm/(K·mol) * 293 K) / 244.583 L
- P_benzene = 0.098 atm

- For toluene:
- P_toluene = (nRT) / V
- P_toluene = (1 mol * 0.0821 L·atm/(K·mol) * 293 K) / 819.192 L
- P_toluene = 0.029 atm

### Step 3: Calculate the mole fraction of benzene in the liquid phase
1. **Using Raoult's Law**:
- Total pressure (P_total) = 46 torr = 46 / 760 atm = 0.06053 atm
- Using the equation:
\[ P_{total} = P^0_{benzene} \cdot X_{benzene} + P^0_{toluene} \cdot (1 - X_{benzene}) \]
- Substitute known values:
\[ 0.06053 = 0.098 \cdot X_{benzene} + 0.029 \cdot (1 - X_{benzene}) \]
- Rearranging gives:
\[ 0.06053 = 0.098 X_{benzene} + 0.029 - 0.029 X_{benzene} \]
\[ 0.06053 - 0.029 = (0.098 - 0.029) X_{benzene} \]
\[ 0.03153 = 0.069 X_{benzene} \]
\[ X_{benzene} = \frac{0.03153}{0.069} \approx 0.456 \]

### Step 4: Calculate the mole fraction of benzene in the vapor phase
1. **Using the formula**:
- Mole fraction of benzene in vapor (Y_benzene):
\[ Y_{benzene} = \frac{P_{benzene}}{P_{total}} = \frac{0.098 \cdot X_{benzene}}{P_{total}} \]
- Substitute values:
\[ Y_{benzene} = \frac{0.098 \cdot 0.456}{0.06053} \]
- Calculate:
\[ Y_{benzene} = \frac{0.044688}{0.06053} \approx 0.739 \]

### Final Answer
The mole fraction of benzene in the vapor above the solution is approximately **0.739**.

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