Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(@)`C is given by `P` (mm Hg) `=180chi_(B)+90`, where `chi_(B)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing `936` g benzene and `736` g toluene and if the vapour over this solution is removed and condensed into liquid and again brought to the temperature of `50^(@)`C. What would be the new mole fraction of `C_(6)H_(6)` in the vapour state ?

Explanation

To solve the problem, we need to determine the new mole fraction of benzene (C6H6) in the vapor state after the vapor over the solution is condensed and brought back to 50°C. We will follow these steps:

### Step 1: Calculate the number of moles of benzene and toluene.
- **Molar mass of benzene (C6H6)**:
\[ M_{C6H6} = 12 \times 6 + 1 \times 6 = 72 + 6 = 78 \text{ g/mol} \]
- **Moles of benzene**:
\[ n_{C6H6} = \frac{936 \text{ g}}{78 \text{ g/mol}} = 12 \text{ moles} \]

- **Molar mass of toluene (C7H8)**:
\[ M_{C7H8} = 12 \times 7 + 1 \times 8 = 84 + 8 = 92 \text{ g/mol} \]
- **Moles of toluene**:
\[ n_{C7H8} = \frac{736 \text{ g}}{92 \text{ g/mol}} = 8 \text{ moles} \]

### Step 2: Calculate the mole fraction of benzene in the solution.
- **Total moles**:
\[ n_{total} = n_{C6H6} + n_{C7H8} = 12 + 8 = 20 \text{ moles} \]
- **Mole fraction of benzene**:
\[ \chi_{C6H6} = \frac{n_{C6H6}}{n_{total}} = \frac{12}{20} = 0.6 \]

### Step 3: Calculate the vapor pressure of the mixture.
Using the given equation for vapor pressure:
\[ P = 180 \chi_{C6H6} + 90 \]
Substituting \(\chi_{C6H6} = 0.6\):
\[ P = 180 \times 0.6 + 90 = 108 + 90 = 198 \text{ mm Hg} \]

### Step 4: Calculate the partial pressures of benzene and toluene.
- **Vapor pressure of pure benzene (P°B)**:
\[ P°_{C6H6} = 270 \text{ mm Hg} \quad (\text{from } P°_{C7H8} = 90 \text{ mm Hg and } P°_{C6H6} - P°_{C7H8} = 180) \]
- **Partial pressure of benzene**:
\[ P_{C6H6} = \chi_{C6H6} \times P°_{C6H6} = 0.6 \times 270 = 162 \text{ mm Hg} \]
- **Partial pressure of toluene**:
\[ P_{C7H8} = \chi_{C7H8} \times P°_{C7H8} = (1 - 0.6) \times 90 = 0.4 \times 90 = 36 \text{ mm Hg} \]

### Step 5: Calculate the total pressure.
\[ P_{total} = P_{C6H6} + P_{C7H8} = 162 + 36 = 198 \text{ mm Hg} \]

### Step 6: Calculate the mole fraction of benzene in the vapor phase.
\[ \chi_{C6H6}^{vapor} = \frac{P_{C6H6}}{P_{total}} = \frac{162}{198} = \frac{81}{99} = \frac{9}{11} \approx 0.818 \]

### Step 7: Conclusion
The new mole fraction of benzene in the vapor state is approximately \(0.818\).

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