Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(@)C` is given by `P(mm Hg) = 179 X_(B) +92`, where `X_(R)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of `50^(@)C`. what would be mole fraction of `C_(6)H_(6)` in the vapour state?

Explanation

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

### Step 1: Calculate the vapor pressures of pure components
Given the equation for vapor pressure:
\[ P(mmHg) = 179X_B + 92 \]
Where \( X_B \) is the mole fraction of benzene (C₆H₆).

1. **Calculate the vapor pressure of pure benzene (C₆H₆)**:
- When \( X_B = 1 \):
\[ P^0_B = 179(1) + 92 = 271 \, mmHg \]

2. **Calculate the vapor pressure of pure toluene (C₇H₈)**:
- When \( X_B = 0 \):
\[ P^0_T = 179(0) + 92 = 92 \, mmHg \]

### Step 2: Calculate the total moles of each component in the mixture
1. **Calculate moles of benzene (C₆H₆)**:
- Molar mass of benzene = 78 g/mol
\[ \text{Moles of C₆H₆} = \frac{936 \, g}{78 \, g/mol} = 12 \, moles \]

2. **Calculate moles of toluene (C₇H₈)**:
- Molar mass of toluene = 92 g/mol
\[ \text{Moles of C₇H₈} = \frac{736 \, g}{92 \, g/mol} = 8 \, moles \]

### Step 3: Calculate the mole fractions of each component
1. **Total moles in the mixture**:
\[ \text{Total moles} = 12 + 8 = 20 \, moles \]

2. **Mole fraction of benzene (C₆H₆)**:
\[ X_B = \frac{12}{20} = 0.6 \]

3. **Mole fraction of toluene (C₇H₈)**:
\[ X_T = \frac{8}{20} = 0.4 \]

### Step 4: Calculate the total vapor pressure of the mixture
Using Raoult's Law:
\[ P_m = P^0_B \cdot X_B + P^0_T \cdot X_T \]
Substituting the values:
\[ P_m = (271 \, mmHg \cdot 0.6) + (92 \, mmHg \cdot 0.4) \]
Calculating each term:
\[ P_m = 162.6 + 36.8 = 199.4 \, mmHg \]

### Step 5: Calculate the mole fraction of benzene in the vapor phase
Using the formula for the mole fraction of benzene in the vapor phase:
\[ X_B^{vapor} = \frac{P_B}{P_m} \]
Where \( P_B \) is the partial pressure of benzene:
\[ P_B = P^0_B \cdot X_B = 271 \cdot 0.6 = 162.6 \, mmHg \]
Now substituting into the mole fraction formula:
\[ X_B^{vapor} = \frac{162.6}{199.4} \approx 0.815 \]

### Step 6: Calculate the new vapor pressure after condensation
When the vapor is condensed and brought back to 50°C, the new total pressure \( P_m' \) is calculated using the new mole fractions:
\[ P_m' = P^0_B \cdot X_B^{vapor} + P^0_T \cdot X_T^{vapor} \]
Where \( X_T^{vapor} = 1 - X_B^{vapor} \approx 0.185 \) (since \( 1 - 0.815 = 0.185 \)):
\[ P_m' = 271 \cdot 0.815 + 92 \cdot 0.185 \]
Calculating:
\[ P_m' = 220.865 + 17.02 \approx 237.885 \, mmHg \]

### Step 7: Calculate the final mole fraction of benzene in the vapor phase
Using the new total pressure:
\[ X_B^{final} = \frac{P_B}{P_m'} = \frac{220.865}{237.885} \approx 0.928 \]

### Final Answer
The mole fraction of benzene (C₆H₆) in the vapor state is approximately **0.928**.

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