At `88^(@)C` benzene has a vapour pressure of 900 torr and toluene has a vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will boil at `88^(@)C` at 1 atm pressure? (Consider that benzene toluene form an ideal solution):

Options

A. 0.416
B. 0.588
C. 0.68
D. 0.74

Answer

To find the mole fraction of benzene in a mixture with toluene that will boil at 88°C and 1 atm pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Vapor pressure of benzene (P0B) at 88°C = 900 torr
- Vapor pressure of toluene (P0T) at 88°C = 360 torr
- Total pressure (P_total) at boiling point = 1 atm = 760 torr

2. **Apply Raoult's Law**:
According to Raoult's Law, the total vapor pressure of an ideal solution is the sum of the partial pressures of each component:
\[ P_{total} = P_B + P_T \]
where \( P_B \) and \( P_T \) are the partial pressures of benzene and toluene, respectively.

3. **Express Partial Pressures**:
The partial pressures can be expressed in terms of the mole fractions:
\[ P_B = P0B \cdot X_B \]
\[ P_T = P0T \cdot X_T \]
where \( X_B \) is the mole fraction of benzene and \( X_T \) is the mole fraction of toluene.

4. **Relate Mole Fractions**:
Since the sum of the mole fractions is equal to 1:
\[ X_B + X_T = 1 \implies X_T = 1 - X_B \]

5. **Substitute into the Total Pressure Equation**:
Substitute the expressions for \( P_B \) and \( P_T \) into the total pressure equation:
\[ P_{total} = P0B \cdot X_B + P0T \cdot (1 - X_B) \]
Plugging in the values:
\[ 760 = 900 \cdot X_B + 360 \cdot (1 - X_B) \]

6. **Simplify the Equation**:
Expanding the equation:
\[ 760 = 900X_B + 360 - 360X_B \]
Combine like terms:
\[ 760 = (900 - 360)X_B + 360 \]
\[ 760 - 360 = 540X_B \]
\[ 400 = 540X_B \]

7. **Solve for Mole Fraction of Benzene**:
\[ X_B = \frac{400}{540} = \frac{20}{27} \approx 0.7407 \]

8. **Final Answer**:
The mole fraction of benzene in the mixture is approximately \( 0.740 \).