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 solve the problem, we need to find the mole fraction of benzene in a mixture with toluene that will boil at 88°C and 1 atm pressure. We will use the given vapor pressures of benzene and toluene at that temperature to calculate the mole fraction.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Vapor pressure of benzene (P_benzene) = 900 torr
- Vapor pressure of toluene (P_toluene) = 360 torr
- Boiling pressure (P_total) = 1 atm = 760 torr

2. **Use Raoult's Law:**
For an ideal solution, the total vapor pressure (P_total) is given by:
\[ P_{total} = X_{benzene} \cdot P_{benzene} + X_{toluene} \cdot P_{toluene} \]
where \(X_{benzene}\) and \(X_{toluene}\) are the mole fractions of benzene and toluene, respectively.

3. **Express Mole Fractions:**
Since \(X_{benzene} + X_{toluene} = 1\), we can express \(X_{toluene}\) as:
\[ X_{toluene} = 1 - X_{benzene} \]

4. **Substitute into Raoult's Law:**
Substitute \(X_{toluene}\) into the equation:
\[ 760 = X_{benzene} \cdot 900 + (1 - X_{benzene}) \cdot 360 \]

5. **Simplify the Equation:**
Expanding the equation gives:
\[ 760 = 900X_{benzene} + 360 - 360X_{benzene} \]
Combine like terms:
\[ 760 = (900 - 360)X_{benzene} + 360 \]
\[ 760 - 360 = 540X_{benzene} \]
\[ 400 = 540X_{benzene} \]

6. **Solve for \(X_{benzene}\):**
\[ X_{benzene} = \frac{400}{540} = \frac{20}{27} \approx 0.7407 \]

7. **Final Result:**
The mole fraction of benzene in the mixture is approximately \(0.74\).

### Conclusion:
The mole fraction of benzene in the mixture that will boil at 88°C at 1 atm pressure is approximately **0.74**.