At `90^(@)C` th vapour pressure fo toluene is 400 mm and that of xylene is 150 mm. The mole of fraction of toluene in liquid mixture that will boil at `90(@)C` when the pressure of mixture is 0.5 atm will be :

Options

A. 0.08
B. 0.45
C. 0.92
D. 0.55

Answer

To solve the problem, we need to find the mole fraction of toluene in a liquid mixture that will boil at 90°C when the total pressure of the mixture is 0.5 atm. We are given the vapor pressures of pure toluene and xylene at this temperature.

### Step-by-Step Solution:

1. **Convert the Total Pressure from atm to mm Hg:**
\[ \text{Total Pressure} = 0.5 \, \text{atm} \times 760 \, \text{mm Hg/atm} = 380 \, \text{mm Hg} \]

2. **Identify the Given Vapor Pressures:**
- Vapor pressure of toluene (\(P^0_T\)) = 400 mm Hg
- Vapor pressure of xylene (\(P^0_X\)) = 150 mm Hg

3. **Apply Raoult's Law:**
According to Raoult's Law, the total vapor pressure (\(P\)) of the solution is given by:
\[ P = P^0_T \cdot x_T + P^0_X \cdot x_X \]
where \(x_T\) is the mole fraction of toluene and \(x_X\) is the mole fraction of xylene.

4. **Express \(x_X\) in terms of \(x_T\):**
Since \(x_T + x_X = 1\), we can write:
\[ x_X = 1 - x_T \]

5. **Substitute into the Raoult's Law Equation:**
\[ 380 = 400 \cdot x_T + 150 \cdot (1 - x_T) \]

6. **Expand and Rearrange the Equation:**
\[ 380 = 400x_T + 150 - 150x_T \]
\[ 380 = 250x_T + 150 \]
\[ 380 - 150 = 250x_T \]
\[ 230 = 250x_T \]

7. **Solve for \(x_T\):**
\[ x_T = \frac{230}{250} = 0.92 \]

8. **Conclusion:**
The mole fraction of toluene in the liquid mixture that will boil at 90°C under a pressure of 0.5 atm is \(x_T = 0.92\).

### Final Answer:
The mole fraction of toluene in the liquid mixture is **0.92**.