The total vapour pressure of equimolar solution of benzene and toluene is given by `P_(T)` (in mm Hg) `=600-400X_("toluene")` then:

Options

A. `p_("benzene")=200mmHg`
B. `p_("benzene")^(0)=600mmHg`
C. `p_("toluene")^(0)=200mmHg`
D. `p_("toluene")^(0)=600mmHg`

Explanation

To solve the problem, we need to find the vapor pressures of benzene and toluene in an equimolar solution where the total vapor pressure \( P_T \) is given by the equation:

\[ P_T = 600 - 400X_{\text{toluene}} \]

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( P_T = 600 - 400X_{\text{toluene}} \) indicates that the total vapor pressure decreases linearly with increasing mole fraction of toluene. Here, \( X_{\text{toluene}} \) is the mole fraction of toluene in the solution.

2. **Case 1: Pure Toluene**:
- When the solution is pure toluene, the mole fraction of toluene \( X_{\text{toluene}} = 1 \).
- Substitute \( X_{\text{toluene}} = 1 \) into the equation:
\[ P_T = 600 - 400(1) = 600 - 400 = 200 \text{ mm Hg} \]
- Thus, the vapor pressure of pure toluene \( P^0_{\text{toluene}} = 200 \text{ mm Hg} \).

3. **Case 2: Pure Benzene**:
- When the solution is pure benzene, the mole fraction of toluene \( X_{\text{toluene}} = 0 \).
- Substitute \( X_{\text{toluene}} = 0 \) into the equation:
\[ P_T = 600 - 400(0) = 600 - 0 = 600 \text{ mm Hg} \]
- Thus, the vapor pressure of pure benzene \( P^0_{\text{benzene}} = 600 \text{ mm Hg} \).

4. **Conclusion**:
- The vapor pressure of pure toluene is \( 200 \text{ mm Hg} \).
- The vapor pressure of pure benzene is \( 600 \text{ mm Hg} \).

### Final Answer:
- Vapor pressure of pure toluene: **200 mm Hg**
- Vapor pressure of pure benzene: **600 mm Hg**