If an ideal solution is made by mixing 2 moles of benzene `(p^(@)=266mm)` and 3 moles of another liquid `(p^(@)=236 mm)`. The total vapour pressure of the solution at the same temperature would be

Options

A. 502 mm-Hg
B. 248mm-Hg
C. 600 mm-Hg
D. 250.6 mm -Hg

Answer

To find the total vapor pressure of the solution made by mixing 2 moles of benzene and 3 moles of another liquid, we can use Raoult's Law, which states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Moles of benzene (C₆H₆) = 2 moles
- Moles of another liquid = 3 moles
- Vapor pressure of pure benzene (P₁°) = 266 mmHg
- Vapor pressure of pure liquid (P₂°) = 236 mmHg

2. **Calculate the total number of moles in the solution:**
\[ \text{Total moles} = \text{Moles of benzene} + \text{Moles of another liquid} = 2 + 3 = 5 \text{ moles} \]

3. **Calculate the mole fraction of benzene (X₁):**
\[ X_1 = \frac{\text{Moles of benzene}}{\text{Total moles}} = \frac{2}{5} \]

4. **Calculate the mole fraction of the other liquid (X₂):**
\[ X_2 = \frac{\text{Moles of another liquid}}{\text{Total moles}} = \frac{3}{5} \]

5. **Calculate the partial vapor pressure of benzene (P₁):**
\[ P_1 = X_1 \cdot P_1° = \left(\frac{2}{5}\right) \cdot 266 \text{ mmHg} \]
\[ P_1 = \frac{532}{5} = 106.4 \text{ mmHg} \]

6. **Calculate the partial vapor pressure of the other liquid (P₂):**
\[ P_2 = X_2 \cdot P_2° = \left(\frac{3}{5}\right) \cdot 236 \text{ mmHg} \]
\[ P_2 = \frac{708}{5} = 141.6 \text{ mmHg} \]

7. **Calculate the total vapor pressure of the solution (P_total):**
\[ P_{\text{total}} = P_1 + P_2 = 106.4 \text{ mmHg} + 141.6 \text{ mmHg} \]
\[ P_{\text{total}} = 248 \text{ mmHg} \]

### Final Answer:
The total vapor pressure of the solution is **248 mmHg**.