Benzene and toulene form an ideal solution. 3 mole benzene and 2 mole toulene are added. V.P. of pure benzene and toulene are 300 & 200 mm of Hg respectively. The V.P of the solution (in mm of Hg) is

Options

A. 500
B. 250
C. 260
D. 440

Answer

To find the vapor pressure of the solution formed by benzene and toluene, we will use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of its components.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Moles of Benzene (A) = 3 moles
- Moles of Toluene (B) = 2 moles
- Vapor Pressure of Pure Benzene (P₀A) = 300 mm Hg
- Vapor Pressure of Pure Toluene (P₀B) = 200 mm Hg

2. **Calculate the Total Moles in the Solution:**
\[ \text{Total moles} = \text{moles of benzene} + \text{moles of toluene} = 3 + 2 = 5 \text{ moles} \]

3. **Calculate the Mole Fraction of Benzene (xA):**
\[ x_A = \frac{\text{moles of benzene}}{\text{total moles}} = \frac{3}{5} = 0.6 \]

4. **Calculate the Mole Fraction of Toluene (xB):**
\[ x_B = \frac{\text{moles of toluene}}{\text{total moles}} = \frac{2}{5} = 0.4 \]

5. **Apply Raoult's Law to Find the Vapor Pressure of the Solution (P):**
\[ P = P₀A \cdot x_A + P₀B \cdot x_B \]
Substituting the values:
\[ P = (300 \, \text{mm Hg} \cdot 0.6) + (200 \, \text{mm Hg} \cdot 0.4) \]

6. **Calculate Each Term:**
- For Benzene:
\[ 300 \, \text{mm Hg} \cdot 0.6 = 180 \, \text{mm Hg} \]
- For Toluene:
\[ 200 \, \text{mm Hg} \cdot 0.4 = 80 \, \text{mm Hg} \]

7. **Sum the Partial Pressures:**
\[ P = 180 \, \text{mm Hg} + 80 \, \text{mm Hg} = 260 \, \text{mm Hg} \]

### Final Answer:
The vapor pressure of the solution is **260 mm Hg**.

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