Benzene and toulene form an ideal solution. 3 mole benzene and 2 mole toulene are added. V.P. of pure benezene 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 can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component in the solution.

### Step-by-Step Solution:

1. **Identify the components and their moles**:
- Moles of benzene (A) = 3
- Moles of toluene (B) = 2

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

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. **Use Raoult's Law to calculate the vapor pressure of the solution (Ps)**:
\[ P_s = P^0_A \cdot x_A + P^0_B \cdot x_B \]
where:
- \(P^0_A\) = vapor pressure of pure benzene = 300 mm Hg
- \(P^0_B\) = vapor pressure of pure toluene = 200 mm Hg

Plugging in the values:
\[ P_s = (300 \, \text{mm Hg} \cdot 0.6) + (200 \, \text{mm Hg} \cdot 0.4) \]

6. **Calculate each term**:
\[ P_s = 180 \, \text{mm Hg} + 80 \, \text{mm Hg} \]

7. **Add the results**:
\[ P_s = 260 \, \text{mm Hg} \]

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

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