At `40^(@)C` the vapour pressure of pure liquids, benzene and toluene, are `160 mm Hg` and `60 mm Hg` respectively. At the same temperature, the vapour pressure of an equimolar solution of the liquids, assuming the ideal solution will be:

Options

A. `140 mm Hg`
B. `110 mm Hg`
C. `220 mm Hg`
D. `100 mm Hg`

Explanation

To solve the problem, we need to calculate the vapor pressure of an equimolar solution of benzene and toluene at 40°C, given their pure vapor pressures.

### Step-by-Step Solution:

1. **Identify the given data**:
- Vapor pressure of pure benzene (P₀B) = 160 mm Hg
- Vapor pressure of pure toluene (P₀T) = 60 mm Hg
- Since the solution is equimolar, we can assume we have 1 mole of benzene and 1 mole of toluene.

2. **Calculate the total moles in the solution**:
- Moles of benzene (nB) = 1
- Moles of toluene (nT) = 1
- Total moles (n_total) = nB + nT = 1 + 1 = 2 moles

3. **Calculate the mole fractions of benzene and toluene**:
- Mole fraction of benzene (xB) = nB / n_total = 1 / 2 = 0.5
- Mole fraction of toluene (xT) = nT / n_total = 1 / 2 = 0.5

4. **Apply Raoult's Law to find the total vapor pressure (P_total)**:
- According to Raoult's Law, the total vapor pressure of the solution is given by:
\[ P_{total} = P_0B \cdot xB + P_0T \cdot xT \]
- Substitute the values:
\[ P_{total} = (160 \, \text{mm Hg} \cdot 0.5) + (60 \, \text{mm Hg} \cdot 0.5) \]
- Calculate each term:
\[ P_{total} = 80 \, \text{mm Hg} + 30 \, \text{mm Hg} \]
- Combine the results:
\[ P_{total} = 110 \, \text{mm Hg} \]

5. **Final Answer**:
- The vapor pressure of the equimolar solution of benzene and toluene at 40°C is **110 mm Hg**.