Benzene `(C_(6)H_(6))` and toluene `(C_(7)H_(8))` from a nearly ideal solution at 313 K. The vapour pressure of pure benzene and toluene are 160 mm of Hg and 60 mm of Hg respectively. Calculate the partial pressure of benzene and toluene and the total pressure over the following solutions : (ii) containing 1 mole of benzene and 4 moles of toluene.

Explanation

To solve the problem, we need to calculate the partial pressures of benzene and toluene in a solution containing 1 mole of benzene and 4 moles of toluene, and then find the total pressure.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Vapor pressure of pure benzene, \( P_{B}^0 = 160 \, \text{mm Hg} \)
- Vapor pressure of pure toluene, \( P_{T}^0 = 60 \, \text{mm Hg} \)
- Moles of benzene, \( n_B = 1 \)
- Moles of toluene, \( n_T = 4 \)

2. **Calculate the Total Moles in the Solution:**
\[ n_{total} = n_B + n_T = 1 + 4 = 5 \, \text{moles} \]

3. **Calculate the Mole Fraction of Benzene (\( X_B \)):**
\[ X_B = \frac{n_B}{n_{total}} = \frac{1}{5} = 0.2 \]

4. **Calculate the Mole Fraction of Toluene (\( X_T \)):**
\[ X_T = \frac{n_T}{n_{total}} = \frac{4}{5} = 0.8 \]

5. **Calculate the Partial Pressure of Benzene (\( P_B \)):**
\[ P_B = P_B^0 \times X_B = 160 \, \text{mm Hg} \times 0.2 = 32 \, \text{mm Hg} \]

6. **Calculate the Partial Pressure of Toluene (\( P_T \)):**
\[ P_T = P_T^0 \times X_T = 60 \, \text{mm Hg} \times 0.8 = 48 \, \text{mm Hg} \]

7. **Calculate the Total Pressure (\( P_{total} \)):**
\[ P_{total} = P_B + P_T = 32 \, \text{mm Hg} + 48 \, \text{mm Hg} = 80 \, \text{mm Hg} \]

### Final Results:
- Partial pressure of benzene, \( P_B = 32 \, \text{mm Hg} \)
- Partial pressure of toluene, \( P_T = 48 \, \text{mm Hg} \)
- Total pressure, \( P_{total} = 80 \, \text{mm Hg} \)