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 : (i) containing equal weights of benzene and toluene.

Explanation

To solve the problem, we need to calculate the partial pressures of benzene and toluene in a solution containing equal weights of both substances. We will use Raoult's Law, which states that the partial pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

### Step-by-Step Solution

1. **Determine the weights of benzene and toluene:**
Let's assume we have 100 g of each component for simplicity.
- Weight of benzene (C₆H₆) = 100 g
- Weight of toluene (C₇H₈) = 100 g

2. **Calculate the number of moles of benzene and toluene:**
- Molar mass of benzene (C₆H₆) = 78 g/mol
- Molar mass of toluene (C₇H₈) = 92 g/mol

\[ \text{Moles of benzene} = \frac{100 \, \text{g}}{78 \, \text{g/mol}} \approx 1.28 \, \text{mol} \]

\[ \text{Moles of toluene} = \frac{100 \, \text{g}}{92 \, \text{g/mol}} \approx 1.09 \, \text{mol} \]

3. **Calculate the total number of moles in the solution:**
\[ \text{Total moles} = \text{Moles of benzene} + \text{Moles of toluene} = 1.28 + 1.09 \approx 2.37 \, \text{mol} \]

4. **Calculate the mole fractions of benzene and toluene:**
\[ X_{\text{benzene}} = \frac{\text{Moles of benzene}}{\text{Total moles}} = \frac{1.28}{2.37} \approx 0.54 \]

\[ X_{\text{toluene}} = \frac{\text{Moles of toluene}}{\text{Total moles}} = \frac{1.09}{2.37} \approx 0.46 \]

5. **Use Raoult's Law to calculate the partial pressures:**
- Vapor pressure of pure benzene (Pₐ⁰) = 160 mmHg
- Vapor pressure of pure toluene (Pᵦ⁰) = 60 mmHg

\[ P_{\text{benzene}} = P_a^0 \cdot X_{\text{benzene}} = 160 \, \text{mmHg} \cdot 0.54 \approx 86.4 \, \text{mmHg} \]

\[ P_{\text{toluene}} = P_b^0 \cdot X_{\text{toluene}} = 60 \, \text{mmHg} \cdot 0.46 \approx 27.6 \, \text{mmHg} \]

6. **Calculate the total pressure over the solution:**
\[ P_{\text{total}} = P_{\text{benzene}} + P_{\text{toluene}} = 86.4 \, \text{mmHg} + 27.6 \, \text{mmHg} \approx 114 \, \text{mmHg} \]

### Final Results
- Partial pressure of benzene: **86.4 mmHg**
- Partial pressure of toluene: **27.6 mmHg**
- Total pressure: **114 mmHg**