What are partial and total vapour pressures at `25^(@)C` above the solution having equal numbers of molecules of benzene and toluene ? The vapour pressures of benzene and toluene at this temperature are `95.1` and `28.4`mmHg respectively. What is the composition of the vapour in equilibrium with benzene toluene solution ?

Explanation

To solve the problem step by step, we will calculate the partial and total vapor pressures of a solution containing equal numbers of molecules of benzene and toluene at `25°C`. We will also determine the composition of the vapor in equilibrium with the solution.

### Step 1: Determine the mole fractions of benzene and toluene
Since we have equal numbers of molecules of benzene and toluene, we can assume:
- Number of moles of benzene (n_b) = 1 mole
- Number of moles of toluene (n_t) = 1 mole

The total number of moles in the solution is:
\[ n_{total} = n_b + n_t = 1 + 1 = 2 \]

Now, we can calculate the mole fractions:
- Mole fraction of benzene (X_b):
\[ X_b = \frac{n_b}{n_{total}} = \frac{1}{2} = 0.5 \]

- Mole fraction of toluene (X_t):
\[ X_t = \frac{n_t}{n_{total}} = \frac{1}{2} = 0.5 \]

### Step 2: Calculate the partial vapor pressures
Using Raoult's Law, the partial vapor pressure of each component can be calculated as:
- Partial vapor pressure of benzene (P_b):
\[ P_b = P_{b0} \times X_b \]
Where \( P_{b0} = 95.1 \, \text{mmHg} \)
\[ P_b = 95.1 \, \text{mmHg} \times 0.5 = 47.55 \, \text{mmHg} \]

- Partial vapor pressure of toluene (P_t):
\[ P_t = P_{t0} \times X_t \]
Where \( P_{t0} = 28.4 \, \text{mmHg} \)
\[ P_t = 28.4 \, \text{mmHg} \times 0.5 = 14.2 \, \text{mmHg} \]

### Step 3: Calculate the total vapor pressure
The total vapor pressure (P_total) above the solution is the sum of the partial vapor pressures:
\[ P_{total} = P_b + P_t \]
\[ P_{total} = 47.55 \, \text{mmHg} + 14.2 \, \text{mmHg} = 61.75 \, \text{mmHg} \]

### Step 4: Determine the composition of the vapor
To find the mole fraction of benzene in the vapor phase (Y_b), we use the formula:
\[ Y_b = \frac{P_b}{P_{total}} \]
\[ Y_b = \frac{47.55 \, \text{mmHg}}{61.75 \, \text{mmHg}} \approx 0.769 \]

The mole fraction of toluene in the vapor phase (Y_t) can be calculated as:
\[ Y_t = 1 - Y_b = 1 - 0.769 \approx 0.231 \]

### Summary of Results
- Partial vapor pressure of benzene (P_b) = 47.55 mmHg
- Partial vapor pressure of toluene (P_t) = 14.2 mmHg
- Total vapor pressure (P_total) = 61.75 mmHg
- Composition of vapor: Mole fraction of benzene (Y_b) ≈ 0.769, Mole fraction of toluene (Y_t) ≈ 0.231