Which of the following statements about the composition of the vapour over an ideal `1:1` mol mixture of benzene and toluene is correct? Assume that the temperature is constant at `25^(@)C`. (Given: vapour pressure Date at `25^(@)C`, benzene=12.8 kP, toluene=3.85 kPa)

Options

A. The vapour will contain a higher percentage of toluene
B. The vapour will contain equal amounts of benzene toluene
C. Not enough information is given to make a prediction
D. The vapour will contain a higher percentage of benzene

Explanation

To solve the problem regarding the composition of the vapor over an ideal 1:1 mol mixture of benzene and toluene, we will use Raoult's Law. Here are the steps to arrive at the solution:

### Step 1: Identify the Given Data
- Vapor pressure of pure benzene (\(P^0_{A}\)) = 12.8 kPa
- Vapor pressure of pure toluene (\(P^0_{B}\)) = 3.85 kPa
- Mole fraction of benzene (\(X_A\)) = 0.5 (since the mixture is 1:1)
- Mole fraction of toluene (\(X_B\)) = 0.5 (since the mixture is 1:1)

### Step 2: Apply Raoult's Law
According to Raoult's Law, the partial pressure of each component in the vapor phase can be calculated as:
- Partial pressure of benzene (\(P_A\)) = \(P^0_{A} \times X_A\)
- Partial pressure of toluene (\(P_B\)) = \(P^0_{B} \times X_B\)

### Step 3: Calculate the Partial Pressures
Using the mole fractions:
1. **For benzene:**
\[ P_A = P^0_{A} \times X_A = 12.8 \, \text{kPa} \times 0.5 = 6.4 \, \text{kPa} \]

2. **For toluene:**
\[ P_B = P^0_{B} \times X_B = 3.85 \, \text{kPa} \times 0.5 = 1.925 \, \text{kPa} \]

### Step 4: Determine the Total Pressure
The total pressure (\(P_{total}\)) of the vapor phase is the sum of the partial pressures:
\[ P_{total} = P_A + P_B = 6.4 \, \text{kPa} + 1.925 \, \text{kPa} = 8.325 \, \text{kPa} \]

### Step 5: Calculate the Mole Fractions in the Vapor Phase
Now, we can find the mole fractions of benzene and toluene in the vapor phase:
1. **Mole fraction of benzene in vapor (\(Y_A\)):**
\[ Y_A = \frac{P_A}{P_{total}} = \frac{6.4 \, \text{kPa}}{8.325 \, \text{kPa}} \approx 0.769 \]

2. **Mole fraction of toluene in vapor (\(Y_B\)):**
\[ Y_B = \frac{P_B}{P_{total}} = \frac{1.925 \, \text{kPa}}{8.325 \, \text{kPa}} \approx 0.231 \]

### Step 6: Conclusion
Since the mole fraction of benzene in the vapor phase (\(Y_A \approx 0.769\)) is greater than that of toluene (\(Y_B \approx 0.231\)), this indicates that the vapor above the mixture contains a higher proportion of benzene compared to toluene.

### Final Answer
The correct statement about the composition of the vapor over the ideal 1:1 mol mixture of benzene and toluene is that the vapor will contain more moles of benzene than toluene.

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