When 1 mole of benzene is mixed with 1 mole of toluene The vapour will contain: (Given : vapour of benzene = 12.8kPa and vapour pressure of toluene = 3.85 kPa).

Options

A. equal amount of benzene and toluene as it forms an ideal solution
B. unequal amount of benzene and toluene as it forms a non ideal solution
C. higher percentage of benzene
D. higher percentage of toluene

Answer

To solve the problem of determining the vapor composition when 1 mole of benzene is mixed with 1 mole of toluene, we can use Raoult's Law. Here’s a step-by-step solution:

### Step 1: Understand the Components
We have two components in the mixture:
- Benzene (C6H6) with a vapor pressure of 12.8 kPa.
- Toluene (C7H8) with a vapor pressure of 3.85 kPa.

### Step 2: Calculate the Total Vapor Pressure
According to Raoult's Law, the total vapor pressure (P_total) of a solution is the sum of the partial pressures of each component:
\[ P_{total} = P_{benzene} + P_{toluene} \]

### Step 3: Calculate the Partial Pressures
Since we have equal moles of benzene and toluene (1 mole each), we can calculate the partial pressures using their respective vapor pressures and mole fractions.

- Mole fraction of benzene (X_benzene):
\[ X_{benzene} = \frac{n_{benzene}}{n_{benzene} + n_{toluene}} = \frac{1}{1 + 1} = \frac{1}{2} \]

- Mole fraction of toluene (X_toluene):
\[ X_{toluene} = \frac{n_{toluene}}{n_{benzene} + n_{toluene}} = \frac{1}{1 + 1} = \frac{1}{2} \]

Now, we can calculate the partial pressures:
- Partial pressure of benzene (P_benzene):
\[ P_{benzene} = X_{benzene} \times P^{0}_{benzene} = \frac{1}{2} \times 12.8 \, \text{kPa} = 6.4 \, \text{kPa} \]

- Partial pressure of toluene (P_toluene):
\[ P_{toluene} = X_{toluene} \times P^{0}_{toluene} = \frac{1}{2} \times 3.85 \, \text{kPa} = 1.925 \, \text{kPa} \]

### Step 4: Calculate the Total Vapor Pressure
Now, we can find the total vapor pressure:
\[ P_{total} = P_{benzene} + P_{toluene} = 6.4 \, \text{kPa} + 1.925 \, \text{kPa} = 8.325 \, \text{kPa} \]

### Step 5: Determine the Composition of the Vapor
To find the composition of the vapor, we can calculate the mole fractions of each component in the vapor phase:
- Mole fraction of benzene in vapor (Y_benzene):
\[ Y_{benzene} = \frac{P_{benzene}}{P_{total}} = \frac{6.4 \, \text{kPa}}{8.325 \, \text{kPa}} \approx 0.769 \]

- Mole fraction of toluene in vapor (Y_toluene):
\[ Y_{toluene} = \frac{P_{toluene}}{P_{total}} = \frac{1.925 \, \text{kPa}}{8.325 \, \text{kPa}} \approx 0.231 \]

### Final Result
The vapor will contain approximately:
- 76.9% benzene
- 23.1% toluene