At 293 K, vapour pressure of pure benzene is 75mm of Hg and
that of pure toluene is 22 mm of Hg. The vapour pressure of the
solution which contains 20 mol% benzene and
80 mol % toluene is

Options

A. 32.6 mm Hg
B. 64.4 mm Hg
C. 97 mm Hg
D. 3.26 mm Hg

Explanation

To solve the problem of finding the vapor pressure of a solution containing 20 mol% benzene and 80 mol% toluene, we will use Raoult's Law. Here’s a step-by-step solution:

### Step 1: Determine the mole fractions of benzene and toluene.
- Given:
- Mole percent of benzene = 20%
- Mole percent of toluene = 80%

- Convert mole percentages to mole fractions:
- Mole fraction of benzene (X_benzene) = 20% = 0.20
- Mole fraction of toluene (X_toluene) = 80% = 0.80

### Step 2: Identify the vapor pressures of pure benzene and pure toluene.
- Given:
- Vapor pressure of pure benzene (P°_benzene) = 75 mm Hg
- Vapor pressure of pure toluene (P°_toluene) = 22 mm Hg

### Step 3: Apply Raoult's Law to calculate the vapor pressure of the solution.
- According to Raoult's Law:
\[ P_{solution} = P°_{benzene} \times X_{benzene} + P°_{toluene} \times X_{toluene} \]

### Step 4: Substitute the values into the equation.
- Substitute the known values:
\[ P_{solution} = (75 \, \text{mm Hg} \times 0.20) + (22 \, \text{mm Hg} \times 0.80) \]

### Step 5: Calculate the contributions from each component.
- Calculate the contribution from benzene:
\[ P_{benzene} = 75 \, \text{mm Hg} \times 0.20 = 15 \, \text{mm Hg} \]
- Calculate the contribution from toluene:
\[ P_{toluene} = 22 \, \text{mm Hg} \times 0.80 = 17.6 \, \text{mm Hg} \]

### Step 6: Add the contributions to find the total vapor pressure of the solution.
- Total vapor pressure:
\[ P_{solution} = 15 \, \text{mm Hg} + 17.6 \, \text{mm Hg} = 32.6 \, \text{mm Hg} \]

### Final Answer:
The vapor pressure of the solution is **32.6 mm Hg**.

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