Water and chlorobenzene are immiscible liquids. Their mixture boils at `89^(@)C` under a redued pressure of `7.7 xx10^(4)Pa`. The vapour pressure of pure water at `89^(@)C` is `7xx10^(4)Pa`. Weight per ccent of chlorobenzene in the distillate is `:`

Options

A. 50
B. 60
C. `78.3`
D. `38.46`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify Given Data
- Boiling point of the mixture: \( T = 89^\circ C \)
- Reduced pressure of the mixture: \( P_{\text{total}} = 7.7 \times 10^4 \, \text{Pa} \)
- Vapor pressure of pure water at \( 89^\circ C \): \( P_{B0} = 7 \times 10^4 \, \text{Pa} \)

### Step 2: Calculate Vapor Pressure of Pure Chlorobenzene
Using the formula for total pressure:
\[ P_{\text{total}} = P_{A0} + P_{B0} \]
where \( P_{A0} \) is the vapor pressure of pure chlorobenzene. Rearranging gives:
\[ P_{A0} = P_{\text{total}} - P_{B0} \]
Substituting the values:
\[ P_{A0} = 7.7 \times 10^4 \, \text{Pa} - 7 \times 10^4 \, \text{Pa} = 0.7 \times 10^4 \, \text{Pa} \]

### Step 3: Use the Formula for Weight Ratio
The weight ratio of chlorobenzene to water can be calculated using the formula:
\[ \frac{W_A}{W_B} = \frac{P_{A0} \cdot M_A}{P_{B0} \cdot M_B} \]
where:
- \( W_A \) = weight of chlorobenzene
- \( W_B \) = weight of water
- \( M_A \) = molar mass of chlorobenzene = 112.5 g/mol
- \( M_B \) = molar mass of water = 18 g/mol

Substituting the known values:
\[ \frac{W_A}{W_B} = \frac{(0.7 \times 10^4) \cdot 112.5}{(7 \times 10^4) \cdot 18} \]

### Step 4: Calculate the Weight Ratio
Calculating the right side:
\[ \frac{W_A}{W_B} = \frac{0.7 \times 112.5}{7 \times 18} \]
Calculating the numerator:
\[ 0.7 \times 112.5 = 78.75 \]
Calculating the denominator:
\[ 7 \times 18 = 126 \]
Thus:
\[ \frac{W_A}{W_B} = \frac{78.75}{126} \approx 0.625 \]

### Step 5: Calculate Weight Percent of Chlorobenzene
To find the weight percent of chlorobenzene in the distillate:
\[ \text{Weight percent of chlorobenzene} = \frac{W_A}{W_A + W_B} \times 100 \]
From the ratio \( W_A = 0.625 W_B \), we can express \( W_A + W_B \) as:
\[ W_A + W_B = 0.625 W_B + W_B = 1.625 W_B \]
Thus:
\[ \text{Weight percent of chlorobenzene} = \frac{0.625 W_B}{1.625 W_B} \times 100 \]
The \( W_B \) cancels out:
\[ \text{Weight percent of chlorobenzene} = \frac{0.625}{1.625} \times 100 \approx 38.46\% \]

### Final Answer
The weight percent of chlorobenzene in the distillate is approximately **38.46%**.