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. 70
D. 80

Explanation

To solve the problem, we will use Raoult's law and the concept of weight percent. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- The boiling point of the mixture is \( 89^\circ C \).
- The reduced pressure at which the mixture boils is \( 7.7 \times 10^4 \, \text{Pa} \).
- The vapor pressure of pure water at \( 89^\circ C \) is \( 7 \times 10^4 \, \text{Pa} \).

### Step 2: Apply Raoult's Law
According to Raoult's law, the partial vapor pressure of each component in a mixture is proportional to its mole fraction in the liquid phase. The total vapor pressure of the mixture can be expressed as:
\[ P_{total} = P_{A} + P_{B} \]
Where:
- \( P_{A} = \text{Vapor pressure of component A (water)} \)
- \( P_{B} = \text{Vapor pressure of component B (chlorobenzene)} \)

### Step 3: Set Up the Equation
Using the given data, we can write:
\[ P_{A} = P^0_{A} \cdot x_A \]
\[ P_{B} = P^0_{B} \cdot x_B \]
Where:
- \( P^0_{A} = 7 \times 10^4 \, \text{Pa} \) (vapor pressure of pure water)
- \( P^0_{B} \) is the vapor pressure of pure chlorobenzene (not given directly, but we can find it using the total pressure).

Since the total pressure \( P_{total} = 7.7 \times 10^4 \, \text{Pa} \):
\[ P_{total} = P_{A} + P_{B} = P^0_{A} \cdot x_A + P^0_{B} \cdot x_B \]

### Step 4: Calculate Mole Fractions
Let \( x_A \) be the mole fraction of water and \( x_B \) be the mole fraction of chlorobenzene. Since \( x_A + x_B = 1 \), we can express \( x_B \) as:
\[ x_B = 1 - x_A \]

### Step 5: Substitute and Rearrange
Substituting \( x_B \) in the total pressure equation:
\[ 7.7 \times 10^4 = (7 \times 10^4) \cdot x_A + P^0_{B} \cdot (1 - x_A) \]

### Step 6: Calculate the Weight Ratio
Using the molar masses:
- Molar mass of water \( M_A = 18 \, \text{g/mol} \)
- Molar mass of chlorobenzene \( M_B = 112.5 \, \text{g/mol} \)

The weight ratio \( \frac{W_A}{W_B} \) can be expressed as:
\[ \frac{W_A}{W_B} = \frac{P^0_A}{P^0_B} \cdot \frac{M_B}{M_A} \]

### Step 7: Solve for Weight Percent of Chlorobenzene
The weight percent of chlorobenzene in the distillate is given by:
\[ \text{Weight percent of } B = \frac{W_B}{W_A + W_B} \times 100 \]
Where \( W_A \) and \( W_B \) can be found from the weight ratio calculated in the previous step.

### Final Calculation
After substituting the values and solving, we find:
\[ \frac{W_A}{W_B} = 0.625 \]
Thus, if \( W_A = 0.625 W_B \), then:
\[ W_B = 1 \quad \text{(assuming a unit mass for simplicity)} \]
\[ W_A = 0.625 \]
Now, calculating the weight percent:
\[ \text{Weight percent of chlorobenzene} = \frac{W_B}{W_A + W_B} \times 100 = \frac{1}{0.625 + 1} \times 100 \approx 38.46\% \]

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