Water and chlorobenzene are immiscible liquids. Their mixture boils at `89^(@)C` under reduced pressure of `7.7 xx10^(4) pa`. The vapour pressure of pure water at `89^(@)C` is `7 xx10^(4) pa` Mass per cent of chlorobenzene in the distillate is :

Options

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

Answer

To solve the problem of finding the mass percent of chlorobenzene in the distillate, we can follow these steps:

### Step 1: Identify Given Data
- Boiling point of the mixture: \( 89^\circ C \)
- Reduced pressure of the mixture: \( P_{\text{mix}} = 7.7 \times 10^4 \, \text{Pa} \)
- Vapor pressure of pure water at \( 89^\circ C \): \( P^0_{\text{water}} = 7.0 \times 10^4 \, \text{Pa} \)

### Step 2: Calculate the Vapor Pressure of Chlorobenzene
Using the relationship:
\[ P_{\text{CB}} = P_{\text{mix}} - P^0_{\text{water}} \]
Substituting the values:
\[ P_{\text{CB}} = 7.7 \times 10^4 \, \text{Pa} - 7.0 \times 10^4 \, \text{Pa} = 0.7 \times 10^4 \, \text{Pa} = 7.0 \times 10^3 \, \text{Pa} \]

### Step 3: Apply Raoult's Law
According to Raoult's Law:
\[ \frac{P_{\text{CB}}}{P^0_{\text{water}}} = \frac{x_{\text{CB}}}{x_{\text{water}}} \]
Where \( x_{\text{CB}} \) and \( x_{\text{water}} \) are the mole fractions of chlorobenzene and water respectively.

Substituting the pressures:
\[ \frac{7.0 \times 10^3}{7.0 \times 10^4} = \frac{x_{\text{CB}}}{x_{\text{water}}} \]
This simplifies to:
\[ \frac{1}{10} = \frac{x_{\text{CB}}}{x_{\text{water}}} \]
This implies:
\[ x_{\text{CB}} = 0.1 \cdot x_{\text{water}} \]

### Step 4: Express Mole Fractions
Let \( n_{\text{CB}} \) be the number of moles of chlorobenzene and \( n_{\text{water}} \) be the number of moles of water. Then:
\[ x_{\text{CB}} = \frac{n_{\text{CB}}}{n_{\text{CB}} + n_{\text{water}}} \]
\[ x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{CB}} + n_{\text{water}}} \]

From the previous equation:
\[ 0.1 = \frac{n_{\text{CB}}}{n_{\text{water}}} \]
This implies:
\[ n_{\text{CB}} = 0.1 \cdot n_{\text{water}} \]

### Step 5: Calculate Masses
Let the mass of chlorobenzene be \( m_{\text{CB}} \) and the mass of water be \( m_{\text{water}} \).
Using molecular weights:
- Molecular weight of chlorobenzene \( M_{\text{CB}} = 112.56 \, \text{g/mol} \)
- Molecular weight of water \( M_{\text{water}} = 18 \, \text{g/mol} \)

Then:
\[ m_{\text{CB}} = n_{\text{CB}} \cdot M_{\text{CB}} = 0.1 \cdot n_{\text{water}} \cdot 112.56 \]
\[ m_{\text{water}} = n_{\text{water}} \cdot M_{\text{water}} = n_{\text{water}} \cdot 18 \]

### Step 6: Find Mass Percent of Chlorobenzene
The total mass is:
\[ m_{\text{total}} = m_{\text{CB}} + m_{\text{water}} = (0.1 \cdot n_{\text{water}} \cdot 112.56) + (n_{\text{water}} \cdot 18) \]
Factoring out \( n_{\text{water}} \):
\[ m_{\text{total}} = n_{\text{water}} \cdot (0.1 \cdot 112.56 + 18) \]

Now, the mass percent of chlorobenzene is:
\[ \text{Mass percent of CB} = \frac{m_{\text{CB}}}{m_{\text{total}}} \times 100 = \frac{0.1 \cdot n_{\text{water}} \cdot 112.56}{n_{\text{water}} \cdot (0.1 \cdot 112.56 + 18)} \times 100 \]
This simplifies to:
\[ = \frac{0.1 \cdot 112.56}{0.1 \cdot 112.56 + 18} \times 100 \]
Calculating the values:
\[ = \frac{11.256}{11.256 + 18} \times 100 = \frac{11.256}{29.256} \times 100 \approx 38.46\% \]

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