A liquid is immiscible in water was steam distilled at `95.2^(@)C` at a pressure of `0.983` atm. What is the mass of the liquid present per gram of water in the distullate. Molar mass of the liquid is `134.3` g/mol and the vapour pressure of water is `0.84` atm. Also, Vapour pressure of pure liquid is `0.143` atm.

Options

A. `1 g`
B. `1.27 g`
C. `0.787 g`
D. `13.43 g`

Answer

To solve the problem, we need to determine the mass of the immiscible liquid present per gram of water in the distillate. We will use the given data and apply Dalton's law of partial pressures.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Molar mass of the liquid (M_l) = 134.3 g/mol
- Molar mass of water (M_w) = 18 g/mol
- Vapor pressure of water (P_w) = 0.84 atm
- Vapor pressure of the liquid (P_l) = 0.143 atm
- Total pressure (P_total) = 0.983 atm

2. **Apply Dalton's Law of Partial Pressures:**
According to Dalton's law, the total pressure is the sum of the partial pressures of the components:
\[ P_{total} = P_w + P_l \]

3. **Calculate Mole Fractions:**
The mole fraction of the liquid (X_l) can be expressed as:
\[ \frac{P_l}{P_w} = \frac{n_l}{n_l + n_w} \]
Where \( n_l \) is the number of moles of the liquid and \( n_w \) is the number of moles of water.

4. **Express Moles in Terms of Mass:**
The number of moles can be calculated using the formula:
\[ n = \frac{mass}{molar\ mass} \]
For water (1 g):
\[ n_w = \frac{1 \, \text{g}}{18 \, \text{g/mol}} = \frac{1}{18} \, \text{mol} \]
For the liquid, let the mass of the liquid be \( m_l \):
\[ n_l = \frac{m_l}{134.3} \]

5. **Substitute into the Mole Fraction Equation:**
Now substituting \( n_l \) and \( n_w \) into the mole fraction equation:
\[ \frac{0.143}{0.84} = \frac{\frac{m_l}{134.3}}{\frac{m_l}{134.3} + \frac{1}{18}} \]

6. **Cross Multiply and Simplify:**
Cross-multiplying gives:
\[ 0.143 \left( \frac{m_l}{134.3} + \frac{1}{18} \right) = 0.84 \cdot \frac{m_l}{134.3} \]
Expanding this:
\[ 0.143 \cdot \frac{m_l}{134.3} + 0.143 \cdot \frac{1}{18} = 0.84 \cdot \frac{m_l}{134.3} \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ 0.143 \cdot \frac{1}{18} = (0.84 - 0.143) \cdot \frac{m_l}{134.3} \]

8. **Calculate the Mass of the Liquid:**
Now, calculate \( m_l \):
\[ m_l = \frac{0.143 \cdot \frac{1}{18} \cdot 134.3}{0.84 - 0.143} \]

9. **Perform the Calculation:**
- Calculate \( 0.143 \cdot \frac{1}{18} \approx 0.00794 \)
- Calculate \( 0.84 - 0.143 = 0.697 \)
- Substitute these values:
\[ m_l = \frac{0.00794 \cdot 134.3}{0.697} \approx 1.27 \, \text{g} \]

10. **Final Result:**
The mass of the liquid present per gram of water in the distillate is approximately **1.27 g**.