The vapour pressure of `C Cl_4` at `25^@C` is 143 mm Hg. If 0.5 gm of a non-volatile solute (mol.weight=65) is dissolved in 100g `C Cl_4`, the vapour pressure of the solution will be

Options

A. 199.34 mm Hg
B. 143.99 mm Hg
C. 141.43 mm Hg
D. 94.39 mm Hg

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solution (P_s) is related to the vapor pressure of the pure solvent (P_0) and the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure CCl₄ (P₀) = 143 mm Hg
- Mass of non-volatile solute (m_solute) = 0.5 g
- Molar mass of solute (M_solute) = 65 g/mol
- Mass of CCl₄ (m_solvent) = 100 g
- Molar mass of CCl₄ (M_solvent) = 153.82 g/mol (Note: The molar mass of CCl₄ is approximately 153.82 g/mol, not 154 g/mol as mentioned in the transcript.)

2. **Calculate Moles of Solute:**
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{0.5 \text{ g}}{65 \text{ g/mol}} = 0.007692 \text{ mol} \]

3. **Calculate Moles of Solvent (CCl₄):**
\[ \text{Moles of solvent} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{100 \text{ g}}{153.82 \text{ g/mol}} \approx 0.649 \text{ mol} \]

4. **Calculate the Mole Fraction of Solvent (CCl₄):**
\[ \text{Mole fraction of solvent} (X_{solvent}) = \frac{\text{moles of solvent}}{\text{moles of solute} + \text{moles of solvent}} = \frac{0.649}{0.007692 + 0.649} \approx \frac{0.649}{0.656} \approx 0.9903 \]

5. **Apply Raoult's Law:**
\[ P_s = X_{solvent} \times P_0 \]
\[ P_s = 0.9903 \times 143 \text{ mm Hg} \approx 141.43 \text{ mm Hg} \]

### Final Answer:
The vapor pressure of the solution (P_s) is approximately **141.43 mm Hg**.