Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)

Options

A. `143.99`mm
B. `94.39`mm
C. `199.34`mm
D. `14.197`mm

Explanation

To find the vapor pressure of the solution, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure CCl₄ (P₀) = 143 mmHg
- Mass of non-volatile solute (m) = 0.05 g
- Molar mass of solute (M) = 65 g/mol
- Volume of CCl₄ = 100 mL
- Density of CCl₄ = 1.58 g/cm³

2. **Calculate the Mass of CCl₄:**
\[ \text{Mass of CCl₄} = \text{Volume} \times \text{Density} = 100 \, \text{mL} \times 1.58 \, \text{g/mL} = 158 \, \text{g} \]

3. **Calculate the Moles of Solute:**
\[ \text{Moles of solute} (n_{solute}) = \frac{\text{mass}}{\text{molar mass}} = \frac{0.05 \, \text{g}}{65 \, \text{g/mol}} = 0.000769 \, \text{mol} \]

4. **Calculate the Moles of CCl₄:**
\[ \text{Moles of CCl₄} (n_{CCl₄}) = \frac{\text{mass}}{\text{molar mass}} = \frac{158 \, \text{g}}{153.8 \, \text{g/mol}} \approx 1.027 \, \text{mol} \]

5. **Calculate the Total Moles in the Solution:**
\[ n_{total} = n_{CCl₄} + n_{solute} = 1.027 \, \text{mol} + 0.000769 \, \text{mol} \approx 1.027769 \, \text{mol} \]

6. **Calculate the Mole Fraction of CCl₄:**
\[ X_{CCl₄} = \frac{n_{CCl₄}}{n_{total}} = \frac{1.027}{1.027769} \approx 0.9993 \]

7. **Calculate the Vapor Pressure of the Solution (P):**
\[ P = P₀ \times X_{CCl₄} = 143 \, \text{mmHg} \times 0.9993 \approx 142.7 \, \text{mmHg} \]

### Final Answer:
The vapor pressure of the solution is approximately **142.7 mmHg**.