`0.5g` of a nonvolatile organic compound (mol.wt. 65) is dissolved in 100mL of `"CC"l_(4)`. If the vapour pressure of pure `"CC"l_(4)` is 143 mm, what would be the vapour pressure of the solution? (Density of `"CC"l_(4)` solution is `1.58g//"cc"`)

Answer

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 its mole fraction in the solution.

### Step-by-Step Solution:

1. **Calculate the moles of the solute (non-volatile organic compound)**:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{0.5 \, \text{g}}{65 \, \text{g/mol}} = 0.007692 \, \text{mol} \]

2. **Calculate the weight of the solvent (CCl₄)**:
- Given the volume of CCl₄ = 100 mL and its density = 1.58 g/mL:
\[ \text{Weight of CCl₄} = \text{Volume} \times \text{Density} = 100 \, \text{mL} \times 1.58 \, \text{g/mL} = 158 \, \text{g} \]

3. **Calculate the moles of the solvent (CCl₄)**:
\[ \text{Moles of CCl₄} = \frac{\text{mass of CCl₄}}{\text{molar mass of CCl₄}} = \frac{158 \, \text{g}}{153.8 \, \text{g/mol}} \approx 1.025 \, \text{mol} \]

4. **Calculate the total moles in the solution**:
\[ \text{Total moles} = \text{Moles of solute} + \text{Moles of solvent} = 0.007692 \, \text{mol} + 1.025 \, \text{mol} \approx 1.0327 \, \text{mol} \]

5. **Calculate the mole fraction of the solvent (CCl₄)**:
\[ \text{Mole fraction of CCl₄} = \frac{\text{Moles of CCl₄}}{\text{Total moles}} = \frac{1.025 \, \text{mol}}{1.0327 \, \text{mol}} \approx 0.991 \]

6. **Calculate the vapor pressure of the solution (Pₛ)**:
- Using Raoult's Law:
\[ Pₛ = P₀ \times \text{Mole fraction of CCl₄} \]
Where \(P₀\) (vapor pressure of pure CCl₄) = 143 mmHg:
\[ Pₛ = 143 \, \text{mmHg} \times 0.991 \approx 141.9 \, \text{mmHg} \]

### Final Answer:
The vapor pressure of the solution is approximately **141.9 mmHg**.