The vapour pressure of `CCl_4` at `25^@C` is 143 mm Hg. If 0.5 gm of a non-volatile solute (mol.weight=65) is dissolved in 100g `CCl_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 of finding the vapor pressure of a solution when a non-volatile solute is dissolved in carbon tetrachloride (CCl₄), we can follow these steps:

### Step 1: Calculate the moles of CCl₄
First, we need to calculate the number of moles of CCl₄ using its mass and molecular weight.

- **Given:**
- Mass of CCl₄ = 100 g
- Molecular weight of CCl₄ = 12 (C) + 4 × 35.5 (Cl) = 12 + 142 = 154 g/mol

- **Calculation:**
\[ \text{Moles of CCl}_4 = \frac{\text{Mass}}{\text{Molecular Weight}} = \frac{100 \text{ g}}{154 \text{ g/mol}} \approx 0.6494 \text{ mol} \]

### Step 2: Calculate the moles of the solute
Next, we calculate the number of moles of the non-volatile solute.

- **Given:**
- Mass of solute = 0.5 g
- Molecular weight of solute = 65 g/mol

- **Calculation:**
\[ \text{Moles of solute} = \frac{\text{Mass}}{\text{Molecular Weight}} = \frac{0.5 \text{ g}}{65 \text{ g/mol}} \approx 0.0077 \text{ mol} \]

### Step 3: Calculate the mole fraction of the solute
Now, we can calculate the mole fraction of the solute in the solution.

- **Mole fraction of solute (X_solute):**
\[ X_{\text{solute}} = \frac{\text{Moles of solute}}{\text{Moles of solute} + \text{Moles of CCl}_4} = \frac{0.0077}{0.0077 + 0.6494} \approx \frac{0.0077}{0.6571} \approx 0.0117 \]

### Step 4: Apply Raoult's Law
According to Raoult's Law, the vapor pressure of the solution (P) can be calculated using the formula:
\[ P = P^0_{\text{CCl}_4} \times (1 - X_{\text{solute}}) \]
Where \( P^0_{\text{CCl}_4} \) is the vapor pressure of pure CCl₄.

- **Given:**
- \( P^0_{\text{CCl}_4} = 143 \text{ mmHg} \)

- **Calculation:**
\[ P = 143 \text{ mmHg} \times (1 - 0.0117) \approx 143 \text{ mmHg} \times 0.9883 \approx 141.44 \text{ mmHg} \]

### Conclusion
The vapor pressure of the solution is approximately **141.44 mmHg**.

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