Vapour pressure of `"CCl"_(4)` at `25^(@)C` is 143 mm Hg . 0.5 g of a non-volatile solute ( molar mass = `65 mol^(-1)`) is dissolved in 100 mL of `"CCl"_(4)` (density = 1.538g `mL^(-1)` ) Vapour pressure of solution is :

Options

A. `141.9` mm Hg
B. `94.4` mmHg
C. `99.3` mm Hg
D. `144.1` mm Hg

Answer

To find the vapor pressure of the solution, 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. The formula is given by:

\[ \frac{P_0 - P_s}{P_0} = X_{solute} \]

where \(X_{solute}\) is the mole fraction of the solute.

### Step 1: Calculate the number of moles of the solute
Given:
- Mass of solute (m) = 0.5 g
- Molar mass of solute (M) = 65 g/mol

Using the formula for moles:

\[ n_{solute} = \frac{m}{M} = \frac{0.5 \text{ g}}{65 \text{ g/mol}} = 0.007692 \text{ mol} \]

### Step 2: Calculate the mass of the solvent (CCl₄)
Given:
- Volume of CCl₄ = 100 mL
- Density of CCl₄ = 1.538 g/mL

Using the formula for mass:

\[ \text{mass of CCl}_4 = \text{density} \times \text{volume} = 1.538 \text{ g/mL} \times 100 \text{ mL} = 153.8 \text{ g} \]

### Step 3: Calculate the number of moles of the solvent (CCl₄)
Molar mass of CCl₄ can be calculated as follows:
- Molar mass of C = 12 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of CCl₄ = 12 + (4 × 35.5) = 12 + 142 = 154 g/mol

Using the formula for moles:

\[ n_{solvent} = \frac{\text{mass}}{\text{molar mass}} = \frac{153.8 \text{ g}}{154 \text{ g/mol}} = 0.99935 \text{ mol} \]

### Step 4: Calculate the mole fraction of the solute
The total number of moles in the solution is:

\[ n_{total} = n_{solute} + n_{solvent} = 0.007692 + 0.99935 = 1.007042 \text{ mol} \]

Now, the mole fraction of the solute \(X_{solute}\) is given by:

\[ X_{solute} = \frac{n_{solute}}{n_{total}} = \frac{0.007692}{1.007042} = 0.00764 \]

### Step 5: Apply Raoult's Law to find the vapor pressure of the solution
Given:
- Vapor pressure of pure CCl₄ (P₀) = 143 mm Hg

Using Raoult's Law:

\[ \frac{P_0 - P_s}{P_0} = X_{solute} \]

Rearranging gives:

\[ P_s = P_0 (1 - X_{solute}) = 143 \text{ mm Hg} \times (1 - 0.00764) = 143 \text{ mm Hg} \times 0.99236 = 141.9 \text{ mm Hg} \]

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