Elevation in boiling point of a solution of non-electrolyte in `CCl_(4)` is `0.60^(@)C`. What is the depression in freezing point for the same solution? `K_(f)(CCl_(4)) = 30.00 K kg mol^(-1), k_(b)(CCl_(4)) = 5.02 k kg mol^(-1)`

Options

A. `0^(@)C`
B. `5.39^(@)C`
C. `3.59^(@)C`
D. `2.49^(@)C`

Answer

To solve the problem of finding the depression in freezing point for a solution of non-electrolyte in carbon tetrachloride (CCl₄), we will follow these steps:

### Step 1: Understand the relationship between boiling point elevation and molality
The elevation in boiling point (ΔT_b) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( K_b \) = ebullioscopic constant of the solvent (CCl₄ in this case)
- \( m \) = molality of the solution
- \( i \) = van 't Hoff factor (which is 1 for non-electrolytes)

Given that \( \Delta T_b = 0.60^\circ C \) and \( K_b = 5.02 \, \text{K kg mol}^{-1} \), we can find the molality \( m \).

### Step 2: Calculate the molality
Rearranging the formula to solve for molality:
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the known values:
\[ m = \frac{0.60}{5.02} \]
Calculating this gives:
\[ m \approx 0.119 \, \text{mol/kg} \]

### Step 3: Use the molality to find the depression in freezing point
The depression in freezing point (ΔT_f) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( K_f \) = cryoscopic constant of the solvent (CCl₄ in this case)

Given that \( K_f = 30.00 \, \text{K kg mol}^{-1} \), we can substitute the values to find ΔT_f:
\[ \Delta T_f = 1 \cdot K_f \cdot m \]
Substituting the values:
\[ \Delta T_f = 30.00 \cdot 0.119 \]
Calculating this gives:
\[ \Delta T_f \approx 3.57^\circ C \]

### Conclusion
The depression in freezing point for the solution is approximately \( 3.57^\circ C \).