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, we will follow these steps:

### Step 1: Understand the relationship between boiling point elevation and molality
The elevation in boiling point (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the ebullioscopic constant and \( m \) is the molality of the solution.

### Step 2: Rearrange the formula to find molality
We can rearrange the formula to find the molality:
\[ m = \frac{\Delta T_b}{K_b} \]

### Step 3: Substitute the known values
We know that:
- \( \Delta T_b = 0.60^\circ C \)
- \( K_b = 5.02 \, \text{K kg mol}^{-1} \)

Substituting these values into the equation gives:
\[ m = \frac{0.60}{5.02} \]

### Step 4: Calculate the molality
Calculating the above expression:
\[ m = \frac{0.60}{5.02} \approx 0.1195 \, \text{mol/kg} \]

### Step 5: Use the molality to find the depression in freezing point
The depression in freezing point (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot m \]
where \( K_f \) is the cryoscopic constant.

### Step 6: Substitute the known values for freezing point depression
We know that:
- \( K_f = 30.00 \, \text{K kg mol}^{-1} \)
- \( m \approx 0.1195 \, \text{mol/kg} \)

Substituting these values into the equation gives:
\[ \Delta T_f = 30.00 \cdot 0.1195 \]

### Step 7: Calculate the depression in freezing point
Calculating the above expression:
\[ \Delta T_f \approx 30.00 \cdot 0.1195 \approx 3.585 \, \text{K} \]

### Step 8: Final answer
Thus, the depression in freezing point is approximately:
\[ \Delta T_f \approx 3.59^\circ C \]

### Summary
The depression in freezing point for the solution is \( 3.59^\circ C \).

---