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 a non-electrolyte in carbon tetrachloride (CCl₄), we can 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:
\[ \Delta T_b = K_b \cdot m \cdot i \]
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)

### Step 2: Rearrange the formula to find molality
Given that \( \Delta T_b = 0.60^\circ C \) and \( K_b = 5.02 \, \text{kg} \cdot \text{mol}^{-1} \):
\[ 0.60 = 5.02 \cdot m \cdot 1 \]
Rearranging gives:
\[ m = \frac{0.60}{5.02} \]

### Step 3: Calculate the molality
Calculating the molality:
\[ m = \frac{0.60}{5.02} \approx 0.119 \, \text{mol/kg} \]

### Step 4: Use the molality to find depression in freezing point
The depression in freezing point (ΔT_f) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \( K_f = 30.00 \, \text{K kg mol}^{-1} \)

Substituting the values we have:
\[ \Delta T_f = 30.00 \cdot 0.119 \cdot 1 \]

### Step 5: Calculate the depression in freezing point
Calculating ΔT_f:
\[ \Delta T_f = 30.00 \cdot 0.119 \approx 3.57^\circ C \]

### Step 6: Round the result
Rounding to two decimal places, we find:
\[ \Delta T_f \approx 3.59^\circ C \]

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

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