The boiling point of an aqueous solution of a non - electrolyte is `100.52^@C` . Then freezing point of this solution will be [ Given : `k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1)` for water]

Options

A. `0^@C`
B. `-1.86^@C`
C. `1.86^@C`
D. None of the above

Answer

To solve the problem, we need to find the freezing point of an aqueous solution given its boiling point and the constants for freezing point depression and boiling point elevation.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Boiling point of the solution, \( T_b = 100.52^\circ C \)
- Boiling point of pure solvent (water), \( T_{b, \text{solvent}} = 100^\circ C \)
- Freezing point depression constant, \( K_f = 1.86 \, \text{K kg mol}^{-1} \)
- Boiling point elevation constant, \( K_b = 0.52 \, \text{K kg mol}^{-1} \)

2. **Calculate the Elevation in Boiling Point (\( \Delta T_b \)):**
\[ \Delta T_b = T_b - T_{b, \text{solvent}} = 100.52^\circ C - 100^\circ C = 0.52^\circ C \]

3. **Use the Relationship Between Freezing Point Depression and Boiling Point Elevation:**
The relationship between the depression in freezing point (\( \Delta T_f \)) and elevation in boiling point (\( \Delta T_b \)) can be expressed as:
\[ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b} \]

4. **Substitute Known Values:**
\[ \frac{\Delta T_f}{0.52} = \frac{1.86}{0.52} \]

5. **Calculate \( \Delta T_f \):**
\[ \Delta T_f = 0.52 \times \frac{1.86}{0.52} = 1.86^\circ C \]

6. **Determine the Freezing Point of the Solution:**
The freezing point depression is related to the freezing point of the solvent and the freezing point of the solution:
\[ \Delta T_f = T_{f, \text{solvent}} - T_f \]
Here, \( T_{f, \text{solvent}} = 0^\circ C \):
\[ 1.86 = 0 - T_f \implies T_f = -1.86^\circ C \]

### Final Answer:
The freezing point of the solution is \( -1.86^\circ C \).

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