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 the Given Data:**
- Boiling point of the solution, \( T_b = 100.52^\circ C \)
- Boiling point of pure water, \( T_{b, \text{water}} = 100^\circ C \)
- Freezing point of pure water, \( T_f = 0^\circ C \)
- 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 (\( \Delta T_b \)):**
\[ \Delta T_b = T_b - T_{b, \text{water}} = 100.52^\circ C - 100^\circ C = 0.52^\circ C \]

3. **Use the Formula for Boiling Point Elevation:**
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot m \cdot k_b \]
Since the solute is a non-electrolyte, \( i = 1 \). Therefore:
\[ 0.52 = 1 \cdot m \cdot 0.52 \]
From this, we can see that \( m = 1 \, \text{mol/kg} \).

4. **Calculate the Freezing Point Depression (\( \Delta T_f \)):**
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot m \cdot k_f \]
Again, since the solute is a non-electrolyte, \( i = 1 \). Thus:
\[ \Delta T_f = 1 \cdot 1 \cdot 1.86 = 1.86^\circ C \]

5. **Determine the Freezing Point of the Solution:**
The freezing point of the solution can be calculated using:
\[ T_f = T_f, \text{water} - \Delta T_f \]
Substituting the values:
\[ T_f = 0^\circ C - 1.86^\circ C = -1.86^\circ C \]

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

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