An aqueous solution of a non-electrolyte boils at `100.52^@C` . The freezing point of the solution will be

Options

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

Answer

To find the freezing point of an aqueous solution of a non-electrolyte that boils at 100.52°C, we can follow these steps:

### Step 1: Determine the elevation in boiling point
The elevation in boiling point (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = T_b - T_{b, \text{solvent}} \]
Where:
- \( T_b \) is the boiling point of the solution (100.52°C)
- \( T_{b, \text{solvent}} \) is the boiling point of the pure solvent (water, which is 100°C)

Substituting the values:
\[ \Delta T_b = 100.52°C - 100°C = 0.52°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( K_b \) is the molal elevation constant for water (approximately 0.52°C/m)
- \( i \) is the van 't Hoff factor (for non-electrolytes, \( i = 1 \))
- \( m \) is the molality of the solution

Since we are considering a non-electrolyte and assuming molality \( m = 1 \):
\[ 0.52°C = 1 \cdot 0.52°C/m \cdot m \]
This confirms that the elevation in boiling point is consistent with a molality of 1.

### Step 3: Set up the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) is the molal depression constant for water (approximately 1.86°C/m)
- \( i \) is the van 't Hoff factor (for non-electrolytes, \( i = 1 \))

Using \( m = 1 \):
\[ \Delta T_f = 1 \cdot 1.86°C/m \cdot 1 = 1.86°C \]

### Step 4: Calculate the freezing point of the solution
The freezing point depression (\( \Delta T_f \)) is defined as:
\[ \Delta T_f = T_{f, \text{solvent}} - T_{f, \text{solution}} \]
Where:
- \( T_{f, \text{solvent}} \) is the freezing point of the pure solvent (water, which is 0°C)

Rearranging gives us:
\[ T_{f, \text{solution}} = T_{f, \text{solvent}} - \Delta T_f \]
Substituting the values:
\[ T_{f, \text{solution}} = 0°C - 1.86°C = -1.86°C \]

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

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