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 boiling point elevation (ΔTb)
The normal boiling point of water is 100°C. The boiling point of the solution is given as 100.52°C.

\[ \Delta T_b = T_{b,\text{solution}} - T_{b,\text{pure}} = 100.52°C - 100°C = 0.52°C \]

### Step 2: Use the boiling point elevation formula
The boiling point elevation can be calculated using the formula:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \( K_b \) is the ebullioscopic constant (for water, \( K_b = 0.52°C \cdot kg/mol \))
- \( m \) is the molality of the solution.

Rearranging the formula to find molality \( m \):

\[ m = \frac{\Delta T_b}{K_b} = \frac{0.52°C}{0.52°C \cdot kg/mol} = 1 \, mol/kg \]

### Step 3: Calculate the freezing point depression (ΔTf)
Now, we will use the molality to find the freezing point depression using the formula:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \( K_f \) is the cryoscopic constant (for water, \( K_f = 1.86°C \cdot kg/mol \))

Substituting the values:

\[ \Delta T_f = 1.86°C \cdot kg/mol \cdot 1 \, mol/kg = 1.86°C \]

### Step 4: Determine the freezing point of the solution
The normal freezing point of water is 0°C. Since the freezing point is depressed, we subtract the freezing point depression from the normal freezing point:

\[ T_f = T_{f,\text{pure}} - \Delta T_f = 0°C - 1.86°C = -1.86°C \]

### Final Answer
The freezing point of the solution is **-1.86°C**.

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