An aqueous solution of `0.1` molal concentration of sucrose should have freezing point `(K_(f)=1.86K mol^(-1)kg)`

Options

A. `0.186^(@)C`
B. `1.86^(@)C`
C. `-1.86^(@)C`
D. `-0.186^(@)C`

Explanation

To solve the problem, we need to determine the freezing point of an aqueous solution of sucrose with a molality of 0.1 mol/kg, given that the freezing point depression constant (Kf) is 1.86 K kg/mol.

### Step-by-Step Solution:

1. **Identify the given values:**
- Molality (m) = 0.1 mol/kg
- Freezing point depression constant (Kf) = 1.86 K kg/mol

2. **Use the formula for freezing point depression (ΔTf):**
\[ \Delta T_f = K_f \times m \]
Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.1 \, \text{mol/kg} \]

3. **Calculate ΔTf:**
\[ \Delta T_f = 1.86 \times 0.1 = 0.186 \, \text{K} \]

4. **Determine the freezing point of the solution (Tf):**
The freezing point of pure water is 0°C. The freezing point of the solution can be calculated using:
\[ T_f = T_{f, \text{water}} - \Delta T_f \]
Substituting the values:
\[ T_f = 0 \, \text{°C} - 0.186 \, \text{K} \]
Since 0.186 K is equivalent to 0.186 °C, we have:
\[ T_f = -0.186 \, \text{°C} \]

5. **Final Result:**
The freezing point of the aqueous solution of sucrose is:
\[ T_f = -0.186 \, \text{°C} \]

### Conclusion:
The correct option for the freezing point of the solution is **Option D**.