If `K_f "value of " H_2 O` is `1.86 ^@C` /molal. The value of `DeltaT_f` for 0.1 m solution of non-volatile solute is

Options

A. 18.6
B. 0.186
C. 1.86
D. 0.0186

Answer

To solve the problem, we need to calculate the depression in freezing point (ΔTf) for a 0.1 m solution of a non-volatile solute in water, given that the cryoscopic constant (Kf) for water is 1.86 °C/molal.

### Step-by-Step Solution:

1. **Understand the Formula**:
The formula for calculating the depression in freezing point is:
\[ \Delta T_f = K_f \times m \]
where:
- \( \Delta T_f \) is the depression in freezing point,
- \( K_f \) is the cryoscopic constant of the solvent (water in this case),
- \( m \) is the molality of the solution.

2. **Identify Given Values**:
From the problem, we have:
- \( K_f = 1.86 \, °C/\text{molal} \)
- \( m = 0.1 \, \text{molal} \)

3. **Substitute the Values into the Formula**:
Now we can substitute the values into the formula:
\[ \Delta T_f = 1.86 \, °C/\text{molal} \times 0.1 \, \text{molal} \]

4. **Perform the Calculation**:
\[ \Delta T_f = 1.86 \times 0.1 = 0.186 \, °C \]

5. **Conclusion**:
The value of \( \Delta T_f \) for the 0.1 m solution of the non-volatile solute is \( 0.186 \, °C \).

### Final Answer:
\[ \Delta T_f = 0.186 \, °C \]