The molal freezing point constant of water is 1.86°C/m. Therefore, the freezing point of 0.1M NaCl solution in water is expected to be?
[A] -1.86°C
[B] -0.186°C
[C] -0.372°C
[D] +0.372°C

To solve this question, you need to understand how the freezing point depression of a solution is calculated. The formula for freezing point depression is given by:

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

where:
- \(\Delta T_f\) is the freezing point depression,
- \(K_f\) is the molal freezing point constant of the solvent (water in this case),
- \(m\) is the molality of the solution, and
- \(i\) is the van 't Hoff factor (which accounts for the number of particles the solute dissociates into).

For NaCl, it dissociates into two ions: Na\(^+\) and Cl\(^-\). So, the van 't Hoff factor \(i\) for NaCl is 2.

Here’s how you solve the problem step by step:

1. **Determine the van 't Hoff factor**: For NaCl, \(i = 2\).

2. **Convert the molarity to molality**: For dilute solutions, molarity (M) is approximately equal to molality (m). So, \(m = 0.1\) mol/kg.

3. **Use the freezing point depression formula**:

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

Substitute the values:

\[
\Delta T_f = 1.86 \, ^\circ \text{C/m} \times 0.1 \, \text{mol/kg} \times 2
\]

\[
\Delta T_f = 1.86 \times 0.1 \times 2
\]

\[
\Delta T_f = 0.372 \, ^\circ \text{C}
\]

4. **Determine the freezing point of the solution**: Since \(\Delta T_f\) is the amount by which the freezing point decreases, the freezing point of the solution is:

\[
\text{Freezing point} = 0 \, ^\circ \text{C} - 0.372 \, ^\circ \text{C} = -0.372 \, ^\circ \text{C}
\]

Therefore, the freezing point of the 0.1M NaCl solution in water is expected to be \(-0.372 \, ^\circ \text{C}\).

So the correct answer is: [C] \(-0.372 \, ^\circ \text{C}\).