The molal freezing points constant of water is 1.86 K kg `"mol"^(-1)`. Therefore, the freezing point of 0.1 M NaCl in water is expected to be
1) `-1.86^(@)C`
2) `-0.372^(@)C`
3) `-0.186^(@)C`
4) `+0.372^(@)C`

Options

A. `-1.86^(@)C`
B. `-0.372^(@)C`
C. `-0.186^(@)C`
D. `+0.372^(@)C`

Explanation

To solve the problem, we need to calculate the depression in the freezing point of water when 0.1 M NaCl is dissolved in it. We will use the formula for freezing point depression:

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

Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_f\) = molal freezing point constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 1: Determine the van 't Hoff factor (i)
For NaCl, it dissociates into two ions:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Thus, the van 't Hoff factor \(i\) for NaCl is:
\[ i = 2 \]

### Step 2: Identify the molal freezing point constant (K_f)
The molal freezing point constant for water is given as:
\[ K_f = 1.86 \, \text{K kg} \, \text{mol}^{-1} \]

### Step 3: Calculate the molality (m)
Given that the solution is 0.1 M NaCl, we can assume that the density of the solution is approximately that of water (1 kg/L). Therefore, the molality (m) can be approximated as:
\[ m = 0.1 \, \text{mol/kg} \]

### Step 4: Substitute the values into the freezing point depression formula
Now we can substitute the values into the formula:
\[ \Delta T_f = i \cdot K_f \cdot m = 2 \cdot 1.86 \, \text{K kg} \, \text{mol}^{-1} \cdot 0.1 \, \text{mol/kg} \]

### Step 5: Calculate \(\Delta T_f\)
Calculating the above expression:
\[ \Delta T_f = 2 \cdot 1.86 \cdot 0.1 = 0.372 \, \text{K} \]

### Step 6: Determine the freezing point of the solution
The normal freezing point of water is 0°C. Since we are calculating the depression, we subtract:
\[ T_f = T_f^0 - \Delta T_f = 0°C - 0.372°C = -0.372°C \]

### Conclusion
Thus, the freezing point of 0.1 M NaCl in water is expected to be:
\[ \boxed{-0.372°C} \]