The molal freezing point constant of water is 1.86C/M. Therefore the freezing point of 0.1 M NaCl solution in water is expected to be

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}} \times \mathrm{m}\times \mathrm{i} (\mathrm{i}=2 ,\mathrm{for} \mathrm{Nacl} \to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-} \\ \mathrm{i} =0.1 \mathrm{molal} \\ {\mathrm{K}}_{\mathrm{f}} =1.86{}^0\mathrm{C}/\mathrm{molal} \end{gathered}$$

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}=1.86\times 2\times 0.1 \\ ∆{\mathrm{T}}_{\mathrm{f}} =0.372 {}^0\mathrm{C} \end{gathered}$$

Therefore, now freezing point = 0−0.372^oC=−0.372^oC
where i is vant hoff constant/factor.