The molal elevation constant of water is 0.51.The boiling point of 0.1 molal aqueous NaCl solution is nearly:

$$\begin{gathered} \mathrm{Molality} \mathrm{of} \mathrm{NaCl} =0.1 \\ \mathrm{Molality} \mathrm{elevation} \mathrm{constant} \left({\mathrm{K}}_{\mathrm{b}}\right) =0.51, \\ \mathrm{Boiling} \mathrm{point} \mathrm{of} \mathrm{NaCl}=? \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that}, \\ ∆\mathrm{Tb}=K_b\times \mathrm{molality} \\ ∆\mathrm{Tb}=\mathrm{i}\times K_b\times \mathrm{molality} \\ (\mathrm{NaCl} \mathrm{in} \mathrm{aqueous} \mathrm{phase} \mathrm{broken} \mathrm{into} {\mathrm{Na}}^{+} \mathrm{and} {\mathrm{Cl}}^{-} \mathrm{ions}.\mathrm{So},\mathrm{i}=2) \\ ∆\mathrm{Tb}=2\times 0.51\times 0.1 \\ =0.2\times 0.5 \\ =0.10 \\ \left({\mathrm{T}}_{\mathrm{b}}\right)\mathrm{solution}-({\mathrm{T}}_{\mathrm{b}}°)\mathrm{water}=0.1 \\ \mathrm{Boiling} \mathrm{point} \mathrm{of} \mathrm{pure} \mathrm{water} \mathrm{is} 100°\mathrm{C} \\ \left({\mathrm{T}}_{\mathrm{b}}\right) \mathrm{solution}=100+0.1 \\ \left({\mathrm{T}}_{\mathrm{b}}\right)\mathrm{solution}=100.1°\mathrm{C} \end{gathered}$$

Hence, correct option is: (B) 100.1°C .