What will be the molecular weight of $\mathrm{NaCl}$ determined experimentally from elevation in the boiling point or depression in freezing point method?

$\mathrm{NaCl}\to {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}$

$\mathrm{NaCl}$ furnishes two ions on complete dissociation.
so i=2 and
Normal molecular weight of electrolyte >Experimental molecular weight.

$i=\frac{M_{\text{Theo }}}{M_{\text{Obs }}}$[Put $i\mathit{=}\mathit{2}$]

$$\begin{gathered} \Rightarrow M_{\text{Obe }}<M_{\text{Theo }} \\ =58.5 \end{gathered}$$