What is the degree of dissociation of sodium chloride, if the molar mass determined by a cryoscopic method was found to $\text{ be }31.80\mathrm{g}{\mathrm{mol}}^{-1}\left[\text{ Atomic mass }\mathrm{Na}=23\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{Cl}=35.5\mathrm{g}\right.\left.{\mathrm{mol}}^{-1}\right]?$

Let $\alpha$ be the give of dissociation then  
$$\begin{gathered} \underset{1-\mathrm{\alpha }}{\mathrm{NaCl}\to }\underset{\mathrm{\alpha }}{{\mathrm{Na}}^{+}}+\underset{\mathrm{\alpha }}{{\mathrm{Cl}}^{-}} \\ \text{ Van't Hoff's factor }\mathrm{i}=\frac{1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha }}{1}=1+\mathrm{\alpha } \\ \mathrm{Again} \mathrm{Van}’\mathrm{t} \mathrm{Hoff}’ \mathrm{s} \mathrm{factor} \\ =\frac{\text{ Normal mol }.\mathrm{wt}}{\text{ Observed mol.wt }}=\frac{58.5}{31.8}=1.83 \\ \text{ Equating to both values of }\mathrm{i},\therefore 1+\mathrm{\alpha }=1.83 \\ \therefore \mathrm{\alpha }=0.83 \end{gathered}$$