A solution containing $$0.85\,g$$  of $$ZnC{l_2}$$  in $$125.0\,g$$  of water freezes at $$ - {0.23^ \circ }C.$$  The apparent degree of dissociation of the salt is ( $${K_f}$$ for water $$ = 1.86K\,kg\,mo{l^{ - 1}},$$    atomic mass : $$Zn = 65.3$$   and $$Cl = 35.5$$   )

Solution :
$$\eqalign{ & Mol.wt. = \frac{{{k_f} \times w \times 1000}}{{\Delta {T_f} \times W}} \cr & = \frac{{1.86 \times 0.85 \times 1000}}{{0.23 \times 125}} \approx 55\,g \cr & {\text{Where}} \cr & w = 0.85\,g \cr & W = 125\,g \cr & \Delta {T_f} = {0^ \circ }C - \left( { - {{23}^ \circ }C} \right) = {23^ \circ }C \cr & {\text{Now,}}\,i = \frac{{{M_{normal}}}}{{{M_{observed}}}} = \frac{{136.3}}{{55}} = 2.47 \cr & \mathop {ZnC{l_2}}\limits_{1 - \alpha } \rightleftharpoons \mathop {Z{n^{ + + }}}\limits_\alpha + \mathop {2C{l^ - }}\limits_{2\alpha } \cr & {\text{Van't Hoff factor }}(i) \cr & = \frac{{1 - \alpha + \alpha + 2\alpha }}{1} = 2.47 \cr & \therefore \,\,\alpha = 0.735 = 73.5\% \cr} $$