$0.01 \mathrm{M}$ solution of $\mathrm{NaI}$ is found to be isotonic with $0.016 \mathrm{M}$ solution of glucose. The apparent degree of ionization of $\mathrm{NaI}$ is:

$\begin{array}{ccccc}\text{ Na I }& \rightleftharpoons & {\mathrm{Na}}^{+}& +& {\mathrm{I}}^{-}\\ 1-\alpha & & \alpha & & \alpha \end{array}$

Total no. of mol $=1-\alpha +\alpha +\alpha$

$=1+\alpha ;$ Observed osmotic pressure,

$\begin{array}{l}{\pi }_{\text{obs }}={\pi }_{\text{Theoretical }}\times (1+\alpha )\\ =0.01\text{ RT }(1+\alpha );\end{array}$

${\mathrm{\pi }}_{\mathrm{glucose}}=0.016\mathrm{RT}$. 

So, $0.01\mathrm{RT}(1+\alpha )$$=0.016\mathrm{RT}$. Hence , 

$\mathrm{\alpha }$= 0.6 or 0.6 x 100 = 60 %.