0.004 M solution of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is isotonic with 0.01 M solution of glucose at same temperature. The approximate degree of dissociation of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is

For ${\mathrm{Na}}_2{\mathrm{SO}}_4,{\pi }_1=i\mathrm{MRT}$=$\mathrm{i}\times 0.004\times \mathrm{RT}$

For glucose, ${\pi }_2=\mathrm{CRT}$=$0.01\mathrm{RT}$

${\pi }_1={\pi }_2$ (isotonic)
or 0.004i=0.01
i=2.5
${\mathrm{Na}}_2{\mathrm{SO}}_4\rightleftharpoons$  $2{\mathrm{Na}}^{\oplus }$ +  ${\mathrm{SO}}_4^{2-}$ 

1                             0         0     

1 -  a                       2a         a

Total particles = 1-a+2a+a

$\therefore \mathrm{i}=\frac{1+2\mathrm{a}}{1}=2.5$
a = 0.75 or 75% dissociated

Hence, the correct option  C. 75%