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

Here the given data is, 

Molarity of ${Na}_2{So}_4$ = 0.004

Molarity of glucose = 0.01 

Temperature is same 

Degree of dissociation of ${Na}_2{So}_4$ = ? 

 As given in que. That the solution is isotonic type so, 

            π${\mathrm{\pi }}_{{\mathrm{Na}}_2{\mathrm{So}}_4}={\mathrm{\pi }}_{\mathrm{glucose}}$  

 ( π = osmotic pressure)

              $M_1$×R×T×$i_1$ = $M_2$×R×T×$i_2$

( R = constant

  T = same , so both cancel out.)

$M_1\times i_1=M_2\times i_2$       

  ( Glucose is a non polar solute so i = 1)

             0.004×$i_1$ = 0.01×1

                  $i_1$ = 2.5

 i = 1+(n-1) degree of dissociation ($\alpha$)

   (n = 3 as ${Na}_2{So}_4$  brokes into $2{Na}^{+}$and ${So}_4^{2-}$)

   2.5 = 1+(3-1) degree of dissociation

   Degree of dissociation = 1.5/2 

                                        = 0.75 

   Degree of dissociation in % = 75% .

Hence, the correct option is: (C) 75% .