Explanation

(a)
When 0.004 M $Na_{2}SO_{4}$ solution is isotonic with 0.01 M solution of glucose, so their osmotic pressures are equal to each other.
Osmotic pressure of 0.01 M glucose ($\pi)_{glucose\ }$ =CST
$\therefore$ C= concentration of solution =0.01 M
S= solution constant =0.0821 L atm/K/mol
T= absolute temperature
$\therefore$ $\pi_{glucose} = 0.01\ \times 0.0821\ T$ ---(i)
$\pi_{glucose} = \pi_{Na_{2}SO_{4}}$
$\ Na_{2}SO_{4}$ is present in ionic state in solution
So, $Na_{2}SO_{4}\ \ \ \ \ \rightleftharpoons \ \ \ \ 2Na^{+}\ \ \ + \ \ \ \ SO_{4}^{2 -}$
At t=0 1 0 0
At equilibrium $1 - \alpha\ \ \ \ \ \ \ \ \ \ \ \ \ 2\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha$
(where, $\alpha$ is the degree of dissociation of $Na_{2}SO_{4})$
($\pi_{cal})_{Na_{2}SO_{4}\ \ } = \ \ C \times S \times T$= 0.004 $\times$0.0821$\times$ T ---(ii)
By van’t Hoff facter
$\frac{(\pi_{obs})_{Na_{2}SO_{4}}}{(\pi_{cal})_{Na_{2}SO_{4}\ \ }} = \frac{number\ of\ particles\ after\ dissociation\ }{number\ of\ particles\ before\ dissociation\ }$
$= \frac{1 - \ \propto + 2 \propto + \propto}{1}$
$\because\ \ \ \ \ \ \ \ (\pi_{obs})_{Na_{2}SO_{4}} = \pi_{glucose}\ \$
$\therefore\ \ \ \ \ \ \ \frac{0.01\ \times 0.0821\ T\ \ \ }{0.004\ \times 0.0821 \times \ T} = \frac{1 + 2\alpha}{1}$
Or $\frac{10}{4} = \frac{1 + 2 \propto}{1}$ or 10=4+8$\propto$
$\propto = \frac{10 - 4}{8} = 0.75$
%of$\ \ \propto$= 75%