Explanation

(b)
Degree of dissociation=0.4
$PCl_{5}\ \ \ \ \ \ \ \ \ \ \ \ \rightleftharpoons PCl_{3} + Cl_{2}$
$a(1 - x)$ $ax$ $ax$
$a = 2,\ x = 0.4,\ V = 2$
$\left\lbrack PCl_{5} \right\rbrack = \frac{2(1 - 0.4)}{2} = 0.6$ mol/L
$\left\lbrack PCl_{3} \right\rbrack = \frac{2 \times 0.4}{2} = 0.4$ mol/L
$\left\lbrack Cl_{2} \right\rbrack = \frac{2 \times 0.4}{2} = 0.4$ mol/L
$\therefore$ $K_{c} = \frac{\left\lbrack PCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack}{\left\lbrack PCl_{5} \right\rbrack}$
$= \frac{0.4 \times 0.4}{0.6} = 0.267$