c
(c) \(\mathop {N{a_2}S{O_4}}\limits_{(0.004 - x)} \rightleftharpoons \mathop {2N{a^ + }}\limits_{2x} + \mathop {SO_4^{2 - }}\limits_x \) Since both the solution are isotonic \(0.004 + 2x = 0.01\) \(\therefore \;\;x = 3 \times {10^{ - 3}}\) \(\therefore \;\;\)Percent dissociation \( = \frac{{3 \times {{10}^{ - 3}}}}{{0.004}} \times 100 = 75\% \).