Explanation

(b)
Let the solubility of $Ag_{2}S\ is\ s$.
$Ag_{2}S \rightleftharpoons 2Ag^{+} + S^{2 -}$
$s$ $2s$ $s$
$Na_{2}S$, being a strong electrolyte, dissociates completely and provides
[$S^{2 -}$]=0.1 M
$\therefore$ $\left\lbrack Ag^{+} \right\rbrack = s$
$\left\lbrack S^{2 -} \right\rbrack = (s + 0.1)\ M$
$K_{sp} = \left\lbrack Ag^{+} \right\rbrack^{2}\left\lbrack S^{2 -} \right\rbrack$
$= (2s)^{2}(s + 0.1)$
$= 4s^{3} + 0.4s^{2}$
$\Rightarrow$ $10^{- 17} = 0.4s^{2}$
$\lbrack\because Higher\ powers\ are\ neglected.\rbrack$
$\therefore$ $s = 5 \times 10^{- 9}\ mol\ L^{- 1}$