Explanation

(c)
$BaCl_{2} \longrightarrow {Ba}^{2 +} + 2{Cl}^{-}$
Let the solubility of $BaCl_{2}\ is$ $x$ mol/L
$\therefore$ $K_{sp} = \left\lbrack {Ba}^{2 +} \right\rbrack\left\lbrack {Cl}^{-} \right\rbrack^{2}$
$= (x) \times (2x)^{2}$
$= x \times 4x^{2} = 4x^{3}$
or solubility of $BaCl_{2} = \frac{\left( solubility\ product\ of\ BaCl_{2} \right)^{\frac{1}{3}}}{4}$
$= \frac{\left( 4 \times 10^{- 9} \right)^{\frac{1}{3}}}{4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{- 3}$ mol/L