The minimum volume of water required to dissolve 0.1 g lead (II)...

${{( K}_{sp} )}_{Pb{Cl}_{2}} = 32 \times {10}^{- 9}$

$\begin{cases}{PbCl}_{2} & = & {Pb}^{2 +} \\ & & s\end{cases}$\begin{cases}+ & 2{Cl}^{-} \\ & 2s\end{cases}$$<br /><br />${K}_{sp} = \left[P{b}^{2 +}\right]{\left[2C{l}^{-}\right]}^{2}$<br /><br />${K}_{sp} = 4{s}^{3} = 32 \times {10}^{- 9}$<br /><br />${s}^{3} = 8 \times {10}^{- 9}$<br /><br />$s = 2 \times {10}^{- 3}M$<br /><br />$\dfrac{w}{M . w} \times \dfrac{1}{{V}_{L}} = 2 \times {10}^{- 3}$<br /><br />$\dfrac{0.1}{278} \times \dfrac{1}{{V}_{L}} = 2 \times {10}^{- 3}$<br /><br />${V}_{L} = \dfrac{0.1 \times 1000}{278 \times 2} = 0.18L$