Suppose solubility of PbCl₂ in water is s mol L^-1.

$\begin{array}{ccc}{\text{PbCl}}_2\left(\text{s}\right)& \rightleftharpoons & {\text{Pb}}^{2+}\text{(aq)}+{\text{2Cl}}^{-}\text{(aq)}\\ \left(1-\text{s}\right)& & \text{s}\text{2s}\end{array}$

K_sp = [Pb²⁺] · [Cl^-]²

K_sp = [s] [2s]² = 4s³

3.2 × 10^-8 = 4s³

${\text{s}}^3=\frac{\text{3.2}\times {\text{10}}^{-8}}{4}=\text{0.8}\times {\text{10}}^{-8}$

s³ = 8.0 × 10^-9

Solubility of PbCl₂, = s = 2 × 10^-3 mol L^-1

Solubility of PbCl₂ in gL^-1 = 278 × 2 × 10^-3 = 0.556 g L^-1

                     (∵ Molar mass of PbCl₂ = 207 + (2 × 35.5) = 278)

0.556 g of PbCl₂ dissolve in 1 L of water.

$\therefore {\text{0.1 g of PbCl}}_2\text{will dissolve in}=\frac{1\times \text{0.1}}{\text{0.556}}\text{L of water}$

​= 0.1798 L = 179.8 mL