K_sp of PbBr₂ (Molar mass= 367 ) is 3 .2 x 10^-5 • If the salt is 80% dissociated in solution, calculate the solubility of salt in gram per litre. (in litre^-1)

Let solubility of PbBr₂ is S mol litre^-1 

$PbB{r}_2\left(s\right)+aq{\rightleftharpoons }_{\frac{S\times 80}{100}}P{b}^{2+}+\underset{\frac{2s\times 80}{100}}{2B{r}^{-}}$

Since, PbBr₂ ionizes to 80% only. 

Now, $K_{sp}=\left[P{b}^{2+}\right]{\left[B{r}^{-}\right]}^2$

or $\begin{array}{l}3.2\times {10}^{-5}=\left[\frac{S\times 80}{100}\right]{\left[\frac{2S\times 80}{100}\right]}^2\\ S=0.025\text{ mollitre }{e}^{-1}\\ =0.025\times 367{\mathrm{glitre}}^{-1}\\ =9.175{\mathrm{glitre}}^{-1}\end{array}$