The solubility product of BaSO₄ in water is 1.5x10^-9. The molar solubility of BaSO₄ in 0.1M solution of Ba(NO₃)₂ in

Let the solubility of BaSO₄=s mole
$$\begin{gathered} \begin{array}{c}{\mathrm{BaSO}}_4\rightleftharpoons {\mathrm{Ba}}^{+2}+{\mathrm{SO}}_4^{2-}\\ \mathrm{Ba}{\left({\mathrm{NO}}_3\right)}_2\to {\mathrm{Ba}}^{+2}+2{\mathrm{NO}}_3^{-}\\ \left[{\mathrm{Ba}}^{+2}\right]=\mathrm{s}+0.1;\left[{\mathrm{SO}}_4^{-2}\right]=\mathrm{s}\end{array} \\ \begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}={\mathrm{BaSO}}_4\\ 1.5\times {10}^{-9}=[\mathrm{s}+0.1]\left[\mathrm{s}\right]={\mathrm{s}}^2+0.1\mathrm{s}\end{array} \end{gathered}$$
s is very small; neglect s²
$\begin{array}{l}1.5\times {10}^{-9}=0.1\mathrm{s}\\ \mathrm{S}=1.5\times {10}^{-8}\mathrm{M}\end{array}$