$\mathrm{CaSO}_{4(s)}$ ⇌ $\mathrm{Ca}^{2+}_{(aq)} + \mathrm{SO}^{2-}_{4(aq)}$ If S is the solubility of $\mathrm{CaSO}_4$ in $\mathrm{mol\,L^{-1}}$, then $K_{sp}$ = $[\mathrm{Ca}^{2+}] \times [\mathrm{SO}^{2-}_4]$ = $S^2$ or , S = $\sqrt{K_{sp}}$ = $\sqrt{0.1 \times 10^{-6}}$ = 3.02 × $10^{-3}\,\mathrm{mol\,L^{-1}}$ = 3.02 × $10^{-3}$ × 136 $\mathrm{g\,L^{-1}}$ = 0.411 $\mathrm{g\,L^{-1}}$ (Molar mass of $\mathrm{CaSO}_4$ = 136 $\mathrm{g\,mol^{-1}}$) Thus, for dissolving 0.411 g, water required = 1 L ∴ For dissolving 1 g, water required = $\frac{1}{0.411}$ L = 2.43 L