Explanation:
\[\begin{gathered}
{\text{CaS}}{{\text{O}}_{\text{4}}} \rightleftharpoons {\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{ + SO}}_{\text{4}}^{{\text{2 - }}}{\text{; }}{{\text{K}}_{{\text{SP}}}}{\text{ = 9}}{\text{.1}} \times {\text{1}}{{\text{0}}^{{\text{ - 6}}}} \hfill \\
{\text{S}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{S}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,{\mkern 1mu} {\text{S}} \hfill \\
\end{gathered} \]
where \(\mathrm {\mathrm{S}}\) is the solubility of \(\mathrm {\mathrm{CaSO}_{4}}\).
\(\mathrm {K_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}^{2-}\right]=S . S=S^{2}}\)
\(\mathrm {S=\sqrt{K_{s p}}=\sqrt{9.1 \times 10^{-6}}}\)
\(\mathrm {S=3.017 \times 10^{-3} \mathrm{M}}\)
\(\mathrm {\mathrm{CaSO}_{4}}\) solubility \(\mathrm {=3.017 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}}\)
The solubility in \(\mathrm {g / L=3 \times 10^{-3} \times 136}\)
\(\mathrm {=0.410 \mathrm{~g} / \mathrm{L}}\)
Hence, \(\mathrm {0.41 \mathrm{~g}}\) dissolve in \(\mathrm {1 \mathrm{~L}}\)
\(\mathrm {1 \mathrm{~g}}\) will dissolve in \(\mathrm {\dfrac{1}{0.41}=2.43 \mathrm{~L}}\).