What is the minimum volume of water required to dissolve $1 \, g$ of calcium sulphate at $298 \, K$? (For calcium sulphate,$K_{sp}$ is $9.1 \times 10^{-6}$).

(D) $CaSO_{4(s)} \leftrightarrow Ca^{2+}_{(aq)} + S{O_{4}}^{2-}_{(aq)}$
$K_{sp} = [Ca^{2+}][SO_{4}^{2-}] = s^2$
$s = \sqrt{K_{sp}} = \sqrt{9.1 \times 10^{-6}} = 3.0166 \times 10^{-3} \, mol/L$
Molar mass of $CaSO_{4} = 40 + 32 + (4 \times 16) = 136 \, g/mol$
Solubility in $g/L = s \times \text{Molar mass} = 3.0166 \times 10^{-3} \times 136 \approx 0.4103 \, g/L$
Volume required for $1 \, g = \frac{1 \, g}{0.4103 \, g/L} \approx 2.44 \, L$.