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

$CaS{O}_4$

$K_{sp}=S^2=9.1\times {10}^{-6}$

$S=3\times {10}^{-3}mole/L$

Solubility in grams $=3\times {10}^{-3}\times 136g/L$

$=0.408g/L$

$0.408g\to 1L$

$1g\to \frac{1}{0.408}=2.45L$