What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`?
(For calcium sulphate , `K_(sp) is 9.1xx10^(-6))`.

Explanation

For `CaSO_(4)`, Let solubility be S mol/litre
Thus, `3.02xx10^(-3)` mole of `CaSO_(4)` is soluble in water = 1 litre
`:. (1)/(136)` mole (1 g) of `CaSO_(4)` is soluble in
`= (1)/(136xx3.02xx10^(-3))`
`=2.43` litre water