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

`CaSO_(4)(s) overset(aq)(hArr) Ca^(2+) (aq) + SO_(4)^(2-) (aq)`
Let the solubility of `CaSO_(4) be_(x) S`
`S^(2) =K_(sp) or S =(K_(sp))^(1//2) =(9.0 xx 10^(-6))^(1//2)`
`S=(3.0 xx 10^(-3))M =3.0 xx10^(-3) mol L^(-1).`
No. of moles in 1 g of `CaSO_(4) =("Mass")/("Molar mass") = ((1g))/((136 g mol^(-1)))= (1)/(36) mol`
`3xx10^(-3)` moles of `CaSO_(4)` dissolve in water =1 L
`1//136` moles of `CaSO_(4)` dissolve in water `=((1L) xx(1//136 mol))/((3xx10^(-3) mol))=2.45 L`