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))`.

Answer

`2.43` litre of water

Explanation

`CaSO_(4(s)) harr Ca_((aq))^(2+) + SO_(4(aq))^(2-)`
`K_(sp) = [Ca^(2+)][SO_(4)^(2-)]`
Let the solubility of `CaSO_(4)` be s.
Then, `K_(sp) = s^(2)`
`9.1 xx 10^(-6) = s^(2)`
`s = 3.02 xx 10^(-3) "mol"//L`
Molecular mass of `CaSO_(4) = 136 g//"mol"`
Solubility of `CaSO_(4)` in gram/L `= 3.2 xx 10^(-3) xx 136`
`= 0.41 g//L`
This means that we need 1L of water to dissolve 0.41g of `CaSO_(4)`
Therefore, to dissolve 1g of `CaSO_(4)` we require `= (1)/(0.41) L = 2.44 L` of water.