What is the minimum volume of water requires to dissolve 1g of calcium sulphate at 298K ? (For calcium sulphate, `K_(sp)` is `9.1xx10^(-6)`).

Explanation

`CaSO_(4)(s)hArr Ca^(2+)(aq)+SO_(4)^(2-)(aq)`
`K_(sp)=[Ca^(2+)][SO_(4)^(2-)]=9.1xx10^(-6)`
or, `s^(2)=9.1xx10^(-6)` [s=solublity of `CaSO_(4)`]
`therefore s=3.02xx10^(-3)mol*L^(-1)`
Therefore, `3.02xx10^(-3)mol=3.02xx10^(-3)mol*L^(-1)`
Therefore, `3.02xx10^(-3)mol=3.02xx10^(-3)xx(40+32+64)g` of
`CaSO_(4)` can remain dissolved in 1L of solution.
To dissolve 1g of `CaSO_(4)`, the required volume of water
`=(1L)/(3.02xx10^(-3)xx136)=2.43L` [As the solution is very dilute, we assume of solution=volume of water].