1.

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K, For calcium sulphate, K_(sp)=9.1 times 10^-6.

Answer»

Solution :`CaSO_4(s) leftrightarrowCa^2(aq)+SO_4^(2-)(aq)`
It’s the solubility of `CaSO_4` in moles `L^-1` then `K_(sp)=[Ca^(2-)] times [SO_4^(2-)]=s^2`
`s=sqrt(K_(sp))=sqrt(9.1 times 10^-6)=3.02 times 10^-3 mol L^-1`
`=3.02 times 10^-3 times 136 G L^-1=0.411 gL^-1`
(Molar mass of `CaSO_4=136g mol^-1`)
Thus, for dissolving 0.441 g, WATER required =1 L
`therefore` For dissolving 1g, water required =`1/0.411 L=2.43L`


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