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

What is the minimum volume of water required to dissolve 1 g of calciu

1.	What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K ? (For calcium sulphate , K_(sp)=9.1xx10^(-6) )
Answer» Solution :Suppose solubility in water of `CaSO_4` is S mol `L^(-1)` `{:(CaSO_(4(s)) hArr, Ca_((aq))^(2+)+ , SO_(4(aq))^(2-)),(S, S M, S M):}` `therefore K_(SP)(CaSO_4)=[Ca^(2+)][SO_4^(2-)]=(S)(S)=S^2` `therefore S=sqrt(K_(sp))` `=sqrt(9.1xx10^(-6))=3.0167xx10^(-3) "mol L"^(-1)` `therefore "mol L"^(-1) = "mol"/L = "weight in gm"/"molecular X VOLUME in litre"` `therefore 3.0167xx10^(-3) = "1 G"/("136 g mol"^(-1) xx "volume")` molecular mass of `CaSO_4` =40+32+64=136 g `mol^(-1)` `therefore` Volume =`1/(3.0167xx136xx10^(-3))=1000/(3.0167xx136)` =2.4374 L =2437.4 mL 1 g `CaSO_4` soluble in minimum 2437.4 mL solution.