What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K. K_(sp) for CaSO_(4) is 9.0xx10^(-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. K_(sp) for CaSO_(4) is 9.0xx10^(-6).
Answer» 2.45 L 4.08 L 4.90 L 3.00 L Solution :`CaSO_(4)(s)+AQ hArr Ca^(2+)(aq)+SO_(4)^(2-)(aq)` `K_(sp)=[Ca^(2+)][SO_(4)^(2-)]` Suppose thesolubility of `CaSO_(4) = s "mol" L^(-1)` `K_(sp)=sxxs=s^(2)` `9.0xx10^(-6) = s^(2) or s=3xx10^(-3) "mol" L^(-1)` Molar MASS of `CaSO_(4) = 40 + 32 + 64 = 136 g "mol" ^(-1)` `:. 1g CaSO_(4) = (1)/(136) ` mol `3xx10^(-3)` mol of `CaSO_(4)` require minimum WATER to dissolve = 1 L `:. (1)/(136) ` molof `CaSO_(4)` will require water `=(1)/(3xx10^(-3))xx(1)/(136) ~=2.45 L`