7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).

CaSO4(s)Ca2+(aq)+SO42-(aq)
Ksp=[Ca2+][SO42-]
LetthesolubilityofCaSO4bes.
Then,Ksp=s2
9.1×10-6=s2
s=3.02×10-3mol/L
MolecularmassofCaSO4=136g/mol
SolubilityofCaSO4ingram/L
=3.02×10-3×136
=0.41g/L
Thismeansthatweneed1Lofwatertodissolve0.41gofCaSO4
Therefore,todissolve1gofCaSO4werequire=10.41L=2.44Lofwater.