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-]Let the solubility of CaSO4 be s. Then, Ksp=s29.1×10-6=s2s=3.02×10-3 mol/LMolecular mass of CaSO4 = 136 g/molSolubility of CaSO4 in gram/L = 3.02 × 10-3 × 136= 0.41 g/LThis means that we need 1L of water to dissolve 0.41g of CaSO4Therefore, to dissolve 1g of CaSO4 we require =10.41L=2.44L  of water.