Question

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, $K_{sp}$ is 9.1 x 10⁻⁶)

Answer 1

Ksp = [Ca2+][SO42-]

Let the solubility is S

Then

Ksp = s2

9,1 x 10-6 = s2

S = 3.02 x 10-3 mol/L

Molecular mass of CuSO4 = 136 g

Solubility in Gram/L = 3.02 x 10-3 x 136 = 0.41

Therefore to dissolve 1 g of salt we require 1/0.41 = 2.44 L water