Given:

For calcium sulphate, K_sp is 9.1 × 10^–6

Ionization of calcium sulphate

CaSO₄Ca²⁺ + SO₄^2-

As we know that, K_sp = [A⁺] [B^-]

In the above reaction,

[A⁺] = Ca²⁺

[B^-] = SO₄^2-

∴ K_sp = [Ca²⁺] [SO₄^2-]

Let the solubility of calcium sulphate is x.

So, K_sp = x²

∴9.1×10^-6=x²

∴x=3.02×10^-3

As molar mass of CaSO₄ is :

⇒40 +32 + 4× 16

⇒40+32+64

⇒136g/mol.

Solubility of calcium sulphate in g/mol is

=3.02×10^-3×136

=0.41 g/L

Thus, for dissolving 0.41g, water required = 1L

∴ for dissolving 1g, water required = L

⇒2.44L

Thus, minimum volume of water required to dissolve 1 g of CaSO₄ is 2.44L