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). ?

Hello Student
Ksp = [Ca²⁺][SO₄^2-]
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

Dear Student

Let s be the solubility ofCaSO₄​.
CaSO₄​ ⇌ Ca²⁺ SO₄²⁻​
[Ca²⁺] = [SO₄²⁺​] =s
Ksp​[Ca²⁺][SO₄²⁻​] = s×s = s² =9.1×10−6
s = 0.003016 M
The molar mass ofCaSO₄​is40 + 32 + 64 = 136 g.
The solubility in g/L is0.003017 × 136 = 0.410 g/L.
Hence, 0.41 g dissolves in 1 L.
1 g will dissolve in 1/0.41 ​= 2.43 L

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya