Question

Question 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)).

Class XI Equilibrium Page 230

Answer 1

CaSO4 <=> Ca²+ + SO4²-
Let s is the solubility of CaSO4
then, we know, Ksp = [Ca²+][SO4²-]
Ksp = s.s = s²
A/C to question,
Ksp = 9.1 × 10^-6
so, 9.1 × 10^-6 = s²
s = √{9.1 × 10^-6} = 3.017 × 10^-3 M

so, solubility of CaSO4 = 3.017 × 10^-3 M
= 3.017 × 10^-3 mol/L
= 3.017 × 10^-3 × 136g/L [ weight = molecule weight × mole . and molar weight = 136g/mol]
= 410.3 × 10^-3 g/L
= 0.4103 g/L
it means 0.4103 g CaSO4 is dissolved in 1 L
Therefore, 1g CaSO4 is dissolved in 1/0.4103
= 2.437 L

Answer 2

Answer:

Explanation:

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