What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10-6).
CaSO4(s) ↔ Ca 2+ (aq) + SO2-4(aq)
Ksp = [ Ca 2+ ] [ SO2- ]
Let the solubility of CaSO4 be s.
Then, Ksp = s2
9.1 x 10-6 = s2
s = 3.02 x 10-3 mol/L
Molecular mass of CaSO4 = 136 g/mol
Solubility of CaSO4 in gram/L = 3.02 × 10-3 × 136 = 0.41 g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4
Therefore, to dissolve 1g of CaSO4 we require = 1/0.41 L = 2.44 Lof water.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 72: What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calci....
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