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Question:

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).

Answer:

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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