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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 9.1×106).

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Solution

CaSO4(S)Ca2+(aq)+SO24(aq)Ksp=[Ca2+][SO24]
Let the solubility of CaSO4 be s.
Then, kSP=s29.1×106=s2s=3.02×103mol/L
Molecular mass of CaSO4=136 g/mol
Solubility of in gram/L
=3.02×103×136=0.41g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4
Therefore, to dissolve 1g of CaSO4 we require =10.41L=2.44L of water.


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