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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 ). 

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Let the solubility of CaSO4 be s. 

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 CaSO

Therefore, to dissolve 1g of CaSO4 we require  = 1/0.41L=2.44L of water

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