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InterviewSolution
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What is the minimum volume of water required to dissolve `1.0 g` of calcium sulphate at `298 K`? (For calcium sulphate , `K_(sp) is 9.1xx10^(-6))`. |
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Answer» Correct Answer - `2.43` litre of water `CaSO_(4(s)) harr Ca_((aq))^(2+) + SO_(4(aq))^(2-)` `K_(sp) = [Ca^(2+)][SO_(4)^(2-)]` Let the solubility of `CaSO_(4)` be s. Then, `K_(sp) = s^(2)` `9.1 xx 10^(-6) = s^(2)` `s = 3.02 xx 10^(-3) "mol"//L` Molecular mass of `CaSO_(4) = 136 g//"mol"` Solubility of `CaSO_(4)` in gram/L `= 3.2 xx 10^(-3) xx 136` `= 0.41 g//L` This means that we need 1L of water to dissolve 0.41g of `CaSO_(4)` Therefore, to dissolve 1g of `CaSO_(4)` we require `= (1)/(0.41) L = 2.44 L` of water. |
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