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50g of a saturated aqueous solution of potassium chloride at `30^(@)C` is evaporated to dryness, when 13.2 g of dry `KCl` was obtained. Calculate the solubility of `KCl` in water at `30^(@)C`.

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Mass of water in solution `= (50-13.2) = 36.8g`
Solubility of KCl `= ("Mass of KCl")/("Mass of water")xx100=(13.2)/(36.8)xx100 =35.87g`.

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