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A K2Cr2O7 solution containing 16 g of K2Cr2O7 per 100.0 g of water is cooled from 35°C to 0°C. What happens during cooling? I know from the solubility graph that: at 35 C, the the solution can contain up to 20 g of K2Cr2O7 per 100 g H2O & at 0 C, the solution can contain up to 5 g of K2Cr2O7 per 100 g H2O. How many grams of K2Cr2O7 will precipitate out of the solution? I don't want an answer, just an equation or method to solve the problem. Thanks :)
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Assuming we neglect supersaturation, all the solute in excess of the solubility at the cooler temperature will precipitate.
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