A solid substance X is soluble in water to the extent of 1.70 mg/mL water at 25

Question

A solid substance X is soluble in water to the extent of 1.70 mg/mL water at 25 degree C and 22.5 mg/mL of water at 100 degree C. You have been provided with a sample that contains 59.5 mg of X and an impurity Y. Assuming that 8.50 mg of the impurity of Y are present along with 59.5 mg of X. By showing all the calculation steps, describe how you would be able to purify X if Y has the same solubility behavior as X in water. Indicate the number of crystallization steps required to purify X, as well as the amount and percent recovery of pure X obtained after the crystallization steps have been completed. Show all calculations. (Assume that are no insoluble impurities present in the sample).

Explanation / Answer

solubility of X in H2O = 22.5 mg/ml at 100 oC

So for 59.5 mg of X we would need = 2.644 ml H2O

Assuming same solubility for Y

solubility of Y at this temperature = 0.38 ml of H2O is needed for 8.50 mg of impurity Y

So a solution of 0.76 ml of H2O would have 8.50 mg of Y in it and 8.50 mg of X in it

remainder solid left is all X = 51 mg

Thus X can be separated from Y in great percent yield = 51 x 100/59.5 = 85.71%

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