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

Question

A solid substance X is soluble in water to the extent of 1.70 mg/mL. of 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 an impurity V. 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 purigy X, as well as the amount and percent recovery of pure X obtained after the crystallization steps have been complete Show all calculations. (Assume that there no insoluble impurities present in the sample). Be CLEAR AND SHOW ALL THE STEPS OF THE CALCULATIONS.

Explanation / Answer

Solubility of X=1.70mg/ml

Solubility at 100 deg C=22.5 mg/ml

Amount crystallized after first step==22.5-1.70=20.8 mg/ml

Salt +impurity=59.5mg

If 1 ml of solvent is used for dissolving,

Remaining X,(undissolved or thrown out of solution)=59.5-20.8=38.7 mg

After Second step amount crystallized=38.7-20.8=17.9 mg,

So a third time recrystallisation is required,when the rest amount of X is unrecovered as it has high impurity content.

X recovered=59.5-17.9=41.6

%X recovered=41.6/59.5*100=69.91%

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