# Chemistry

The solubility of KNO3 is 80g/100g water at 50 degrees and 10g/100g water at 0 degrees. The solubility of CuSO4 is 60g/100g water at 50 degrees and 20g/100g water at 0 degrees. A mixture contains 13.8g of KNO3 and 1.94g of CuSO4. A) Calculate the minimum mass of water that would dissolve the mixture at 50 degrees. B)the mixture is dissolved in the mass of water calculated in part (A) and then cooled to 0 degrees will any KNO3 crystallize out of the solution if so how many grams? What percent of the original amount of KNO3 of the original mixture has crysttallized out?

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1. WHAT IS THE MAXIMUM AMOUNT OF cR2S3 IN GRAMS OBTAINABLE

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2. To answer Lu the answer is zero. There is no Cr2S3 in the above problem.
A. The easiest way is to calculate the minimum amount of H2O for BOTH solutes.
At 50C, solubility KNO3 is 80/100. We want to dissolve 13.8g KNO3. How much H2O do we need That is
100 g x 13.8/80 = about 17 g H2O but you need a more accurate answer.

The solubility of CuSO4 at 50 C is 60g/100. We want to dissolve 1.94g. How much water do we need? That's
100 x 1.94/60 = about 3 g H2O.
If we need about 3g H2O to dissolve the Cu and about 17g H2O to dissolve the KNO3, we will need about 17 to dissolve the mixture

B. Now we cool the mess to zero C. We have about 17 g H2O. Solubility at zero C for KNO3 is 10/100. How much will dissolve in the 17g. That's
10 x (17/100) = about 2g; therefore, 13.8-2 = 11.8 will crystallize. % recovery is about (11.8/13.8)*100 = ?%

The problem doesn't ask for it but how much of CuSO4 crystallizes to contaminate the KNO3? We had 1.94 and about 17g H2O so
20 x (17/100) = about 3g CuSO4 will dissolve in that 17 g H2O; we had only 1.94 initially; therefore, none of it crystallizes and the KNO3 is pure KNO3. The lesson here is that you can crystallize a mixture of solids and by using temperature and solvent the two separate the mixture and get at least one of them out pure.

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3. For A wouldnt it be 20 g of water total instead of 17 g

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