Question

5. Consider a case in which the solubilities of two substances (we will call them Substance A and an impurity Substance B) are both 1.000g/ 100.0 mL of solvent at 20.0 C and 10.000 g/100.0 ml of solvent at 100.0 °C (their solubilities are independent of each other, meaning that 1.000 g A and 1.000 g B can dissolve in 100.0 mL solvent at 20.0°C 10.000 g A AND 10.000 g B can dissolve in 100.0 mL solvent at 100.0*c). If the composition of 12.000 grams of impure product is determined to be 9.000 g of A and 3.000 g of B, answer the following questions concerning the recrystallization of this product in an attempt to eventually isolate only pure A-recrystallization involves adding 100.0 mL of solvent to the 12.000 grams of impure A/B, heating to 100.0°C, cooling to 20.0℃, and vacuum filtering the crystals that form. How many grams of A will dissolve in the hot solvent? How many grams of B will dissolve in the hot solvent? How many grams of A will crystallized from the solution when the solvent is cooled to 20 C? How many grams of B will crystallized from the solution when the solvent is cooled to 20°C? If you filter the cooled solution, how many grams of A will be found on the filter paper? If you filter the cooled solution, how many grams of B will be found on the filter paper? NG 2

Answer 1

1. Since the mass of A ( 9 g) is lower than the solubility of A ( 10 g) all 9 g of A will dissolve in hot solvent.

2. Since the mass of B ( 3 g) is lower than the solubility of B ( 10 g) all 9 g of A will dissolve in hot solvent.

3. When cooled to 20 ⁰C, the solubility reduces to 1 g/ 100 ml. So only 1 g can remain in solution and the remaining gets precipitated.

Mass of A crystallised = 9 - 1 = 8 g

4. When cooled to 20 ⁰C, the solubility reduces to 1 g/ 100 ml. So only 1 g can remain in solution and the remaining gets precipitated.

Mass of A crystallised = 3 - 1 = 2 g

5. On filtering the cooled solution, we get 8 g of A on filter paper.

6. On filtering the cooled solution, we get 2 g of B on filter paper.