Question

If the solubility of KCl in 100 mL of H,O is 34 g at 20°C and 43 g at 50 °C, label each of the following solutions as unsaturated, saturated, or supersaturated. If more solid is added than can dissolve in the solvent, assume that undissolved solid remains at the bottom of the flask.

a. adding 30 g to 100 mL of H.O at 20 °C
b. adding 65 g to 100 mL of H2O at 50 °C
c. adding 42 g to 100 mL of H,O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate

Answer 1

a) An unsaturated solution of KCl is obtained by adding 30 g in 100 mL of H₂O at 20°C. b) A saturated solution results from adding 65 g to 100 mL at 50°C, leaving undissolved KCl. c) Adding 42 g to 100 mL at 50°C and cooling to 20°C without precipitate forms a supersaturated solution.

Step-by-step explanation:

If the solubility of KCl in 100 mL of H₂O is 34 g at 20°C and 43 g at 50°C, the following solutions can be classified as:

a. Adding 30 g to 100 mL of H₂O at 20°C: Since the solubility at 20°C is 34 g and only 30 g is added, this solution is unsaturated because the amount of KCl is less than the maximum that could be dissolved at that temperature.

b. Adding 65 g to 100 mL of H₂O at 50°C: As the solubility is 43 g at 50°C, adding 65 g means that not all the KCl can dissolve and the excess will remain undissolved. This solution is saturated, and there will be 65 - 43 = 22 g of KCl that will not dissolve and will remain as a solid at the bottom.

c. Adding 42 g to 100 mL of H₂O at 50°C and slowly cooling to 20°C: Initially at 50°C, this solution is unsaturated. However, if the solution is cooled to 20°C and remains clear with no precipitate, it has become a supersaturated solution, since 42 g is more than the solubility at 20°C, which is 34 g.