Question

A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 21.° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg.Using only the information from above, can you calculate the solubility of X at 21.° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.

Answer 1

Answer:

• Yes, it is 14. g of compound X in 100 ml of solution.

Explanation:

The relevant fact here is:

• the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

• 84. g solute / 600 ml solution = y / 100 ml solution

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.