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?

Answer 1

0.140 g/mL

Step-by-step explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in a quantity of solvent.

At 21°C a quantity of solute is precipitated given a saturated solution (That is the maximum amount of solute that solvent can dissolve). A quantity of solute remains in the solution. This quantity is recovered evaporating the water. You find the quantity is 0.084kg =

84g of solute

The solution had 600.0mL of water. That means solubility of the solute X at 21°C is:

84g X / 600mL = 0.140 g/mL