🤓 Based on our data, we think this question is relevant for Professor Singh's class at HCC.
The solubility of Cr(NO3)3 • 9H2O in water is 208 g per 100 g of water at 15˚C. A solution of Cr(NO3)3 • 9H2O in water at 35˚C is formed by dissolving 313 g in 100 g water. When this solution is slowly cooled to 15˚C, no precipitate forms. Is the solution that has cooled down to 15˚C unsaturated, saturated, or supersaturated?
We’re asked to classify the solution that has cooled down to 15˚C as unsaturated, saturated, or supersaturated.
Recall that solubility is the maximum amount of solute that can be dissolved in a solvent at a specific temperature.
Solutions can be classified depending on the amount of solute present:
We need to compare the amount of solute (Cr(NO3)3 • 9H2O) present in the given solution with its solubility.