# Problem: 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?

###### FREE Expert Solution

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

• Unsaturated solutionA solution that contains an amount of solute less than its solubility.
• Saturated solution: A solution that contains an amount of solute equal to its solubility (at equilibrium)
• Supersaturated solution: A solution that contains an amount of solute more than its solubility (starts to precipitate/crystallize)

We need to compare the amount of solute (Cr(NO3)3 • 9H2O) present in the given solution with its solubility

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###### Problem Details

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?