🤓 Based on our data, we think this question is relevant for Professor Gorman's class at LONESTAR.
The solubility of Cr ( NO3 )3 9 H2O in water is 208 g per 100 g of water at 15 oC. A solution of Cr ( NO3 )3 9 H2O in water at 35 oC is formed by dissolving 313 g in 100 g water. When this solution is slowly cooled to 15 oC, no precipitate forms.
At equilibrium, what mass of crystals do you expect to form?
We’re asked to determine the mass of Cr(NO3)3•9H2O crystals expected to form at equilibrium based on its solubility at 15°C.
Recall that a saturated solution contains the maximum amount of solute that can be dissolved in a solvent at a particular temperature.
We know that 313 g of Cr(NO3)3•9H2O is dissolved in 100 g water at 35°C to make a solution.
When this solution is cooled to 15°C; no precipitate forms.
We're given the solubility of Cr(NO3)3•9H2O at 15°C.
We will now find the mass of Cr(NO3)3•9H2O crystals that will precipitate at equilibrium or in a saturated solution.