# Problem: 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?

###### FREE Expert Solution

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 solutthat can be dissolved in a solvent at a particular temperature.

• When a solution is saturated, it is in equilibrium
• This means that the rate of dissolving is equal to the rate of precipitation (solids are formed)

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

• This means that it is supersaturated → dissolves more than the maximum amount of solute that can be dissolved at a higher temperature w/o precipitation (solids form)

We're given the solubility of Cr(NO3)3•9H2O at 15°C.

• Solubility is the maximum amount of solute that can dissolve in a solvent.
• This means that at 15°C, only 208 g of Cr(NO3)3•9H2O will dissolve in 100 g of water.

We will now find the mass of Cr(NO3)3•9H2O crystals that will precipitate at equilibrium or in a saturated solution.

88% (351 ratings) ###### Problem Details

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?