Chemistry Practice Problems Making Solutions Practice Problems Solution: The solubility of Cr ( NO3 )3 9 H2O in water is 20...

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Solution: 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?

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?

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. 

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