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 322 g in 100 g water. When this solution is slowly cooled to 15 ∘C, no precipitate forms.

The given question is incomplete. The complete question is as follows.

The solubility of in water is 208 g per 100 g of water at . A solution of in water at is formed by dissolving 322 g in 100 g water. When this solution is slowly cooled to , no precipitate forms. At equilibrium, what mass of crystals do you expect to form?

Explanation:

The given data is as follows.

 Solubility of in 100 g of water = 208 g (at )

 Amount of  in 100 g of water = 322 g (at )

At equilibrium, the mass of crystals formed is calculated as follows.

        Mass of crystal = Mass of  at - Mass of

Putting the given values into the above formula, mass of the crystals will be calculated as follows.

    Mass of crystal = Mass of  at - Mass of

                            = 322 g - 208 g

                            = 114 g

Thus, we can conclude that crystals formed is 114 g.    

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