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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.

User NonDucor
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1 Answer

3 votes

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

The solubility of
Cr(NO_(3)).9H_(2)O in water is 208 g per 100 g of water at
15^(o)C. A solution of
Cr(NO_3)_(3)9H_2O in water at
35^(o)C is formed by dissolving 322 g in 100 g water. When this solution is slowly cooled to
15^(o)C, no precipitate forms. At equilibrium, what mass of crystals do you expect to form?

Step-by-step explanation:

The given data is as follows.

Solubility of
Cr(NO_(3))9H_(2)O in 100 g of water = 208 g (at
15^(o)C)

Amount of
Cr(NO_(3))9H_(2)O in 100 g of water = 322 g (at
35^(o)C)

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

Mass of crystal = Mass of
Cr(NO_(3))9H_(2)O at
15^(o)C - Mass of

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

Mass of crystal = Mass of
Cr(NO_(3))9H_(2)O at
35^(o)C - Mass of

= 322 g - 208 g

= 114 g

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

User Akshay Borade
by
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