Question

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.

Answer 1

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.