Respuesta :
The given question is incomplete. The complete question is as follows.
The solubility of [tex]Cr(NO_{3}).9H_{2}O[/tex] in water is 208 g per 100 g of water at [tex]15^{o}C[/tex]. A solution of [tex]Cr(NO_3)_{3}9H_2O[/tex] in water at [tex]35^{o}C[/tex] is formed by dissolving 322 g in 100 g water. When this solution is slowly cooled to [tex]15^{o}C[/tex], no precipitate forms. At equilibrium, what mass of crystals do you expect to form?
Explanation:
The given data is as follows.
Solubility of [tex]Cr(NO_{3})9H_{2}O[/tex] in 100 g of water = 208 g (at [tex]15^{o}C[/tex])
Amount of [tex]Cr(NO_{3})9H_{2}O[/tex] in 100 g of water = 322 g (at [tex]35^{o}C[/tex])
At equilibrium, the mass of crystals formed is calculated as follows.
Mass of crystal = Mass of [tex]Cr(NO_{3})9H_{2}O[/tex] at [tex]15^{o}C[/tex] - Mass of
Putting the given values into the above formula, mass of the crystals will be calculated as follows.
Mass of crystal = Mass of [tex]Cr(NO_{3})9H_{2}O[/tex] at [tex]35^{o}C[/tex] - Mass of
= 322 g - 208 g
= 114 g
Thus, we can conclude that crystals formed is 114 g.
