Question

The solubility of a solute is 35 at ${60}^{o}C$ and 25 to ${30}^{o}C$. If 200 g of solvent is made to saturate with the solute at ${60}^{o}C$, what is the amount of precipitate formed at ${30}^{o}C$?

Answer 1

1. The amount of solute dissolved in a definite quantity of a solvent depends upon the nature of the solute & nature of the solvent.
2. The maximum amount of a solute in grams , which can dissolve in 100g of a solvent at a given temperature is called solubility.
3. For most of the substances, solubility increases with the increase in temperature .
4. It is given that solubility of a solute at ${60}^{o}C$ = 35 and solubility of the solute at ${30}^{0}C$ = 25
5. 100 g of the solvent a t ${60}^{o}C$ dissolves 35 g of the solute to get a saturated solution
6. In 200 g of the solvent at ${60}^{o}C$ dissolve ( 2 x 35 ) = 70g of solute to get a saturated solution.
7. At ${30}^{0}C$, 100g of solvent dissolves 25 g of solute.
8. At ${30}^{o}C$ 200 g of solvent dissolves ( 25 x 2 ) 50 g of solute .
9. The extra salt is precipitated when temperature decreases .
10. The solute precipitated is 70 - 50 = 20 g at ${30}^{o}C$.

Therefore the amount of precipitate formed at ${30}^{o}C$ is 20 g solute .