Question

10) A sample of potassium nitrate (49.0 g) is dissolved in 101 g of water at 100 °C with precautions taken to avoid evaporation of any water. The solution is cooled to 30.0 °C and a small amount of precipitate is observed. This solution is

Answer 1

Answer:

The solubility of potassium nitrate, KNO3 at 25 °C is 24 g/100 mL of water. The solubility at 100 °C is 244 g/100 mL of water.

That means, if we add 49 g of KNO3 to 100 mL water at 25 °C, 24 g dissolve. 25 g of KNO3 solid remain on the bottom. This is called as saturated solution.

Now if you heat the mixture to 100 °C, the remaining 25 g of KNO3 will dissolve. At this temperature, the solubility limit in 100 mL of water is 244 g glucose. The solution is called unsaturated.

Now cool the mixture back to 30 °C, ~ 25 g of KNO3 should precipitate from solution.

So again the solution is called saturated solution.

If precipitate is not formed then the solution is called supersaturated solution.

In the present condition, a small amount of precipitate is formed when the solution is cooked to 30°C so this solution is called saturated solution. (It is also known as dynamic equilibrium between solute and solvent molecules)

Please let me know if you have any doubt. Thanks