A beaker containing 80 g of KNO3, in 100 g of water, has a temperature of 50 ºC. If the solution cools to 30 ºC, approximately how many grams of the salt recrystallizes on the bottom of the beaker?

Answer 1

To determine how many grams of KNO3 recrystallizes when the solution cools from 50 ºC to 30 ºC, we need to consider the solubility of KNO3 at these temperatures.

1. Solubility of KNO3:

   • At 50 ºC, the solubility of KNO3 in water is approximately 109 grams per 100 grams of water.
   • At 30 ºC, the solubility of KNO3 in water is approximately 31 grams per 100 grams of water.

2. Given:

   • We have 80 g of KNO3 dissolved in 100 g of water at 50 ºC.

3. At 50 ºC: Since the solubility at 50 ºC is 109 g per 100 g of water, the solution is unsaturated with a total of 80 g of KNO3 dissolved.

4. At 30 ºC: When the solution cools to 30 ºC, the solubility decreases to 31 g per 100 g of water. This means that only 31 g of KNO3 can remain dissolved in the water at this temperature.

5. Calculating Recrystallization:

   • Amount of KNO3 that can remain in solution at 30 ºC: 31 g
   • Initial amount of KNO3: 80 g

   Therefore, the amount of KNO3 that will recrystallize is: \[ \text{Recrystallized KNO3} = \text{Initial KNO3} - \text{Solubility at 30 ºC} = 80, \text{g} - 31, \text{g} = 49, \text{g} \]

Conclusion: Approximately 49 grams of KNO3 recrystallizes at the bottom of the beaker when the solution cools from 50 ºC to 30 ºC.