Question

A saturated solution is made by dissolving a 36.8 of a solid in 200 ML of water. A second solution is made by dissolving 19.1 g of the same solid in 100 ML of water. How would this solution be classified?

Answer 1

To determine the classification of the second solution, we need to compare the amount of solid dissolved in it to the amount that can be dissolved in water at that temperature.

From the first solution, we know that 36.8 g of the solid can dissolve in 200 mL of water. To convert this to g/mL, we divide 36.8 g by 200 mL to get 0.184 g/mL.

Using this value, we can calculate the maximum amount of solid that can dissolve in 100 mL of water:

0.184 g/mL x 100 mL = 18.4 g

Since the second solution was made by dissolving 19.1 g of the solid in 100 mL of water, it is a supersaturated solution.