Question

If 50.0 g of water saturated with potassium chloride at 80.0°C is slowly evaporator to dryness, how many grams of the dry salt will be recovered

Answer 1

25.0 g.

Step-by-step explanation:

• From the solubility curve that is shown in the attached image, the solubility of KCl per 100.0 g of water is about 50.0 g.

• So, a saturated solution of KCl in 50.0 g of water will contain about 25.0 g.

• Thus, the grams of the dry salt that will be recovered is 25.0 g.