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

Respuesta :

Answer:

25.0 g.

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.
Ver imagen superman1987