Question

A solid mixture consists of 32.3 g of KNO3 (potassium nitrate) and 5.7 g of K2SO4 (potassium sulfate). The mixture is added to 130. g of water.

If the solution described in the introduction is cooled to 0 ∘C, what mass of KNO3 should crystallize?

Answer 1

Given a solid mixture consists of 32.3 g of KNO3 (potassium nitrate) and 5.7 g of K2SO4 (potassium sulphate). The mixture is added to 130 g of water.

Let us assume KNO3 has a solubility of 14 g solute/ 100 g water.

X g of KNO3/ 130 g water= 14 g of KNO3/ 100 g water

Then, X = (14 g KNO3 /100 g H2O) x 130 g H2O = 18.2 g KNO3

Therefore, X = 18.2 g KNO3 i.e., in 130 g H20 the solubility of KNO3 would be 18.2 g KNO3.

If you have 32.3g of KNO3 in 130 g of water at 0 degrees celsius and only 18.2 g can dissolve, then
32.3 g - 18.2g = 14.1 g left undisolved which gives you the amount that would crystallize.

Therefore, mass of KNO3 that would crystallize if the solution described in the introduction is cooled to o degrees celsius is 14.1 g .