Question

A solution containing 74.7 g of NaNO3 in 70.0 g H2O at 50 ∘C is cooled to 20 ∘C.

Use the solubility data from the table below. Solubility (g/100gH2O)

Substance 20∘C 50∘C

KCl            34.0 42.6

NaNO3     88.0 114.0

C12H22O11 (sugar) 203.9   260.4

1)How many grams of NaNO3 remain in solution at 20 ∘C ?

2)How many grams of solid NaNO3 came out of solution after cooling?

Answer 1

According to the table;

88.0 g NaNO3 present in 100 g water at 20.0 C.

114.0 g NaNO3 present in 100 g water at 50.0 C.

Here 74.7 g of NaNO3 in 70.0 g H2O at 50 ∘C is cooled to 20 ∘C.

The table says that at 20∘C 88.0g of NaNO3 will remain dissolved in 100g of H2O.

So at 20∘C ;

70.0 g of H2O will dissolve
(88.0g) x (70.0g / 100g) = 61.6 g NaNO3

61.6 g NaNO3 remain in solution at 20 ∘C
So at 20∘C
74.7 g - 70.4g = 13.1 grams of NaNO3 will come out of solution

Solubility:

When the maximum amount of solute is dissolve in the solvent is called solubility. On the basic of solubility the solution are following types:

Saturate solution:

A solution which contains the maximum amount of dissolved solute at the equilibrium is called saturated solution.

Un-saturate solution:

A solution which contains lesser amount of solute as compare to saturated solution. It will be able to dissolver more solute.

Super -Saturate solution:

A solution which contains the more dissolved amount of solute as compare to saturated solution.