Why is a solution a mixture

Percentage calculation

Proof: The following two calculations are important (only the usual calculation rules are used):
 (c-b) + (b-a) = c-a (Sum of the shares) (c-b) a + (b-a) c = ca-ba + bc-ac = bc-ba = (c-a) b (in the mix contained substance)
For example, if a "portion" is 100 grams, and
• take (c-b) portions of the first solution, i.e. 100 (c-b) grams, and
• one takes (b-a) parts of the second solution, i.e. 100 (b-a) grams,
• so you have a total of 100 (c-b) grams + 100 (b-a) grams = 100 (c-a) grams.
This is to explain the first equation.

Now for the second equation:

• 100 (c-b) grams of the first solution contains a (c-b) grams of the solute
• 100 (b-a) grams of the second solution contains c (b-a) grams of the solute
• So the mixture (according to the second calculation) contains (c-a) b grams of the dissolved substance.

Since the mixture is 100 (c-a) grams, there are b grams of the solute in 100 grams of the mixture.
Exactly what we wanted.

(We have with Mass percent argued - not with Volume percentbecause when mixing substances it is not clear how the volume changes! If you want to work with volumes, you have to know the respective densities.)