Why is a solution a mixture
| (Sum of the shares) || (in the mix |
- 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.
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.)
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