Question 1142665: Three parts of liquid chlorine and two parts of salt are added to each 100 parts of water in a swimming pool. How much chlorine is required if the pool contains 4000 L (including chemicals)?
Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441)   (Show Source):
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
It is not an "in-row" regular word problem on mixtures.
It is SPECIAL. It is designed to check if you know the basics of mixtures from Science,
and if you do not know them, then to teach you a bit.
When the salt solution is of the low concentration, then dissolved salt practically DOES NOT lead to increasing volume.
The masses of water and salt are summed, but the volumes do not.
(Again, it is true for the low concentrated salt mixtures, which is exactly our case).
So, we should add 100 liquid parts of water and 3 parts of liquid chlorine:
100 + 3 = 103 equal liquid parts (by volume).
But dissolving two parts of salt DOES NOT change the volume.
So, in 4000 liters of liquid in the pool we have  = 38.835 standard liquid volumes.
Hence, the chlorine volume required is 3*38.835 = 116.505 liters.
Notice, that NEITHER salt mass NOR the salt volume DO NOT PARTICIPATE in the solution of this problem (!)
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