SOLUTION: I have a chemical that is a 40% concentrate, (40% active ingredient 60% water). I need to make one gallon of product that is 99.50% water and .50% of the active chemical.

Question 755335: I have a chemical that is a 40% concentrate, (40% active ingredient 60% water). I need to make one gallon of product that is 99.50% water and .50% of the active chemical.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
I have a chemical that is a 40% concentrate,
(40% active ingredient 60% water).
I need to make one gallon of product that is 99.50% water and .50% of
the active chemical
:
Let w = amt of pure water required
then
(1-w) = amt of original mixture required
:
Write a mixture equation based on the amt of water, (decimal equiv)
.60(1-w) + w = .995(1)
.6 - .6w + w = .995
-.6w + 1w = .995 - .6
.4w = .395
w = .395/.4
w = .9875 gal of water
and
1 - .9875 = .0125 gal of the original mixture
:
It may be easier to have these two amts in ounces, (32 oz = 1 qt)
.9875(32) = 31.6 oz of water
.0125(32) = .4 oz of original mixture
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