Asked by McKenna Louise
How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?
Answers
Answered by
Reiny
amount of pure acid to be added -- x L
1(x) + .3(60-x) = .5(60)
10x + 3(60-x) = 5(60)
10x + 180 - 3x = 300
7x = 120
x = 120/7 or appr 17.14 L
1(x) + .3(60-x) = .5(60)
10x + 3(60-x) = 5(60)
10x + 180 - 3x = 300
7x = 120
x = 120/7 or appr 17.14 L
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