SOLUTION: A liter of vinegar is 5% acid. How many more liters of pure acid are needed to get a mixture that is 81% acid?

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Question 773669: A liter of vinegar is 5% acid. How many more liters of pure acid are needed to get a mixture that is 81% acid?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Data To Know First:
5% Vinegar is as weight/weight. Density 1.00 gram/ml.
Pure 100% Acetic Acid, assumed weight/weight, Density 1.049 gram/cm^3

Use mass to do the main solving.
Letting a = grams of acetic acid (100% acid),
the mass of acid in the 1 liter of 5% is -------
1%2A%281.00%29%2A%281000%29 L%2A%28g%2Fml%29%2A%28ml%2FL%29%7D%0D%0A=%7B%7B%7B1000 grams, where "g" was an abbreviation for "grams"

Equation first all in mass is then:
highlight%28%281000%2A0.05%2B1.00%2Aa%29%2F%281000%2Ba%29=0.81%29
Note that the 1.00 in the rational equation stands for "100%" for "a".

The algebraic steps, not shown here, will lead to .....
a=4000 grams
from which you must use density to find the result in volume liters:
highlight%284000%2A%281%2F1.049%29%281%2F1000%29=3.81%29 Liters
The units conversion here is g%2A%28cm%5E3%2Fg%29%28L%2Fcm%5E3%29=L