A mixture of acid and water is 35% acid. If the mixture contains a total of 40L, how many liters of pure acid are in the mixture? How many liters of pure water are in the mixture?

As with all solutions, we measure by weight, not by volume.  However we consider an acid solution (which acid(s)) and I presume deionized water (again a dangerous presumption).

Already we have problems with the validity of the question.  But let us devise a calculation anyhow.

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A mixture of acid and water is 35% acid. If the mixture contains a total of 40L, how many liters of pure acid are in the mixture? How many liters of pure water are in the mixture?

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40L (+H-O-H+) + (N-acid [35%])

Calculation: 1/35 = .0287...X 40 = 11.48 L (N-acid) and 28.52 L (+H-O-H+).

I hate to make this sort of calculation because we deviate from format.

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