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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35 HCl/100L water = x HCl/40L

x = 14L

Therefore, there are 16L of water.

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Almost right; 40-14 = 26 liters of water

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?

................

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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