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How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution? 11 L 15 L 25 L
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@sarah786 @ikram002p
@chanelalgr
How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution? : let w = amt of water required to accomplish this : The amt of acid will remain the same only the water will increase, lowering the percent : .25(10) = .10(w+10) 2.5 = .10w + 1 2.5 - 1 = .1w 1.5 = .1w w = 1.5/.1 w = 15 liters of water @Elijah16 Source:Algebra .com
Thanks!
yw ^-^
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