Question

How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?

Answer 1

Original Mixture: 10 liters, 25% acid = 0.25

Added mixture : x liters, 0% acid = 0 since it is pure water.

Total mixture: Volume = (10 + x), % = 10% = 0.1

Using mixtures formula:

Volume₁ * %₁ + Volume₂*%₂ = Total Volume*Total%

10*0.25 + x*0 = (10 + x)*0.10

2.5 + 0 = 10*0.10 + 0.1*x

2.5 = 1 + 0.1x

1 + 0.1x = 0.250

0.1x = 2.5 - 1

0.10 = 1.5

x = 1.5/0.1

x = 15

Amount of pure water to be added is 15 liters.

Hope this helps.

Answer 2

15 liters. that is the answer.