Question

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

Answer 1

10 L of a 25% acid solution contains 0.25 * (10 L) = 2.5 L of acid.

Adding x L of pure water dilutes the solution to a concentration of 10%, such that

(2.5 L)/(10 L + x L) = 0.10

Solve for x :

2.5 = 0.10 * (10 + x)

2.5 = 1 + 0.10x

1.5 = 0.10x

15 = x

so 15 L of pure water are needed.