A chemist mixes a 200 L of a solution that is 60% acid with a 300 L of a solution that is 20% acid. What is the percentage of the mixture?

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Answer 1 · 2015-08-11T14:18:54

The resulting mixture is $36 %$ $36%$ acid.

The resulting mixture's acid proportion is

$\frac{volume of acid in the solution}{volume of the solution}$ $"volume of acid in the solution"/"volume of the solution"$

The volume of the mixture is $200 + 300 = 500 L$ $200+300=500L$

The volume of acid in the first solution is $0.6 \cdot 200 = 120 L$ $0.6*200=120L$

The volume of acid in the second solution is $0.2 \cdot 300 = 60 L$ $0.2*300=60L$

Therefore, the resulting mixture's acid proportion is:

$\frac{120 + 60}{500} = \frac{180}{500} = 36 %$ $(120+60)/500=180/500=36%$