Question

Pure acid is to be added to a 20% acid solution to obtain 28 L of a 40% acid solution. What amounts of each should be used?

Probability and combinatorics
ANSWERED
asked 2021-06-24

Pure acid is to be added to a \(20\%\) acid solution to obtain 28 L of a \(40\%\) acid solution. What amounts of each should be used?

Answers (1)

2021-06-25

Let x be the amount of pure acid to be added so that 28-x is the amount of \(20\%\) acid solution, both in L.
In terms of percentage, a pure acid corresponds to \(100\%\). So, we write: \(20\%\) of \(28-x L+100\%\) of \(x L = 40\%\ of \ 28 L\)
\(0.2(28-x)+1(x)=0.4(28)\)
Solve for \(x: 5.6-0.2x+x=11.2\)
\(5.6+0.8x=11.2\)
\(0.8x=5.6\)
\(x=7\)
So, we need 7 L of pure acid solution and 21 L of \(20\%\) acid solution.

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