Question

7 L of an acid solution was mixed with 3 L of a 15% acid solution to make a 29% acid solution. What is the percent concentration of the first solution?

Answer 1

well, let's say that the first solution is 7 liters and has "x" percent of acid, how much acid total in it? well, 7x, keeping in mind that "x" is a decimal form.

likewise, the second solution is 3 liters and has 15% acid, so how much acid is there in that one? (15/100) * 3 = 0.45, so

`\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&7&x&7x\\ \textit{2nd solution}&3&0.15&0.45\\ \cline{2-4}&\\ mixture&10&0.29&2.9 \end{array}~\hfill \implies 7x~~ + ~~0.45~~ = ~~2.9 \\\\[-0.35em] ~\dotfill\\\\ 7x=2.45\implies x=\cfrac{2.45}{7}\implies x=0.35~\hfill \stackrel{\textit{converting to \%}}{0.35\cdot 100}\implies \stackrel{\%}{35}`

Answer 2

35%

Explanation:

Let c represent the concentration of acid in the unknown solution. Then total amount of acid in the mix is ...

7c +3(0.15) = (7+3)(0.29)

7c +0.45 = 2.9 . . . . . . . . . eliminate parentheses

7c = 2.45 . . . . . . . . . . . . subtract 0.45

c = 0.35 = 35% . . . . . . divide by 7

The concentration of the first solution is 35%.