Question

You need a 75% alcohol solution. On hand, you have a 1125 mL of a 90% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

Answer 1

`\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}\\ &-----&-------&-------\\ \textit{90\% sol'n}&1125&0.9&(1125)0.9\\ \textit{water}&x&0&0x\\ -----&-----&-------&-------\\ mixture&y&0.75&0.75y \end{array}`

so.. whatever "x" is, we know 1125 + x = y
and whatever the alcohol concentration yield may be, we know

(1125)(0.9) = 0.75y

(1125)(0.9) = 0.75y <--- solve for "y", that's how much the final amount of milliliters will be

how much pure water was added? subtract 1125 from that, or y - 1125