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Lilit [14]
3 years ago
13

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?
Mathematics
1 answer:
Alex777 [14]3 years ago
5 0
\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


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arsen [322]
There's two ways to do this.

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