Question #24c97

1 Answer
Stefan V.
Mar 26, 2017

"330 mL"

Explanation:

We use volume by volume percent concentration to show how many milliliters of solute are present in "100 mL" of a given solution.

In your case, the solution is 70.%"v/v" isopropyl alcohol, which means that every "100 mL" of this solution will contain "70. mL" of isopropyl alcohol, the solute.

So regardless of the volume of the solution, you can say for a fact that you will have a ratio of "70. mL" of isopropyl alcohol to "100 mL" of solution.

Now, you can use the solution's volume by volume percent concentration as a conversion factor to calculate the number of grams of isopropyl alcohol present in "473 mL" of solution.

473 color(red)(cancel(color(black)("mL solution"))) * overbrace("70. mL isopropyl alcohol"/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 70.% v/v")) = color(darkgreen)(ul(color(black)("330 mL isopropyl alcohol")))

The answer is rounded to two sig figs, the number of sig figs you have for the percent concentration of the solution.

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