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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