Question

Question #24c97

Answer 1

`"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.