# Question 24c97

Mar 26, 2017

$\text{330 mL}$

#### Explanation:

We use volume by volume percent concentration to show how many milliliters of solute are present in $\text{100 mL}$ of a given solution.

In your case, the solution is 70.%"v/v"# isopropyl alcohol, which means that every $\text{100 mL}$ of this solution will contain $\text{70. mL}$ 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 $\text{70. mL}$ of isopropyl alcohol to $\text{100 mL}$ 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 $\text{473 mL}$ of solution.

$473 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL solution"))) * overbrace("70. mL isopropyl alcohol"/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 70.% v/v")) = color(darkgreen)(ul(color(black)("330 mL isopropyl alcohol}}}}$

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