A solution of rubbing alcohol is 74.3% (v/v) isopropanol in water. How many mililiters of Isopropanol are in 62.4 mL sample of the alcohol solution?

The trick here is to use the known volume by volume percent concentration, #"v/v%"#, to figure out how many milliliters of isopropanol, the solute, are present in your sample.

Now, the thing to remember about volume by volume percent concentrations is that they are meant to represent the number of milliliters of solute present for every #"100 mL"# of solution.

In your case, the solution contains

#color(blue)(74.3)color(red)(%)"v/v isopropanol" = color(blue)("74.3 mL")color(white)(.)"isopropanol in" color(white)(.)color(red)("100 mL")color(white)(.)"solution"#

As you know, solutions are homogeneous mixture, which implies that they have the same composition throughout.

This means that you can use solution's volume by volume percent concentration to calculate the number of milliliters of isopropanol present in #"62.4 mL"# of solution.

Set up the equation as

#(?color(white)(.)"mL solute")/"62.4 mL solution" = (color(blue)("74.3 mL")color(white)(.)"solute")/(color(red)("100 mL")color(white)(.)"solution")#

Rearrange and solve to find

#? = (62.4 color(red)(cancel(color(black)("mL solution"))))/(100color(red)(cancel(color(black)("mL solution")))) * "74.3 mL solute"#

#? = "46.36 mL solute"#

Rounded to three sig figs, the answer will be

#color(darkgreen)(ul(color(black)("volume isopropanol = 46.4 mL")))#