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?

Question

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?

Naomi Aronson · Accepted Answer

#"46.4 mL"#

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")))#

Cooper Adams · Answer

undefined To find the volume of isopropanol in the 62.4 mL sample, multiply the volume of the sample by the percentage of isopropanol in the solution:

Volume of isopropanol = 62.4 mL * (74.3/100) = 46.3536 mL

So, there are approximately 46.35 mL of isopropanol in the 62.4 mL sample of the alcohol solution.

Naomi Aronson · Answer

undefined To find the volume of isopropanol in a 62.4 mL sample of the alcohol solution, you multiply the volume of the sample by the percentage of isopropanol in the solution:

$$ \text{Volume of isopropanol} = \text{Volume of sample} \times \text{Percentage of isopropanol} $$

$$ \text{Volume of isopropanol} = 62.4 \, \text{mL} \times 0.743 $$

$$ \text{Volume of isopropanol} = 46.354 \, \text{mL} $$

Therefore, there are 46.354 milliliters of isopropanol in a 62.4 mL sample of the alcohol solution.