Question #f3a0d

Your target solution must be #2%"v/v"# isopropyl alcohol, which means that it must contain #"2 mL"# of isopropyl alcohol, the solute, for every #"100 mL"# of solution.

Now, your starting solution is #5%"v/v"# isopropyl alcohol, which implies that it contains #"5 mL"# of isopropyl alcohol for every #"100 mL"# of solution.

In order to reduce the concentration of the solution from #5%# to #2%#, you must dilute the sample.

For any dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution will give you the dilution factor, #"DF"#.

In this case, you have

#"DF" = (5 color(red)(cancel(color(black)(%))))/(2color(red)(cancel(color(black)(%)))) = color(blue)(2.5)#

The trick here is to realize that the dilution factor is also equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

#"DF" = V_"diluted"/V_"stock"#

This means that

#V_"diluted" = "DF" * V_"stock"#

In your case, you have

#V_"diluted" = color(blue)(2.5) * "10 mL" = "25 mL"#

So if the volume of the diluted solution is equal to #"25 mL"#, it follows that it must contain

#"25 mL " - " 10 mL" = "15 mL"#

of water. In other words, if you add #"15 mL"# of water to #"10 mL"# of #5%"v/v"# isopropyl solution, you will end up with #"25 mL"# of #2%"v/v"# isopropyl solution.

I'll leave the answer rounded to two sig figs, but keep in mind that your values justify an answer rounded to one significant figure.