Question #682c7

1 Answer
Dec 3, 2016

Answer:

Here's how you could do that.

Explanation:

For starters, you can only dilute the original solution by adding water, the solvent. In any dilution, the amount of solute remains constant.

So, your starting solution is #83%"v/v"#, which means that you get #"83 mL"# of alcohol, the solute, for every #"100 mL"# of solution.

Let's say that the starting solution has a volume #Vcolor(white)(.)"mL"#. This solution will contain

#V color(red)(cancel(color(black)("mL solution"))) * "83 mL alcohol"/(100color(red)(cancel(color(black)("mL solution")))) = (83/100V)color(white)(.)"mL alcohol"#

The diluted solution must have a #"42% v/v"# concentration, which means that the same amount of solute present in the starting solution must now be equivalent to #"42 mL"# for every #"100 mL"# of diluted solution.

The volume of the diluted solution needed to account for this concentration is

#(83/color(blue)(cancel(color(black)(100)))V) color(red)(cancel(color(black)("mL alcohol"))) * (color(blue)(cancel(color(black)(100))) "mL solution")/(42color(red)(cancel(color(black)("mL alcohol")))) = (83/42V)color(white)(.)"mL solution"#

Now, your goal here is to figure out how much water must be added to the starting solution to increase its volume from #V# to #(83/42V)#.

If you take #x# to be the volume of water needed, you can say that

#Vcolor(white)(.)"mL" + x = (83/42V)color(white)(.)"mL"#

This gets you

#x = 83/42Vcolor(white)(.)"mL" - Vcolor(white)(.)"mL" = color(darkgreen)(ul(color(black)((41/42)Vcolor(white)(.)"mL")))#

Therefore, you can say that in order to dilute a solution from #83%"v/v"# to #42%"v/v"#, you must add approximately #41/42# times as much water as the initial volume of the starting solution.