Question

Question #682c7

Answer 1

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.