# Question 682c7

Dec 3, 2016

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 $\text{83 mL}$ of alcohol, the solute, for every $\text{100 mL}$ of solution.

Let's say that the starting solution has a volume $V \textcolor{w h i t e}{.} \text{mL}$. 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 $\text{42% v/v}$ concentration, which means that the same amount of solute present in the starting solution must now be equivalent to $\text{42 mL}$ for every $\text{100 mL}$ 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 $\left(\frac{83}{42} V\right)$.

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

$V \textcolor{w h i t e}{.} \text{mL" + x = (83/42V)color(white)(.)"mL}$

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 $\frac{41}{42}$ times as much water as the initial volume of the starting solution.