# Question #682c7

##### 1 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 **for every** **solution**.

Let's say that the starting solution has a volume

#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 **same amount of solute** present in the starting solution must now be equivalent to **for every**

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

If you take

#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 **times as much water** as the initial volume of the starting solution.