# Question #ba6e5

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to realize that when you *dilute* a solution, you keep the amount of solute **constant** and **increase** the amount of solvent.

Now, you don't actually need to know *how much* solute you have in your initial solution, all you need to know is the **ratio** that exists between the volume of the diluted solution and the volume of the concentrated solution.

This ratio will give you the **dilution factor**, which can be used to determine the ratio that must exist between the **concentration** of the concentrated solution and the concentration of the diluted solution.

So, you know that you have

#(750 color(red)(cancel(color(black)("mL"))))/(325color(red)(cancel(color(black)("mL")))) = color(blue)(2.3077) -># thedilution factor

So, you know that the volume of the solution **increases** by a factor of **decrease** by a factor of

This is the case because when we're diluting a solution, we're only increasing the volume of the **solvent**, meaning that the same amount of solute is now distributed in a **larger volume** of solution.

You can thus say that

#"% concentrated"/"% diluted" = color(blue)(2.3077)#

which will get you

#"% diluted" = "% concentrated"/color(blue)(2.3077)#

Plug in your value to find

#"% diluted" = "17.5% w/v"/color(blue)(2.3077) = color(darkgreen)(ul(color(black)("7.6% w/v")))#

The answer must be rounded to two **sig figs**, the number of sig figs you have for the volume of the diluted solution.