# You have 67.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?

##### 1 Answer

#### Answer:

#### Explanation:

The thing to remember about **dilutions** is that the ratio that exists between the **concentration** of the concentrated solution and that of the diluted solution is **equal** to the ratio that exists between the **volume** of the diluted solution and that of the concentrated solution.

This ratio is called the **dilution factor**. You thus have

#"DF" = "concentration of the concentrated solution"/"concentration of the diluted solution"#

and

#"DF" = "volume of the diluted solution"/"volume of the concentrated solution"#

In your case, the solution must be diluted from an initial concentration of **dilution factor** equal to

#"DF" = (0.400 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = color(blue)(4)#

You can now say that the volume of the *diluted solution* must be **times greater** than the volume of the *concentrated solution*, since

#"DF" = "volume of diluted solution"/"67.0 mL"#

#"volume of diluted solution" = "DF" xx "67.0 mL"#

and therefore

#"volume of diluted solution" = color(blue)(4) xx "67.0 mL" = "268 mL"#

Assuming that the volumes are *addictive*, you will have to add

#"volume of water added" = "268 mL" - "67.0 mL" = color(darkgreen)(ul(color(black)("201 mL")))#

of water to your concentrated solution in order to go from

The answer is rounded to three **sig figs**.