# If a student needs to make exactly 2.5 liters of a 1.25 M solution of acetic acid from the 12.0 M stock solution in the chemistry closet, what volume of the stock solution should be used?

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you can use the **dilution factor** to figure out the volume of *stock solution* needed to make *diluted solution*.

As you know, you can **decrease** the concentration of a solution by **increasing** its volume while keeping the number of moles of solute **constant**, i.e. by *diluting* it.

The important thing to remember here is that if you increase the volume of a solution by a specific factor, the concentration **must decrease** by the same factor.

This factor is called the *dilution factor*

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = V_"diluted"/V_"stock" = c_"stock"/c_"diluted"color(white)(a/a)|)))#

In your case, the concentration of the solution must go from

The dilution factor will thus be

#"D.F." = (12.0 color(red)(cancel(color(black)("M"))))/(1.25color(red)(cancel(color(black)("M")))) = 9.6#

This means that the volume of the stock solution used for the dilution was

#"D.F." = V_"diluted"/V_"stock" implies V_"stock" = V_"diluted"/"D.F."#

#V_"stock" = "2.5 L"/9.6 = "0.26 L"#

Expressed in *milliliters* and rounded to two **sig figs**, the answer will be

#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("260 mL")color(white)(a/a)|)))#