# What volume (in L) of a 3.95 M potassium chloride, #KCl#, solution would be needed to make 325 mL of a 2.76 M solution by dilution?

##### 1 Answer

#### Explanation:

One possible strategy to use here is to find the **dilution factor** first, then use it to find the volume of the stock solution that is needed in order to make your target solution.

The *dilution factor* essentially tells you **how many times more concentrated** the *stock solution* was compared with the target solution.

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

In your case, the dilution factor is equal to

#"D.F." = (3.95 color(red)(cancel(color(black)("M"))))/(2.76color(red)(cancel(color(black)("M")))) = 1.43#

Now, because a dilution implies that the number of moles of solute **remains constant**, the dilution factor also tells you the ratio that exists between the volume of the **target solution**, i.e. the *diluted solution*, and the volume of the **stock solution**, i. e. the *concentrated solution*.

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

In your case, the volume of the stock solution needed to make the target solution would be

#V_"stock" = V_"target"/"D.F."#

#V_"stock" = "325 mL"/1.43 = "227 mL"#

Expressed in *liters*, the answer will be

#V_"stock" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|))) -># rounded to three sig figs

So, in order to make

Notice that the concentration of the solution **decreases** by the same ratio by which the volume **increases**. This is what the dilution factor is useful for.