# If 250.0 mL of a 0.96 M solution of acetic acid (#C_2H_4O_2#) are diluted to 800.0 mL, what will be the approximate molarity of the final solution?

##### 1 Answer

#### Explanation:

One way to approach this problem would be to use the volume of the initial solution and the volume of the target solution to find the **dilution factor**.

This will then allow you to find the **molarity** of the diluted solution.

So, the *dilution factor*, which tells you how **concentrated** the starting solution is compared with the dilution solution, is calculated like this

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

Here you have

**diluted solution**

**starting solution**

In your case, you have

#"D.F." = (800.0 color(red)(cancel(color(black)("mL"))))/(250.0color(red)(cancel(color(black)("mL")))) = 3.2#

This means that the initial solution was **times more concentrated** than the diluted solution. You thus have

#color(blue)(|bar(ul(color(white)(a/a)color(black)(c_"concentrated" = "D.F." xx c_"diluted")color(white)(a/a)|)))#

which gets you

#c_"diluted" = "0.96 M"/3.2 = color(green)(|bar(ul(color(white)(a/a)color(black)("0.30 M")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the molarity of the concentrated solution.