Question

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?

Answer 1

`"0.30 M"`

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

`V_"final"` - the volume of the diluted solution
`V_"initial"` - the volume of the 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 `3.2` 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.