# Question c541c

Oct 25, 2016

Here's how you can do that.

#### Explanation:

The first thing to do here is to figure out the dilution factor used to make your target solution.

The dilution factor essentially tells you the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

IIn other words, the dilution factor is a measure of how concentrated the initial solution was compared to the diluted solution.

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{DF" = c_"initial"/c_"final" = c_"concentrated"/c_"diluted}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the dilution factor is

"DF" = (2.200 color(red)(cancel(color(black)("M"))))/(0.0300color(red)(cancel(color(black)("M")))) = 73.33#

Now, a dilution is performed by increasing the volume of the solution while keeping the number of moles of solute constant.

This essentially means that the dilution factor can also be calculated by looking at the volume of the diluted solution and the volume of the concentrated solution

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{DF" = V_"final"/c_"initial" = V_"diluted"/V_"concentrated}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You know that the diluted solution was $73.33$ times less concentrated than the concentrated solution and that its volume must be equal to $\text{2.500 L}$, so rearrange the above equation and solve for the volume of the concentrated solution

$\text{DF" = V_"diluted"/V_"concentrated" implies V_"concentrated" = V_"diluted"/"DF}$

Plug in your values to find

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{V}_{\text{concentrated" = "2.500 L"/73.33 = "0.03409 L" = "34.1 mL}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the concentration of the diluted solution.

So, what does this all mean?

In a nutshell, this means that $\text{34.1 mL}$ of a $\text{2.200 M}$ acetic acid solution contains the same number of moles of solute as $\text{2.500 L}$ of $\text{0.0300 M}$.

So, to make this solution, measure $\text{34.1 mL}$ of the concentrated acetic acid solution and add enough water to get the total volume of the resulting solution to $\text{2.500 L}$.