# Question 23bd9

May 8, 2016

$333$

#### Explanation:

The dilution factor can be calculated by dividing the final volume of the diluted solution by the initial volume of the aliquot

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F." = V_"final"/V_"initial} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, notice that you're essentially performing two dilutions here, one of the initial $\text{3-mL}$ aliquot and one of a $\text{10-mL}$ aliquot.

The first aliquot is diluted from $\text{3 mL}$ to $\text{100 mL}$, which is equivalent to a dilution factor of

"D.F"_1 = (100 color(red)(cancel(color(black)("mL"))))/(3color(red)(cancel(color(black)("mL")))) = 33.3

Next, you take a $\text{10-mL}$ sample of this diluted solution and dilute it to $\text{100 mL}$. The dilution factor for the Second dilution will be

"D.F"_2 = (100 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 10

The overall dilution factor for the initial solution will be

${\text{D.F"_"overall" = "D.F"_1 xx "D.F}}_{2}$

In your case, you will have

$\text{D.F"_"overall} = 33.3 \times 10 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 333 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, your goal here was to find the dilution factor of the initial solution, i.e. of the $\text{3-mL}$ aliquot.

One way to think about this is by assuming that instead of diluting a $\text{10-mL}$ sample of the solution that resulted from the first dilution, you dilute the entire $\text{100 mL}$ by the same dilution factor you had for your second dilution.

So, you diluted $\text{10 mL}$ to $\text{100 mL}$, which is equivalent to a dilution factor of $10$. Using the same dilution factor for the entire $\text{100 mL}$ sample would get you

$\text{D.F" = V_"final"/V_"initial" implies V_"final" = "D.F." xx V_"initial}$

In your case, you will have

${V}_{\text{final" = 10 xx "100 mL" = "1000 mL}}$

So, if you dilute the $\text{3-mL}$ aliquot to a total volume of $\text{1000 mL}$, you'll once again get a dilution factor of

"D.F"_"overall" = (1000color(red)(cancel(color(black)("mL"))))/(3color(red)(cancel(color(black)("mL")))) = color(green)(|bar(ul(color(white)(a/a)333color(white)(a/a)|)))#