# Question 11820

Nov 5, 2017

Here's my take on this.

#### Explanation:

Assuming that the question is not incomplete, your goal here is to figure out the dilution factor, $\text{DF}$, for the second dilution, and the overall dilution factor for the two dilutions.

For starters, you know that the dilution factor tells you the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

$\text{DF" = V_"diluted"/V_"stock}$

Now, you know that you take $\text{25.00 mL}$ of the stock solution and dilute it to a total volume of $\text{250.00 mL}$ in order to get solution $\text{A}$.

This means that the dilution factor for the first dilution is equal to

"DF"_1 = (250.00 color(red)(cancel(color(black)("mL"))))/(25.00color(red)(cancel(color(black)("mL")))) = color(blue)(10)

Next, you take a $\text{25.00-mL}$ sample of this solution $\text{A}$ and, presumably, dilute it to $\text{100.00 mL}$. This means that the dilution factor for the second dilution is equal to

"DF"_2 = (100.00 color(red)(cancel(color(black)("mL"))))/(25.00color(red)(cancel(color(black)("mL")))) = color(purple)(4)

Now, if you want to figure out the overall dilution factor, you need to multiply the two dilution factors.

${\text{DF"_"overall" = "DF"_1 xx "DF}}_{2}$

$\text{DF"_"overall} = \textcolor{b l u e}{10} \times \textcolor{p u r p \le}{4} = \textcolor{g r e e n}{40}$

What this means is that the initial solution was diluted by a factor of $\textcolor{g r e e n}{40}$. In other words, the solution that you get by diluting solution $\text{A}$ will be $40$ times less concentrated than the solution that you diluted to get the solution $\text{A}$.

This is the case because the dilution factor also tells you the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution.

$\text{DF" = c_"stock"/c_"diluted}$

So if you start with a sulfuric acid solution of concentration ${c}_{1}$, you will have

${c}_{\text{A}} = {c}_{1} / \textcolor{b l u e}{10}$

This tells you that solution $\text{A}$ will be $\textcolor{b l u e}{10}$ timess less concentrated that the initial solution. Next, you have

${c}_{\text{final" = c_"A}} / \textcolor{p u r p \le}{4}$

But since

${c}_{\text{A}} = {c}_{1} / \textcolor{b l u e}{10}$

you can say that you have

${c}_{\text{final}} = \frac{{c}_{1} / \textcolor{b l u e}{10}}{\textcolor{p u r p \le}{4}} = {c}_{1} / \left(\textcolor{b l u e}{10} \cdot \textcolor{p u r p \le}{4}\right) = {c}_{1} / \textcolor{g r e e n}{40}$

This tells you that the final solution will be color(green)(40# times less concentrated than the initial solution.