# Question a9870

Sep 19, 2016

$\text{0.109 M}$

#### Explanation:

All you have to do here is keep track of the number of moles of solute as you progress towards the target solution.

For instance, the initial solution contains

51.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "1.30 moles solute"/(1color(red)(cancel(color(black)("L")))) = "0.0663 moles solute"

When you dilute this solution to a total volume of $\text{288 mL}$, the number of moles of solute remains constant. This means that after the first dilution, the solution will contain

$\text{0.0663 moles solute } \to$ in a total volume of $\text{288 mL}$

Next, you take $\text{144 mL}$ of this diluted solution and use it as a starting point for a second dilution. Notice that

$\frac{\text{144 mL" = "288 mL}}{2}$

This tells you that the $\text{144 mL}$ sample will contain half as many moles of solute as the $\text{288 mL}$ sample.

$\text{0.0663 moles solute"/2 = "0.03315 moles solute } \to$ in a volume of $\text{144 mL}$

When you're diluting this sample by adding $\text{161 mL}$ of water, the total volume of the resulting solution will be

$\text{144 mL " + " 161 mL" = "305 mL}$

This solution will contain the same number of moles of solute as you had in the $\text{144 mL}$ sample. This means that the final concentration, i.e. the concentration of the target solution, is equal to

"0.03315 moles solute"/(305 * 10^(-3)"L") = color(green)(bar(ul(|color(white)(a/a)color(black)("0.109 M")color(white)(a/a)|)))

The answer is rounded to three sig figs.

$\textcolor{w h i t e}{a}$
ALTERNATIVE APPROACH

You can solve this problem by using the dilution factor

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

Here

${V}_{\text{final}}$ - the volume of the diluted solution
${V}_{\text{initial}}$ - the volume of the concentrated solution

In essence, the dilution factor tells you how concentrated the initial solution was compared with the diluted solution.

So, for the first dilution, you have

"DF"_1 = (288 color(red)(cancel(color(black)("mL"))))/(51.0color(red)(cancel(color(black)("mL")))) = 5.647

This means that the $\text{1.30 M}$ solution was $5.647$ times more concentrated than the solution that resulted from the first dilution.

Now look at the second dilution -- this time, the starting volume is $\text{144 mL}$ and the final volume is $\text{305 mL}$

"DF"_2 = (305color(red)(cancel(color(black)("mL"))))/(144color(red)(cancel(color(black)("mL")))) = 2.118

This means that the sample we took from the $\text{288 mL}$ solution was $2.118$ times more concentrated than the target solution.

We've thus performed a serial dilution. The overall dilution factor, $\text{DF"_"serial}$, will be equal to the product of the two dilution factors

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {\text{DF"_ "serial" = "DF"_ 1 xx "DF}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will have

$\text{DF"_"serial} = 5.647 \cdot 2.118 = 11.96$

This means that the initial $\text{51.0 mL}$ solution was $11.96$ times more concentrated than the target solution, i.e. the solution that resulted from the second dilution.

Therefore, the concentration of the target solution is

${c}_{\text{target" = c_"initial"/"DF"_"serial}}$

c_"target" = "1.30 M"/11.96 = color(green)(bar(ul(|color(white)(a/a)color(black)("0.109 M")color(white)(a/a)|)))#