# 58.0 mL of a 1.30 M solution is diluted to a total volume 248 mL. A 124-mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive

Jul 8, 2016

$\text{0.130 M}$

#### Explanation:

You can solve this problem by using the dilution factor, $\text{DF}$, which essentially tells you the factor by which the concentration of an initial solution decreased upon dilution.

The dilution factor is calculated by using either volumes or molarities

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

So, the dilution factor for the first dilution

$\text{58.0 mL " -> " 248 mL}$

will be

"DF"_1 = (248 color(red)(cancel(color(black)("mL"))))/(58.0color(red)(cancel(color(black)("mL")))) = 4.276

The first solution was diluted by a factor of $4.276$, which means that its concentration decreased by a factor of $4.276$. Therefore, the concentration of the diluted solution will be

${c}_{\text{diluted" = c_"concentrated"/"DF}} _ 1$

${c}_{\text{diluted" = "1.30 M"/4.276 = "0.304 M}}$

Now, you take a sample of $\text{124 mL}$ of this diluted solution and add another $\text{165 mL}$ of water. The final volume of the solution will be

${V}_{\text{final" = "124 mL" + "165 mL" = "289 mL}}$

The concentration of the $\text{124 mL}$ sample is equal to the concentration of the first diluted solution, i.e. $\text{0.304 M}$.

The dilution factor for the second dilution will be

$\text{DF"_ 2 = V_"final"/V_"sample}$

"DF"_ 2 = (289 color(red)(cancel(color(black)("mL"))))/(124color(red)(cancel(color(black)("mL")))) = 2.331

Therefore, the concentration of the final solution will be

${c}_{\text{final" = c_"sample"/"DF}} _ 2$

c_"final" = "0.304 M"/2.331 = color(green)(|bar(ul(color(white)(a/a)color(black)("0.130 M")color(white)(a/a)|)))

The answer is rounded to three sig figs.