# Question 8956f

Apr 28, 2016

$\text{0.600 M}$

#### Explanation:

The trick here is to realize that you can use the volume of the diluted solution and the volume of the stock solution to determine the dilution factor.

As you know, in order to dilute a solution, you must increase its volume while keeping the *number of moles of solute * constant.

If you start from the molarity of the solution

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can use the fact that the number of moles of solute remains unchanged to write

color(blue)(overbrace(c_1 xx V_1)^(color(green)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(green)("moles of solute in diluted solution"))

Here

${c}_{1}$, ${V}_{1}$ - the molarity and volume of the concentrated solution
${c}_{2}$, ${V}_{2}$ - the molarity and volume of the diluted solution

The dilution factor, which essentially tells you much concentrated the stock solution was compared with the dilute solution, will be equal to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{D.F.} = {c}_{1} / {c}_{2} = {V}_{2} / {V}_{1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you know that the volume of stock solution was $\text{250.0 mL}$, and tha the final volume of the dilute solution is $\text{2.50 L}$ . This means that the dilution factor is equal to -- do not forget to convert the volume from milliliters to liters or vice versa!

"D.F." = (2.50 color(red)(cancel(color(black)("L"))))/(250.0 * 10^(-3)color(red)(cancel(color(black)("L")))) = 10

This tells you that the stock solution was $10$ times as concentrated as the dilute solution.

This means that the concentration of the dilute solution is

$\text{D.F." = c_1/c_2 implies c_2 = c_1/"D.F.}$

c_2 = "6.00 M"/10 = color(green)(|bar(ul(color(white)(a/a)"0.600 M"color(white)(a/a)|)))#
So, by increasing the volume by a factor of $10$ you decreased the concentration of the stock solution by a factor of $10$, i.e. you performed a $1 : 10$ dilution.