# Question 67370

##### 1 Answer
Jul 21, 2016

$\text{1 mM}$

#### Explanation:

The first thing to do here is use the molar mass of sodium carbonate to calculate the concentration of the $\text{250 mL}$ solution.

In your case, you know that the initial solution contains

2.65 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"CO"_3)/(106.0 color(red)(cancel(color(black)("g")))) = "0.0250 moles Na"_2"CO"_3

Since you know the volume of this solution, use it to calculate its molarity.

color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution" [color(blue)("in liters")]))color(white)(a/a)|)))

In your case, you have

${c}_{\text{initial" = "0.0250 moles"/(250 * 10^(-3)color(blue)("L")) = "0.10 M}}$

Now, the important thing to keep in mind here is that the concentration of the $\text{10 mL}$ sample is equal to the concentration of the $\text{250 mL}$ sample.

This means that you're going to dilute $\text{10 mL}$ of $\text{0.1 M}$ solution to a total volume of

1 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "1000 mL"

As you know, a dilution is aimed at decreasing the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

You can use this principle to calculate the dilution factor

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

In your case, the sample is being diluted by a factor of

"DF" = (1000 color(red)(cancel(color(black)("mL"))))/(10color(red)(cancel(color(black)("mL")))) = 100

This means that the concentration of the concentrated solution is $100$ times higher than the concentration of the diluted solution, and so

${c}_{\text{diluted" = 1/100 * c_"concentrated}}$

This gets you

c_"diluted" = 1/100 * "0.1 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("1 mM")color(white)(a/a)|)))#

Here

$\text{1 mM" = "1 mmol L"^(-1) = 10^(-3)"mol L"^(-1) = 10^(-3)"M}$

The answer is rounded to one sig fig.