Question #67370

1 Answer
Jul 21, 2016

Answer:

#"1 mM"#

Explanation:

The first thing to do here is use the molar mass of sodium carbonate to calculate the concentration of the #"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_"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 #"10 mL"# sample is equal to the concentration of the #"250 mL"# sample.

This means that you're going to dilute #"10 mL"# of #"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

#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(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_"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

#"1 mM" = "1 mmol L"^(-1) = 10^(-3)"mol L"^(-1) = 10^(-3)"M"#

The answer is rounded to one sig fig.