Question #06777

1 Answer
Oct 12, 2017

#"3.8 mg"#

Explanation:

Your goal here is to figure out the mass of sodium carbonate that you need to dissolve in enough water to get a solution that is #160# #mu"mol L"^(-1)# with a volume of #"225 mL"#.

For starters, you should know that

#"1 mole" = 10^(6)color(white)(.)mu"moles" " "# and #" " "1 L" = 10^3color(white)(.)"mL"#

The first thing that you need to do here is to use the volume and the molarity of the solution to determine how many micromoles of sodium carbonate it must contain.

Since your solution must have a molarity of #160# #mu"mol L"^(-1)#, you can say that your sample will contain

#225 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (160color(white)(.)mu"moles Na"_2"CO"_3)/(1color(red)(cancel(color(black)("L")))) = 36color(white)(.)mu"moles Na"_2"CO"_3#

Next, convert the number of micromoles to moles

#36 color(red)(cancel(color(black)(mu"moles"))) * "1 mole"/(10^6color(red)(cancel(color(black)(mu"moles")))) = 3.6 * 10^(-5)color(white)(.)mu"moles"#

Next, to convert the number of moles to grams, use the molar mass of sodium carbonate, which tells you the mass of exactly #1# mole of this compound.

#3.6 * 10^(-5)color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * "105.99 g"/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3)))) = "0.003816 g"#

Finally, to convert the number of grams to milligrams, use the fact that

#"1 g" = 10^3 color(white)(.)"mg"#

You will end up with

#0.003816 color(red)(cancel(color(black)("g"))) * (10^3 color(white)(.)"mg")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("3.8 mg")))#

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.