# Question #06777

##### 1 Answer

#### Answer:

#### 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

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

#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 **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.