Consider a 0.00050 molar solution of methanol, #"CH"_3"OH"#, in water. Express the concentration of methanol in parts per million?

Consider a 0.00050 molar solution of methanol (CH3OH) in water.
Express the concentration of methanol in parts per million.

1 Answer
May 18, 2016


#"16.2 ppm"#


For solutions that contain very, very small amounts of solute, usually called trace amounts, you can use Parts Per Million, or ppm, to express their concentration.

In essence, ppm denotes one part solute for every #10^6# parts solvent. This means that you can find a solution's concentration in ppm by using

#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "grams of solute"/"grams of solvent" xx 10^6color(white)(a/a)|)))#

Now, your methanol solution is said to have a molarity of #"0.00050 mol L"^(-1)#. This means that you get #0.00050# moles of solute per liter of solution.

To keep the calculations simple, let's assume that you have a #"1-L"# sample of this solution.

The first thing to do here is use methanol's molar mass to convert the number of moles to grams

#0.00050 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "32.04 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = "0.0162 g CH"_3"OH"#

Because the solution contains such a small amount of methanol, you can assume that its density will be equal to that of pure water. Since no information about water's density was given, you can assume it to be equal to #"1 g mL"^(-1)#.

This means that the sample will have a mass of

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

As you can see, you can safely assume that the mass of the solvent is approximately equal to that of the solution.

This means that the concentration in ppm will be equal to

#"ppm" = (0.0162color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("g")))) * 10^6 = color(green)(|bar(ul(color(white)(a/a)"16.2 ppm"color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.