If you have 7 molecules of #SO_2# gas in #2.617 * 10^5# molecules of air, what is the concentration of #SO_2# in ppm?

1 Answer
Apr 29, 2016

Answer:

#"27 ppm"#

Explanation:

All you have to do here is use the definition of parts per million, or ppm.

You can use parts per million to express the concentration of a solution that contains very, very small amounts of solute. As its name suggests, this way of expressing concentration uses parts of solute per one million parts of solvent.

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

In order to have a #"1 ppm"# solution, you need to have a solution that contains one part of solute for every #10^6# parts of solvent.

Now, you can consider air to be your solution and sulfur dioxide, #"SO"_2#, to be your solute. Since you have significantly fewer molecules of sulfur dioxide than the total number of molecules of air, you can say that you'll have approximately #2.617 * 10^5# parts of solvent.

This means that the concentration of sulfur dioxide in ppm will be

#"ppm SO"_2 = (7 color(red)(cancel(color(black)("molecules"))))/(2.617 * 10^5color(red)(cancel(color(black)("molecules")))) xx 10^6 = color(green)(|bar(ul(color(white)(a/a)"27 ppm"color(white)(a/a)|)))#

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

So, this tells you that out of #10^6# molecules of air, #27# molecules will be molecules of sulfur dioxide.