Question

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?

Answer 1

`"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.