# 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?

Apr 29, 2016

$\text{27 ppm}$

#### 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.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{ppm" = "parts of solute"/"parts of solvent} \times {10}^{6} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In order to have a $\text{1 ppm}$ 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, ${\text{SO}}_{2}$, 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 $2.617 \cdot {10}^{5}$ 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.