# Question #21b37

##### 1 Answer

Here's how you can do that.

#### Explanation:

**Parts per million** is a measure of the number of parts of solute present **for every** **parts** of solution.

So every time you have a

In your case, you have the volume of the solution. If you're dealing with an *aqueous solution*, i.e. you have water as the solvent, then you can use its density to find the total mass of the solution.

Let's say that you have

Assuming a density of

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

Now, this solution contains **for every**

#10^3 color(red)(cancel(color(black)("g solution"))) * overbrace("15 g solute"/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 15 ppm")) = 15 * 10^(-3)# #"g solute"#

You can thus say that if you have

#15 * 10^(-3) color(red)(cancel(color(black)("g"))) * "1 mg"/(10^(-3)color(red)(cancel(color(black)("g")))) = "15 mg"#

of solute in