Question

Question #21b37

Answer 1

Here's how you can do that.

Explanation:

Parts per million is a measure of the number of parts of solute present for every `10^6` parts of solution.

So every time you have a `"1 ppm"` solution, you essentially have a solution that contains `"1 g"` of solute for every `10^6` `"g"` of solution.

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 `"1 L"` of a solution that is said to have a concentration of `"15 ppm"`.

Assuming a density of `"1 g mL"^(-1)` for water, you can say that the mass of the solution, which you can easily approximate as being equal to the mass of the solvent, will be

`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 `"15 g"` of solute for every `10^6` `"g"` of solution, which means that `10^3` `"g"` of solution will contain

`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 `"1 L"` of solution, you have a `"15 ppm"` solution.