Question

A sample of 300.0 g of drinking water is found to contain 38 mg Pb. What this concentration in parts per million?

Answer 1

`"130 ppm Pb"`

Explanation:

We use parts per million to express the concentrations of solutions that contain very, very small amounts, often called trace amounts, of a given solute.

More specifically, a solution's concentration in parts per millions tells you the number of parts of solute present for every

`10^6 = 1,000,000`

parts of solution. You can thus say that a `"1 ppm"` solution will contain exactly `"1 g"` of solute for every `10^6"g"` of solution.

In your case, you know that you have

`38 color(red)(cancel(color(black)("mg Pb"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 3.8 * 10^(-2)color(white)(.)"g Pb"`

in exactly

`"300.0 g" = 3.000 * 10^2color(white)(.)"g solution"`

This means that you can use this known composition as a conversion factor to scale up the mass of the solution to `10^6"g"`

`10^6 color(red)(cancel(color(black)("g solution"))) * (3.8 * 10^(-2)color(white)(.)"g Pb")/(3.000 * 10^2color(red)(cancel(color(black)("g solution")))) = "130 g Pb"`

Since this represents the mass of lead present in exactly `10^6"g"` of solution, you can say that the solution has a concentration of

`color(darkgreen)(ul(color(black)("concentration"_ "ppm" = "130 ppm Pb")))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of lead present in the sample.