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

##### 1 Answer

#### 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**

**parts of solution**. You can thus say that a **for every**

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 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

#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.