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

Mar 15, 2017

$\text{130 ppm Pb}$

#### 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 $\text{1 ppm}$ solution will contain exactly $\text{1 g}$ of solute for every ${10}^{6} \text{g}$ 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

$\text{300.0 g" = 3.000 * 10^2color(white)(.)"g solution}$

This means that you can use this known composition as a conversion factor to scale up the mass of the solution to ${10}^{6} \text{g}$

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 ${10}^{6} \text{g}$ of solution, you can say that the solution has a concentration of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{concentration"_ "ppm" = "130 ppm Pb}}}}$

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