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