Question #50ff0
1 Answer
Explanation:
You can use Parts Per Million, or ppm, to express the concentration of solutions that contain very, very small amounts, often called trace amounts, of solute.
More specifically, a concentration of
#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "grams of solute"/"grams of solvent" xx 10^6color(white)(a/a)|)))#
This basically tells you that a concentration of
Since you can safely assume that the mass of the solvent is equal to that of the solution, you can say that a con concentration of
Now, the problem tells you this solution contains
#0.050 color(red)(cancel(color(black)("mg solute"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg solute")))) = 5.0 * 10^(-5)"g"#
Now all you have to do is use the
#5.0 * 10^(-5)color(red)(cancel(color(black)("g solute"))) * overbrace((10^6"g solution")/(6color(red)(cancel(color(black)("g solute")))))^(color(purple)("= 6 ppm")) = color(green)(|bar(ul(color(white)(a/a)"8.3 g solution"color(white)(a/a)|)))#
I'll leave the answer rounded to two sig figs.
One interesting thing to notice here is that you can write a concentration of
#"1 g solute"/(10^6"g solution") = (1 * color(red)(cancel(color(black)(10^(3))))"mg solute")/(1 * color(red)(cancel(color(black)(10^(6)))) * color(red)(cancel(color(black)(10^(-3))))"kg solution") = "1 mg solute"/"1 kg solution"#
So, if
Therefore,
#0.050 color(red)(cancel(color(black)("mg solute"))) * overbrace("1 kg solution"/(6color(red)(cancel(color(black)("mg solute")))))^(color(purple)("= 6 ppm")) = "0.0083 kg solution"#
This will once again be
#0.0083 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = color(green)(|bar(ul(color(white)(a/a)"8.3 g solution"color(white)(a/a)|)))#