Question #36d85

The idea here is that a solution's parts per million concentration, ppm, tells you the number of grams of solute present in exactly

#10^6 = 1,000,000#

parts of solution. Since the mass of the solvent is much, much bigger than the mass of the solute, which I'll assume is the chromium(III) cation, #"Cr"^(3+)#, you can assume that the mass of the solution is equal to the mass of the solvent, i.e. of water.

You know that #"1 L"# of water has a mass of #"2.5 kg"#, so you can say that your solution will contain

#2.5 color(red)(cancel(color(black)("L"))) * (1.00color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("L")))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2.5 * 10^3# #"g"#

of water, the solvent. This means that you assume that the mass of the solution is equal to #2.5 * 10^3# #"g"#.

Now, your solution is #"57 ppm"# chromium(III) cations, which means that you get #"57 g"# of chromium(III) cations for every #10^6# #"g"# of solution.

You can thus say that your solution will contain

#2.5 * 10^3 color(red)(cancel(color(black)("g solution"))) * overbrace("57 g Cr"^(3+)/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 57 ppm Cr"^(3+))) = color(darkgreen)(ul(color(black)("0.14 g Cr"^(3+))))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution and for its ppm concentration.