Question #50ff0

1 Answer
May 4, 2016

Answer:

#"8.3 g"#

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 #"1 ppm"# is equivalent to one part solute for every #10^6# parts of solvent. One way to express ppm concentration is to use the mass of solute in grams and the mass of solvent in grams

#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 #"1 ppm"# will contain #"1 g"# of solute for every #10^6"g"# of solvent.

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 #"6 ppm"# will contain #"6 g"# of solute in #10^6"g"# of solution.

Now, the problem tells you this solution contains #"0.050 mg"# of solute, which is equivalent to

#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 #"6 ppm"# concentration as a conversion factor to determine how many grams of solution would contain this many grams of solvent

#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 ppm"# as

#"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 #"1 ppm"# is equivalent to #"1 mg"# of solute in #"1 kg"# of solution, you can say that #"6 ppm"# will be equivalent to #"6 mg"# of solute in #"1 kg"# of solution.

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)|)))#