Question #9a5c5

1 Answer
Nov 5, 2017

Answer:

#"10 g Ca"("HCO"_3)_2#

Explanation:

For starters, you know that we use parts per million, ppm, to express the concentration of a solution that contains very, very small amounts of solute.

A solution's concentration in ppm tells you the number of grams of solute present in exactly

#10^6color(white)(.)"g" = "1,000,000 g"#

of the solution. In your case, you know that hard water has a concentration of #"200 ppm"# in calcium hydrogen carbonate, #"Ca(HCO"_3)_2#. This means that for every #10^3# #"g"# of hard water, you get #"200 g"# of calcium hydrogen carbonate.

Now, when the problem doesn't provide you with a value for the density of water, you can assume that it's equal to #"1 g mL"^(-1)#.

Since

#"1 L" = 10^3# #"mL"#

you can say that your sample has a mass of

#52.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 52.0 * 10^3color(white)(.)"g"#

Since you know that hard water must contain #"200 g"# of calcium hydrogen carbonate for every #10^# #"g"# of the solution, you can say that your sample will contain

#52.0 * color(red)(cancel(color(black)(10^3))) color(red)(cancel(color(black)("g hard water"))) * overbrace(("200 g Ca"("HCO"_3)_2)/(10^3 * color(red)(cancel(color(black)(10^3))) color(red)(cancel(color(black)("g hard water")))))^(color(blue)("= 200 ppm Ca"("HCO"_3)_2)) = color(darkgreen)(ul(color(black)("10 g Ca"("HCO"_3)_2#

The answer is rounded to one significant figure, the number of sig figs that you have for the ppm concentration of hard water.