Question

Question #9a5c5

Answer 1

`"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.