Question 9a5c5

Nov 5, 2017

"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

${10}^{6} \textcolor{w h i t e}{.} \text{g" = "1,000,000 g}$

of the solution. In your case, you know that hard water has a concentration of $\text{200 ppm}$ in calcium hydrogen carbonate, "Ca(HCO"_3)_2. This means that for every ${10}^{3}$ $\text{g}$ of hard water, you get $\text{200 g}$ 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 ${\text{1 g mL}}^{- 1}$.

Since

$\text{1 L} = {10}^{3}$ $\text{mL}$

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 $\text{200 g}$ of calcium hydrogen carbonate for every 10^ $\text{g}$ 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.