Convert 355mL of a solution containing 6.31 ppm CaCO3 to mmol?

2 Answers
Nov 12, 2016

"0.0224 mmol CaCO"_3


A "ppm" is a part-per-million, which is a concentration. We can define that in multiple ways, but one way is "1 ppm" = ("mg solute")/("kg solution"), where "1 mg" = 10^(-3) "g" and "1 kg" = 10^(3) "g". You should notice that (10^(-3) "g")/(10^(3) "g") = 10^(-6), hence "parts per million".

Since we have a solution, let's suppose we are in a solution of water, which can dissolve small amounts of "CaCO"_3 (normally "insoluble" in water, but at the ppm scale it is considered soluble).

Water can be assumed for your purposes to have a density of "1 g/mL", and for this low of a concentration, we can assume that the density of water is equal to the density of the solution.

So, we can convert the volume of the solution to a mass:

355 cancel"mL" xx "1 g"/cancel"mL" = "355 g" = "0.355 kg solution"

So, we currently have:

("6.31 mg CaCO"_3 " solute")/cancel"kg solution" xx 0.355 cancel("kg solution")

= "2.24 mg CaCO"_3 "solute"

Therefore, the "mmol" of "CaCO"_3 can be calculated from its molar mass of 40.08 + 12.011 + 3 xx 15.999 = "100.008 g/mol":

2.24 "m"cancel("g CaCO"_3) xx "1 mol"/(100.008 cancel("g CaCO"_3))

= color(blue)("0.0224 mmol CaCO"_3)

since the "milli" carries through the calculation from "mg" to "mmol".

By the way, this concentration of "6.31 ppm CaCO"_3 would be equal to about 6.31 xx 10^(-5) "M", or molarity concentration. You might want to figure out the one-step calculation on how to get to that concentration after having found the "mmol CaCO"_3.

Nov 12, 2016

Here's what I got.

Explanation:

For starters, you can use the fact that a "1 ppm" solution contains "1 g" of solute for every 10^6"g" of solvent to figure out how much solute you would get for "1 mg" of solvent.

"1 ppm" = "1 g solute"/(10^6 color(red)(cancel(color(black)("g")))"solvent") * (1 color(red)(cancel(color(black)("g"))))/(10^3"mg") = "1 g solute"/(10^9"mg solvent")

Next, convert this to milligrams of solute per milligrams of solvent

(1 color(red)(cancel(color(black)("g")))"solute")/(10^9"mg solvent") * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) = (10^3"mg solute")/(10^9"mg solvent") = "1 mg solute"/(10^6"mg solvent")

Now, a "6.31 ppm" calcium carbonate solution will contain "6.31 mg" of solute for every 10^6"mg" of solvent.

If you take water's density to be equal to "1.0 g mL"^(-1), you can say that the mass of the solvent, which can easily be approximated to be equal to the mass of the solution, is

355 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("mL")))) * (10^3"mg")/(1color(red)(cancel(color(black)("g")))) = 3.55 * 10^5"mg"

So, if you get "6.31 mg" of solute for every 10^6"mg" of solvent, it follows that this solution will contain

3.55 * 10^5 color(red)(cancel(color(black)("mg solvent"))) * "6.31 mg solue"/(10^6color(red)(cancel(color(black)("mg solvent")))) = "2.240 mg solute"

Now, to convert this to millimoles of calcium carbonate, use the fact that calcium carbonate has a molar mass of "100.09 g mol"^(-1)

2.240 color(red)(cancel(color(black)("mg"))) * (1 color(red)(cancel(color(black)("g"))))/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("mg")))) * (1 color(red)(cancel(color(black)("mole CaCO"_3))))/(100.09 color(red)(cancel(color(black)("g")))) * (color(blue)(cancel(color(black)(10^3)))"mmol")/(1color(red)(cancel(color(black)("mole"))))

= color(green)(bar(ul(|color(white)(a/a)color(black)(2.24 * 10^(-2)"mmol CaCO"_3)color(white)(a/a)|)))

To convert this to mmol per liter, calculate the number of mmoles of calcium carbonate you have in one liter of solution

1 color(red)(cancel(color(black)("L"))) * (2.24 * 10^(-2)"mmol CaCO"_3)/(355 * 10^(-3)color(red)(cancel(color(black)("L")))) = 6.31 * 10^(-2)"mmol CaCO"_3

This means that the solution has a molarity of 5.94 * 10^(-2)"mmol L"^(-1).

The answers are rounded to three sig figs.