Question

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

Answer 1

`"0.0224 mmol CaCO"_3`

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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`.

Answer 2

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.