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

Nov 12, 2016

${\text{0.0224 mmol CaCO}}_{3}$

A $\text{ppm}$ 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 $\text{1 mg" = 10^(-3) "g}$ and $\text{1 kg} = {10}^{3}$ $\text{g}$. You should notice that $\left({10}^{- 3} \text{g")/(10^(3) "g}\right) = {10}^{- 6}$, 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 ${\text{CaCO}}_{3}$ (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 $\text{1 g/mL}$, 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:

$355 \cancel{\text{mL" xx "1 g"/cancel"mL" = "355 g" = "0.355 kg solution}}$

So, we currently have:

$\left(\text{6.31 mg CaCO"_3 " solute")/cancel"kg solution" xx 0.355 cancel("kg solution}\right)$

$= {\text{2.24 mg CaCO}}_{3}$ $\text{solute}$

Therefore, the $\text{mmol}$ of ${\text{CaCO}}_{3}$ can be calculated from its molar mass of $40.08 + 12.011 + 3 \times 15.999 = \text{100.008 g/mol}$:

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

$= \textcolor{b l u e}{{\text{0.0224 mmol CaCO}}_{3}}$

since the "milli" carries through the calculation from $\text{mg}$ to $\text{mmol}$.

By the way, this concentration of ${\text{6.31 ppm CaCO}}_{3}$ would be equal to about $6.31 \times {10}^{- 5} \text{M}$, or molarity concentration. You might want to figure out the one-step calculation on how to get to that concentration after having found the ${\text{mmol CaCO}}_{3}$.

Nov 12, 2016

Here's what I got.

#### Explanation:

For starters, you can use the fact that a $\text{1 ppm}$ solution contains $\text{1 g}$ of solute for every ${10}^{6} \text{g}$ of solvent to figure out how much solute you would get for $\text{1 mg}$ 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 $\text{6.31 ppm}$ calcium carbonate solution will contain $\text{6.31 mg}$ of solute for every ${10}^{6} \text{mg}$ of solvent.

If you take water's density to be equal to ${\text{1.0 g mL}}^{- 1}$, 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 $\text{6.31 mg}$ of solute for every ${10}^{6} \text{mg}$ 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 ${\text{100.09 g mol}}^{- 1}$

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"))))

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2.24 \cdot {10}^{- 2} {\text{mmol CaCO}}_{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $5.94 \cdot {10}^{- 2} {\text{mmol L}}^{- 1}$.

The answers are rounded to three sig figs.