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

##### 2 Answers

A **part-per-million**, which is a concentration. We can define that in multiple ways, but one way is

Since we have a solution, let's suppose we are in a solution of water, which can dissolve small amounts of

Water can be assumed for your purposes to have a 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

#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

By the way, this concentration of

Here's what I got.

#### Explanation:

For starters, you can use the fact that a **for every**

#"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 **for every**

If you take water's density to be equal to

#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 **for every**

#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

#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

The answers are rounded to three **sig figs**.