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

#"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#.

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.