Question #cd6f3

For starters, let's assume that the volume of iron(III) chloride solution that must be added to #"1 L"# of water is small enough to allow for the approximation

#V_ "solution" = V_ "water" + V_ ("FeCl"_ 3) ~~ V_ "water"#

Judging by the number of significant figures you have for your values, the approximation will definitely hold.

Now, a #"1 ppm"# solution will contain #"1 g"# of solute for every #10^6# #"g"# of solution. Consequently, a #"800 ppm"# solution will contain #"800 g"# of iron(III) chloride for every #10^6# #"g"# of solution.

If you take water's density to be equal to #"1 g mL"^(-1)#, you can say that your solution will have a mass of

#1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^3# #"g"#

This means that your solution must contain

#10^3 color(red)(cancel(color(black)("g solution"))) * overbrace("800 g FeCl"_3/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 800 ppm")) = "0.800 g FeCl"_3#

Assuming that your iron(III) chloride solution is #40%# mass by volume, i.e. it contains #"40 g"# of iron(III) chloride for every #"100 mL"# of water, you can say that in order to get #"0.800 g"# of iron(III) chloride, you must take a volume of

#0.800 color(red)(cancel(color(black)("g FeCl"_3))) * overbrace("100 mL solution"/(40color(red)(cancel(color(black)("g FeCl"_3)))))^(color(blue)("= 40% m/v")) = color(darkgreen)(ul(color(black)("2 mL solution")))#

The answer is rounded to one significant figure.

As you can see, we have

#V_"800 ppm solution" = 10^3color(white)(.)"mL" + "2 mL" ~~ 10^3# #"mL"#

You can thus say that if you take #"2 mL"# of #"40% m/v"# iron(III) chloride solution and add it to #"1 L"# of water, you will end up with a solution that is #"800 ppm"# iron(II) chloride.