# Question cd6f3

Apr 8, 2017

$\text{2 mL}$

#### Explanation:

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

${V}_{\text{solution" = V_ "water" + V_ ("FeCl"_ 3) ~~ V_ "water}}$

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

Now, a $\text{1 ppm}$ solution will contain $\text{1 g}$ of solute for every ${10}^{6}$ $\text{g}$ of solution. Consequently, a $\text{800 ppm}$ solution will contain $\text{800 g}$ of iron(III) chloride for every ${10}^{6}$ $\text{g}$ of solution.

If you take water's density to be equal to ${\text{1 g mL}}^{- 1}$, 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 $\text{g}$

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 $\text{40 g}$ of iron(III) chloride for every $\text{100 mL}$ of water, you can say that in order to get $\text{0.800 g}$ of iron(III) chloride, you must take a volume of

$0.800 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g FeCl"_3))) * overbrace("100 mL solution"/(40color(red)(cancel(color(black)("g FeCl"_3)))))^(color(blue)("= 40% m/v")) = color(darkgreen)(ul(color(black)("2 mL solution}}}}$

The answer is rounded to one significant figure.

As you can see, we have

${V}_{\text{800 ppm solution" = 10^3color(white)(.)"mL" + "2 mL}} \approx {10}^{3}$ $\text{mL}$

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