Question

Question #cd6f3

Answer 1

`"2 mL"`

Explanation:

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.