Question

What is the molality, molarity, and mole fraction of `FeCl_3` in a 22.2 mass % aqueous solution?

`d = 1.280 g` `/mL`.

Answer 1

`"molality" = 1.76` `m`

`"molarity" = 1.75` `M`

`"mole fraction" = 0.0307`

Explanation:

We're asked to find

• the molality

• the molarity

• the mole fraction (of `"FeCl"_3`)

of a `22.2%` by mass `"FeCl"_3` solution.

`" "`

MOLALITY:

The equation for molality is

`"molality" = "mol solute"/"kg solvent"`

If we assume a `100`-`"g"` sample of solution, we would have

• `22.2` `"g FeCl"_3`

• `77.8` `"g H"_2"O"`

The number of moles of `"FeCl"_3` is found using the molar mass:

`22.2cancel("g FeCl"_3)((1color(white)(l)"mol FeCl"_3)/(162.2cancel("g FeCl"_3))) = color(red)(ul(0.137color(white)(l)"mol FeCl"_3`

The number of kilograms of the solvent (water) is

`77.8cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(green)(ul(0.0778color(white)(l)"kg H"_2"O"`

And so the molality is

`"molality" = (color(red)(0.137color(white)(l)"mol FeCl"_3))/(color(green)(0.778color(white)(l)"kg H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "1.76color(white)(l)m" ")|)`

`" "`

MOLARITY:

The equation for molarity is

`"molarity" = "mol solute"/"L solution"`

To find the molarity of a solution with its molality known, we need to know the density of the solution (which is given to us as `1.280color(white)(l)"g/mL"`).

Since we assumed a `100`-`"g"` sample of solution, we can find the volume of the solution:

`100cancel("g soln")((1cancel("mL soln"))/(1.280cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.0781color(white)(l)"L solution"`

We found the moles of solute earlier, so the molarity is thus

`"molarity" = (0.137color(white)(l)"mol FeCl"_3)/(color(red)(0.0781color(white)(l)"L solution")) = color(blue)(ulbar(|stackrel(" ")(" "1.75color(white)(l)M" ")|)`

`" "`

MOLE FRACTION:

The equation for the mole fraction of `"FeCl"_3` is given by

`chi_("FeCl"_3) = ("mol FeCl"_3)/("mol FeCl"_3 + "mol H"_2"O")`

Again, we already know the number of moles of iron(III) chloride, so we just need to find the moles of water, using its molar mass and gram-mass found earlier:

`77.8cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(ul(4.32color(white)(l)"mol H"_2"O"`

The mole fraction is thus

`chi_("FeCl"_3) = (0.137color(white)(l)"mol FeCl"_3)/(0.0137color(white)(l)"mol FeCl"_3 + color(red)(4.32color(white)(l)"mol H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "0.0307" ")|)`