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

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" ")|)#