What is the molality, molarity, and mole fraction of #FeCl_3# in a 22.2 mass % aqueous solution?
#d = 1.280 g# #/mL# .
1 Answer
Answer:
Explanation:
We're asked to find
of a
MOLALITY:
The equation for molality is
#"molality" = "mol solute"/"kg solvent"#
If we assume a

#22.2# #"g FeCl"_3# 
#77.8# #"g H"_2"O"#
The number of moles of
#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
Since we assumed a
#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
#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 grammass 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" "))#