What is the normal boiling point for an aqueous solution containing a mol fraction of #0.35# for #"CaCl"_2#? Assume complete dissociation and #K_b = 0.512^@ "C/m"# for water.

#T_b = 145.9^@ "C"#

The mol fraction is relative. Thus, one needs to assume certain number of total mols for the molality. An ideal binary mixture has:

#chi_(CaCl_2) = 0.35 = n_(CaCl_2)/(n_(CaCl_2) + n_(H_2O))#

It is convenient to pick #"1 mol solution"#, so that we have #"0.65 mols H"_2"O"# and #"0.35 mols CaCl"_2#. Thus, we have:

#0.65 cancel("mols H"_2"O") xx (18.015 cancel("g H"_2"O"))/cancel("1 mol H"_2"O") xx "1 kg"/(1000 cancel"g")#

#= "0.012 kg H"_2"O"#

And so, the molality is:

#m = ("0.35 mols solute")/("0.012 kg solvent")#

#=# #ul"29.9 mol/kg"#

which is quite a high molality... So we expect a sizeable increase in boiling point. (What if we assumed #"2 mols solution"#, #"0.70 mols H"_2"O"#, and #"1.30 mols CaCl"_2#? Does that matter?)

The change in boiling point #DeltaT_b# is given by:

#DeltaT_b = T_b - T_b^"*" = iK_bm#

where:

• #T_b# is the boiling point, and #"*"# indicates the pure solvent.
• #i# is the van't Hoff factor (i.e. the effective number of particles in solution per formula unit of electrolyte).
• #K_b = 0.512^@ "C"cdot"kg/mol"# is the boiling point elevation constant of water (not #0.52#).
• #m# is the molality of the solution.

Knowing the molality, and assuming that #i = 3# for #"CaCl"_2# (#1 xx "Ca"^(2+)# and #2xx"Cl"^(-)#), the change in boiling point is:

#DeltaT_b = (3)(0.512^@ "C"cdot"kg/mol")("29.9 mol/kg")#

#= 45.9^@ "C"#

And thus, the normal boiling point, i.e. at #"1 atm"# pressure, is:

#color(blue)(T_b) = T_b^"*" + DeltaT_b#

#= 100^@ "C" + 45.9^@ "C" = color(blue)ul(145.9^@ "C")#

Do note though that the actual van't Hoff factor of #"CaCl"_2# is around #2.7#, due to ion pairing in solution decreasing the effective number of solute particles.