What is the molality for an aqueous solution whose freezing point dropped by #15^@ "C"# due to addition of #"CaCl"_2(s)#? #K_f = 1.86^@ "C/m"# for water.

I got #"2.69 m"#.

In general, the decrease in freezing point can be found from the following equation:

#bb(DeltaT_f = T_f - T_f^"*")#

#= bb(iK_fm)#

where:

• #T_f# is the freezing point of the #"CaCl"_2(aq)# solution.
• #"*"# means of pure water. That is, #T_f^"*" = 0^@ "C"#.
• #i# is the van't Hoff factor. A good way to estimate it is to say that the number of ions the solute makes is equal to #i#.
• #K_f = -1.86^@ "C/m"# is the freezing point depression constant of water near #0^@ "C"#. The units are read as degrees celcius per molal.
• #m = "mol solute"/"kg solvent"# is the molality of the solution. The unit is read molal.

To solve for the molality:

#m = (T_f^"*" - T_f)/(iK_f)#

Now, we would estimate that #bb(i ~~ 3)# since #"CaCl"_2(aq)# dissociates as follows:

#"CaCl"_2(s) stackrel("H"_2"O"(l))(->) "Ca"^(2+)(aq) + 2"Cl"^(-)(aq)#

giving #bb3# ions in solution. Finally, we know that #T_f^"*" = 0^@ "C"#. So plugging the numbers in gives:

#color(blue)(m) = (-15^@ cancel"C" - 0^@ cancel"C")/(3*(-1.86^@ cancel"C""/m")#

#=# #color(blue)("2.69 m")# (molals)