# 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.

##### 1 Answer

Nov 17, 2016

I got

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 thefreezing pointof the#"CaCl"_2(aq)# solution.#"*"# means ofpure water. That is,#T_f^"*" = 0^@ "C"# .#i# is thevan't Hoff factor. A good way to estimate it is to say that thenumber of ions the solute makesis equal to#i# .#K_f = -1.86^@ "C/m"# is thefreezing point depression constant ofwaternear#0^@ "C"# . The units are read asdegrees celcius per molal.#m = "mol solute"/"kg solvent"# is themolalityof the solution. The unit is readmolal.

To solve for the molality:

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

Now, we would estimate that

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

giving

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

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