What is the freezing point of an aqueous solution containing #"40.0 g"# of ethylene glycol in #"60.0 g"# of water? #K_f = 1.86^@"C/m"# and #K_b = 0.512^@ "C/m"# for water.
#A)# #20.1^@ "C"#
#B)# #-20.1^@ "C"#
#C)# #5.53^@ "C"#
#D)# #-5.53^@ "C"#
1 Answer
I got
As a note,
#A# is due to getting the wrong sign on#DeltaT_f# .#C# is due to using#K_b# instead of#K_f# and getting the wrong sign.#D# is due to using#K_b# instead of#K_f# and getting the right sign.
INITIAL PREDICTIONS
First off, let's predict the freezing point from thinking about this qualitatively. Pure ethylene glycol has a freezing point of
So, the solution's freezing point should actually be below
We can eliminate all but
FREEZING POINT DEPRESSION FORMULA
Now let's actually calculate it so we can prove it. The High School version of the formula for freezing point depression is:
#\mathbf(DeltaT_f = T_f - T_f^"*" = K_f*m*i)# where:
#T_f# is the freezing point of the solution.#T_f^"*"# is the freezing point of the pure solvent.#K_f# is the freezing point depression constant of the pure solvent.#m# is the molal concentration of the solution, which is the#"mol"# s of solute per#"kg"# of the pure solvent.#i# is the van't Hoff factor, which for ideal solutions is equal to the number of ions that dissociate in solution per formula unit.
FREEZING POINT CALCULATION
Ethylene glycol is otherwise known as ethanediol. In this solution, we have:
#"40.0" cancel("g") xx "1 mol"/("62.0" cancel("g")) = color(green)("0.6452 mols")# ethanediol
#"60.0" cancel("g") xx "1 mol"/("18.015" cancel("g")) = "2.775 mols"# water
Since we clearly have more water than ethanediol, it's safe to say that water is the solvent.
Therefore, we can use the
#color(green)(m) = "mols ethanediol"/"kg water"#
#= "0.6452 mols ethanediol"/"0.0600 kg water"#
#=# #color(green)("10.75 m")# (man, that is high!)
Also, ethanediol is
That means it can dissociate in water pretty easily in its current form (besides, its
Therefore, for ethanediol, we can say that its van't Hoff factor is approximately
Finally, we can get the solution freezing point from the equation we first listed:
#DeltaT_f = T_f - T_f^"*"#
#= T_f - 0^@ "C" = (1.86^@ "C/m")("10.75 m")(1)#
#=> color(blue)(T_f ~~ -20^@ "C")#
As I said before, it would be a freezing point depression, so it makes sense that
So, there you go; the solution was indeed