# 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** *depression* due to colligative properties of adding solutes into a solvent, so the freezing point should drop).

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 thefreezing pointof thesolution.#T_f^"*"# is the freezing point of thepure solvent.#K_f# is thefreezing point depression constantof the pure solvent.#m# is themolal concentration of the solution, which is the#"mol"# s of solute per#"kg"# of the pure solvent.#i# is thevan't Hoff factor, which for ideal solutions is equal to thenumber 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 **molality**

#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 *or* protonated in water).

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