# Find the freezing point (in #""^@ "C"#) for a solution consisting of #"60.0 g"# water and #"40.0 g"# ethylene glycol?

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Find the freezing point (in #""^@ "C"# ) for a solution consisting of #"60.0 g"# water and #"40.0 g"# ethylene glycol? The #K_f# of water is #1.86^@ "C"cdot"kg/mol"# the #K_b# of water is #0.512^@ "C"cdot"kg/mol"# , and the molar masses of water and ethylene glycol are #"18.015 g/mol"# and #"62.0 g/mol"# , respectively.

#a)# #20.1^@ "C"#

#b)# #-20.1^@ "C"#

#c)# #5.50^@ "C"#

#d)# #28.42^@ "C"#

Find the freezing point (in

##### 2 Answers

B -20.1 degrees C

#### Explanation:

The freezing point depression for water is -1.85 degrees per molal solution.

A molal solution is l mole of particles per kg of solution.

As ethylene glycol is a covalent molecule it does not ionize in water. Because of this 1 mole of ethylene glycol produces 1 mole of particles.

To find the number of moles of ethylene glycol divide the mass of ethylene glycol by the molar mass.

There are 1000 grams of water in a Kg. To find the number of Kg of solution define the 60 grams grams of water by 1000 grams.

The molal concentration is the moles divided by the Kg.

,645 moles / .06 Kg = 10.8 m ( molal concentration)

The freezing point depression is -1.85

To find the depression multiply -1.85 x 10.8 molal.

The normal freezing point is

0 + - 19.88 = - 19.88

The closest answer is B.

I got

The

#bb(DeltaT_f = -iK_fm)# where:

#i# is thevan't Hoff factor, which is thenumber of ionsin afullyionic compound. This was not given so we have to work from a different equation or estimate#i# .#K_f# is thefreezing point depression constant.#m# is themolalityof the solution, i.e.#"mols solute"/"kg solvent"# .#DeltaT_f = T_f - T_f^"*"# is the change in freezing point, where#T_f^"*"# is the freezing point of thepuresolvent and#T_f# is thenewfreezing point.

*Ethylene glycol* is not fully ionic, but it is polar, so it is miscible in water. However, its

The **molality** of the solution is based on which substance is the solvent, i.e. which one there is more of.

#"60.0 g"/"18.015 g/mol" = "3.331 mol water"#

#"40.0 g"/"62.0 g/mol" = "0.6452 mol EG"#

Therefore, water is the solvent.

#color(green)(m_"soln") = "mols EG"/"kg water"#

#= "0.6452 mols EG"/(60.0 cancel"g water" xx "1 kg water"/(1000 cancel"g")#

#=# #color(green)("10.75 m solution")#

Therefore, the change in freezing point is:

#DeltaT_f\ = T_f - T_f^"*" ~~ (1)("1.86"^@ "C"cdotcancel"kg/mol")(10.75 cancel"mol/kg") ~~ -20.0^@ "C"#

Since the freezing point must decrease, the final freezing point is:

#color(blue)(T_f) = - 20.0^@ "C" + T_f^"*"#

#= color(blue)(-20.0^@ "C")#

The apparent answer is therefore

Based on this answer, can you see that