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

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.

# 40 / 62 # = ,645 moles.

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.

# 60 g / 1000 g# = .06 Kg.

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 #C^o# per 1 -molal.
To find the depression multiply -1.85 x 10.8 molal.

# -1.85xx 10.8# = - 19.88

The normal freezing point is #0^o C# so the new freezing point is
0 + - 19.88 = - 19.88

The closest answer is B.

I got #"b"#.

#a)# is incorrect because freezing point depression is negative.
#c)# is incorrect because it results from using #K_b# instead of #K_f#, and yet the change in boiling point is added to the freezing point...
#d)# is incorrect because it results from having calculated the molality for water instead of for ethylene glycol, and then used #K_b# instead of #K_f#, and then still added the change in boiling point to the freezing point.

The #K_f# of water is #"1.86"^@ "C/m"#, and recall the freezing point depression equation:

#bb(DeltaT_f = -iK_fm)#

where:

• #i# is the van't Hoff factor, which is the number of ions in a fully ionic compound. This was not given so we have to work from a different equation or estimate #i#.
• #K_f# is the freezing point depression constant.
• #m# is the molality of 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 the pure solvent and #T_f# is the new freezing point.

Ethylene glycol is not fully ionic, but it is polar, so it is miscible in water. However, its #"pKa"# is significantly higher than that of water (#25# vs. #15.7#), so we can say it does not dissociate significantly in water. Therefore, #i ~~ 1#. It's probably a little higher than #1# for a real solution, however.

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 #-20.1^@ "C"#, which is #"B"#. None of the other answers are right because they are positive and not negative.

Based on this answer, can you see that #i ~~ 1.005#? It was a good estimate to say that #i ~~ 1#, but it's not exactly #1#.