For an arbitrary nonelectrolyte, calculate the freezing point when #"0.0192 mols"# of it is dissolved in #"5.00 g"# of water? #K_f = 1.86^@ "C/m"# for water.

1 Answer
Jan 26, 2017

#-7.14^@ "C"#.


The idea is that dissolving anything into a solvent will decrease its freezing point. It is expected that you already know #T_f^"*"# for water is #0^@ "C"#, i.e. the freezing point of the pure solvent water. Therefore, you are solving for the final, lowered freezing point.

Recall:

#DeltaT_f = T_f - T_f^"*" = -iK_fm#

where:

  • #i = 1# is the van't Hoff factor for a non-electrolyte. It simply says that only one nonelectrolyte particle exists for every nonelectrolyte particle that is placed into solution.
  • #K_f = 1.86^@ "C/m"# is the freezing point depression constant for water.
  • #m# is the molality of the solution, i.e. #"mol solute"/"kg solvent"#.
  • #T_f# is the freezing point of the solution.
  • #T_f^"*"# is the freezing point of the pure solvent.

First, calculate what #DeltaT_f# would be:

#DeltaT_f = -(1)(1.86^@ "C/m")(("0.0192 mols solute")/(5.00xx10^(-3) "kg H"_2"O"))#

#= -7.14^@ "C"#

Therefore, the final freezing point is:

#color(blue)(T_f) = DeltaT_f + T_f^"*" = -7.14^@ "C" + 0^@ "C" = color(blue)(-7.14^@ "C")#