If #"0.64 g"# of napthalene (#M = "128.1705 g/mol"#) is dissolved in #"100 g"# a certain solvent (#T_f^"*" = 6^@ "C"#), whose #K_f = 20^@ "C"cdot"kg/mol"#, what is the new freezing point of the solvent?

#a)# #4^@ "C"#
#b)# #5^@ "C"#
#c)# #-1^@ "C"#
#d)# #7^@ "C"#

1 Answer
Truong-Son N.
May 20, 2017

#(b)#

That's a high #K_f#... If it weren't for the low concentration, #T_f# may have gotten down to #-1^@ "C"#. What would the molality be if the answer were #(a)#?

The change in freezing point is given by:

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

where:

  • #T_f# is the freezing point of the solution.
  • #"*"# indicates pure solvent.
  • #i# is the van't Hoff factor of the solute. #i >= 1#, and corresponds to the effective number of particles per formula unit of solute.
  • #K_f# is the freezing point depression constant of the solvent.
  • #m# is the molality of the solution in #"mol solute/kg solvent"#.

#i = 1# for nonelectrolytes, as they essentially do not dissociate in solution. The molality is the next thing we can calculate...

#m = (0.64 cancel"g Napthalene" xx "1 mol"/(128.1705 cancel"g Napthalene"))/(100 cancel"g solvent" xx "1 kg"/(1000 cancel"g solvent")#

#=# #"0.050 mol solute/kg solvent"#

Therefore, the change in freezing point is:

#DeltaT_f = -(1)(20^@ "C"cdot"kg/mol")("0.0499 mol/kg")#

#= -0.9987^@ "C" -> -1^@ "C"#

Thus, the new freezing point is:

#color(blue)(T_f) = DeltaT_f + T_f^"*" = (-0.9987^@ "C") + (6^@ "C")#

#~~ color(blue)(5^@ "C")#

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