Benzene, a solvent, has a freezing point of 5.5°C and a F of 5.12° CM. If 45.6 g of CCl4 are dissolved in 520.0 g of benzene, what will the freezing point of the solution be?

1 Answer
Truong-Son N.
Jun 7, 2018

Well, it'll be #2.9^@ "C"# lower... giving?

Well, let's gather the data with the proper units...

  • #T_f^"*" = 5.5^@ "C"# is the pure freezing point of the solvent.
  • #K_f = 5.12^@ "C"cdot"kg/mol"# is the freezing point depression constant of the solvent, benzene.
  • #"45.6 g"# #"CCl"_4# is dissolved in #"520.0 g benzene"#.

Freezing point depression is given by:

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

where:

  • #"*"# indicates pure solvent.
  • #i# is the van't Hoff factor, and is #1# for NONelectrolytes. It is the number of dissociated solute particles per undissociated solute particle.
  • #m# is the molality, #"mol solute/kg solvent"#.

The molality is:

#m = (45.6 cancel("g CCl"_4) xx "1 mol CCl"_4/(153.823 cancel"g"))/("0.5200 kg benzene")#

#=# #"0.5701 mol/kg"#

Hence:

#DeltaT_f = -(1)(5.12^@ "C"cdotcancel"kg/mol")(0.5701 cancel"mol/kg")#

#= -2.92^@ "C"#

And since this is the change...

#color(blue)(T_f) = T_f^"*" + DeltaT_f#

#= 5.5^@ "C" - 2.92^@ "C"#

#= color(blue)(2.6^@ "C")#

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