What is the molar mass of the solute if #"4.18 g"# of it dissolved in #"36.30 g"# of benzene (#K_f = 5.12^@ "C"cdot"kg/mol"#) generates a solution with a freezing point of #2.70^@ "C"#? The freezing point of benzene is #5.53^@ "C"#.

#"208.33 g/mol"#

Once you recognize this is freezing point depression, you can use the following equation:

#bb(DeltaT_f = T_f - T_f^"*" = -iK_fm)#

where:

• #T_f# and #T_f^"*"# are the freezing points of the solution and pure solvent, respectively.
• #i# is the van't Hoff factor. For non-ionic solutes, i.e. nonelectrolytes, #i = 1#, as there is only one "dissociated" particle per dissolved particle.
• #K_f = 5.12^@ "C"cdot"kg/mol"# is the freezing point depression constant for benzene, the solvent.
• #m# is the molality of the solution, i.e. #"mols solute"/"kg solvent"#.

As mentioned, benzene is the solvent, because its #K_f# is given, and #K_f# is only for solvents. Therefore, we should calculate the mols of the solute and kg of benzene.

Simply from the units of molality, we have:

#m = ("solute mass" xx 1/"molar mass")/("kg solvent")#

#= ("4.18 g"xx"mol"/"g solute")/("0.03630 kg benzene")#

Therefore, we can solve for the molar mass later, as long as we can calculate the molality.

#DeltaT_f = 2.70^@ "C" - 5.53^@ "C" = -(1)(5.12^@ "C"cdot"kg/mol")(m)#

#=> m = "0.5527 mols solute"/"kg solvent"#

Now, we can solve for the molar mass.

#color(blue)("Molar mass") = ("4.18 g solute")/("0.5527 mols solute"/cancel"kg solvent" xx 0.03630 cancel"kg solvent")#

#=# #color(blue)("208.33 g/mol")#