# 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"#.

##### 1 Answer

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 thefreezing pointsof thesolutionandpure solvent, respectively.#i# is thevan'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 thefreezing point depression constantfor benzene, the solvent.#m# is themolalityof the solution, i.e.#"mols solute"/"kg solvent"# .

As mentioned, benzene is the solvent, because its *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")#