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 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
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")#
