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 andT_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")
