# A solution that contains #"13.2 g"# of solute in #"250 g"# of #"CCl"_4# freezes at #-33^@ "C"#. What is the FW of the solute? #T_f^"*"# of #"CCl"_4# is #-22.8^@ "C"# and its #K_f = -29.8^@ "C/m"#.

##### 1 Answer

#M = "154.26 g/mol"#

You can think of this as just an extension of freezing point depression into solvents that aren't water. Thus, refer to the freezing point depression equation:

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

#T_f = -33^@ "C"# is thefreezing point, and#T_f^"*" = -22.8^@ "C"# is for the pure solvent.#DeltaT_f# is thechange in freezing point, which will end up being negative.#i# is the van't Hoff factor, i.e. theeffective number of solute particlesin solution after any potential dissociation.#K_f = 29.8^@ "C/m"# is thefreezing point depression constantof#"CCl"_4# . (Yours is negative, but I put the negative in the equation instead.)#m# is themolalityof the solution, in#"mols solute/kg solvent"# .

Now, the question hasn't specified whether the solute is a nonelectrolyte. However, since *not* likely to dissociate in solution, so

The molality is what we can end up using to get the **formula mass** ("FW") of the solute, so we would solve for the molality.

#m = -(DeltaT_f)/(iK_f)#

#= -(-33^@ "C" - (-22.8^@ "C"))/((1) cdot 29.8^@ "C/m")#

#= 0.342# #"mols/kg"#

Knowing that we have

#"0.342 mols"/"kg" = "0.0856 mols"/"0.250 kg"#

As a result, we have **formula mass** is given by:

#color(blue)(M) = "13.2 g"/("0.0856 mol") = color(blue)("154.26 g/mol")#