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"#.

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

Now, the question hasn't specified whether the solute is a nonelectrolyte. However, since #"CCl"_4# is nonpolar, it dissolves a solute that is also nonpolar, which is not likely to dissociate in solution, so #i = 1#.

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 #"250 g"# of the solvent, or #"0.250 kg"#, we therefore have:

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

As a result, we have #"0.0856 mols"# of solute for every #"13.2 g"# of solute. Therefore, the formula mass is given by:

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