Question

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"`.

Answer 1

`M = "154.26 g/mol"`

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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")`