Question

For an arbitrary nonelectrolyte, calculate the freezing point when `"0.0192 mols"` of it is dissolved in `"5.00 g"` of water? `K_f = 1.86^@ "C/m"` for water.

Answer 1

`-7.14^@ "C"`.

---

The idea is that dissolving anything into a solvent will decrease its freezing point. It is expected that you already know `T_f^"*"` for water is `0^@ "C"`, i.e. the freezing point of the pure solvent water. Therefore, you are solving for the final, lowered freezing point.

Recall:

`DeltaT_f = T_f - T_f^"*" = -iK_fm`

where:

• `i = 1` is the van't Hoff factor for a non-electrolyte. It simply says that only one nonelectrolyte particle exists for every nonelectrolyte particle that is placed into solution.
• `K_f = 1.86^@ "C/m"` is the freezing point depression constant for water.
• `m` is the molality of the solution, i.e. `"mol solute"/"kg solvent"`.
• `T_f` is the freezing point of the solution.
• `T_f^"*"` is the freezing point of the pure solvent.

First, calculate what `DeltaT_f` would be:

`DeltaT_f = -(1)(1.86^@ "C/m")(("0.0192 mols solute")/(5.00xx10^(-3) "kg H"_2"O"))`

`= -7.14^@ "C"`

Therefore, the final freezing point is:

`color(blue)(T_f) = DeltaT_f + T_f^"*" = -7.14^@ "C" + 0^@ "C" = color(blue)(-7.14^@ "C")`