# 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.

##### 1 Answer
Jan 26, 2017

$- {7.14}^{\circ} \text{C}$.

The idea is that dissolving anything into a solvent will decrease its freezing point. It is expected that you already know ${T}_{f}^{\text{*}}$ for water is ${0}^{\circ} \text{C}$, i.e. the freezing point of the pure solvent water. Therefore, you are solving for the final, lowered freezing point.

Recall:

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - i {K}_{f} m$

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.
• ${K}_{f} = {1.86}^{\circ} \text{C/m}$ is the freezing point depression constant for water.
• $m$ is the molality of the solution, i.e. $\text{mol solute"/"kg solvent}$.
• ${T}_{f}$ is the freezing point of the solution.
• ${T}_{f}^{\text{*}}$ is the freezing point of the pure solvent.

First, calculate what $\Delta {T}_{f}$ would be:

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

$= - {7.14}^{\circ} \text{C}$

Therefore, the final freezing point is:

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