# If "0.64 g" of napthalene (M = "128.1705 g/mol") is dissolved in "100 g" a certain solvent (T_f^"*" = 6^@ "C"), whose K_f = 20^@ "C"cdot"kg/mol", what is the new freezing point of the solvent?

## a) ${4}^{\circ} \text{C}$ b) ${5}^{\circ} \text{C}$ c) $- {1}^{\circ} \text{C}$ d) ${7}^{\circ} \text{C}$

May 20, 2017

$\left(b\right)$

That's a high ${K}_{f}$... If it weren't for the low concentration, ${T}_{f}$ may have gotten down to $- {1}^{\circ} \text{C}$. What would the molality be if the answer were $\left(a\right)$?

The change in freezing point is given by:

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

where:

• ${T}_{f}$ is the freezing point of the solution.
• $\text{*}$ indicates pure solvent.
• $i$ is the van't Hoff factor of the solute. $i \ge 1$, and corresponds to the effective number of particles per formula unit of solute.
• ${K}_{f}$ is the freezing point depression constant of the solvent.
• $m$ is the molality of the solution in $\text{mol solute/kg solvent}$.

$i = 1$ for nonelectrolytes, as they essentially do not dissociate in solution. The molality is the next thing we can calculate...

$m = \left(0.64 \cancel{\text{g Napthalene" xx "1 mol"/(128.1705 cancel"g Napthalene"))/(100 cancel"g solvent" xx "1 kg"/(1000 cancel"g solvent}}\right)$

$=$ $\text{0.050 mol solute/kg solvent}$

Therefore, the change in freezing point is:

$\Delta {T}_{f} = - \left(1\right) \left({20}^{\circ} \text{C"cdot"kg/mol")("0.0499 mol/kg}\right)$

$= - {0.9987}^{\circ} \text{C" -> -1^@ "C}$

Thus, the new freezing point is:

color(blue)(T_f) = DeltaT_f + T_f^"*" = (-0.9987^@ "C") + (6^@ "C")

$\approx \textcolor{b l u e}{{5}^{\circ} \text{C}}$