# An organic solute was dissolved in napthalene such that the freezing point dropped by 7.9 degrees. What is the molality of the solution?

Jul 11, 2017

$\text{1.13 mol/kg}$ to three sig figs (although you only have two...).

Well, you either haven't given us the freezing point depression constant, or you weren't given it. I also assume your degrees are $\text{^@ "C}$...

Either way, we could either look that up, or use the enthalpy of fusion from NIST (Sharma, Gupta, et al., 2008) and the following expressions.

From Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3, the freezing point depression for non-electrolytes is given by:

$\setminus m a t h b f \left(\Delta {T}_{\text{f" = T_"f" - T_"f"^"*}} = - {K}_{f} m\right)$

$\setminus m a t h b f \left({K}_{f} = \left({M}_{\text{solvent")/("1000 g/kg")(R(T_"f"^"*")^2)/(DeltabarH_"f}}\right)\right)$

where:

• ${T}_{\text{f"^"*" = "353.41 K}}$ is the freezing point in $\text{K}$ of the pure solution (no solute)
• ${T}_{\text{f}}$ is the freezing point in $\text{K}$ of the solution (with solute)
• ${M}_{\text{solvent}}$ is the molar mass of the solvent in $\text{g/mol}$.
• $R$ is the universal gas constant: $\text{8.314472 J/mol"cdot"K}$
• $\Delta {\overline{H}}_{\text{f" ~~ "19.1 kJ/mol}}$ is the molar enthalpy of fusion, in $\text{kJ/mol}$ (hence, "molar").
• ${K}_{f} > 0$ is known as the freezing point depression constant in $\text{K"cdot"kg/mol}$.
• $m$ is the molality, which is $\text{mols solute"/"kg solvent}$.

So, we could obtain ${K}_{f}$ ourselves:

${K}_{f} = \left(128.1705 \cancel{\text{g""/mol")/(1000 cancel"g""/kg")((0.008314472 cancel"kJ""/"cancel"mol"cdotcancel"K")("353.41 K")^cancel(2))/(19.11 cancel"kJ""/"cancel"mol}}\right)$

$= \text{6.97 K"cdot"kg/mol}$

while the first reference gives $6.94$. The change in freezing point was given:

$- \text{7.9 K" = DeltaT_f = T_f - T_f^"*} = - {K}_{f} m$

This gives a molality of:

$\textcolor{b l u e}{m} = \frac{\Delta {T}_{f}}{- {K}_{f}} = \left(- 7.9 \cancel{\text{K")/(-6.97 cancel"K"cdot"kg/mol}}\right)$

$=$ $\textcolor{b l u e}{\text{1.13 mol/kg}}$

although we only have two sig figs, apparently...