# What mass of cholesterol should be dissolved in "295 g" of benzene to lower the freezing point by "0.450 K"? (M = "386.6 g/mol" for cholesterol and "78.11 g/mol" for benzene.)

Jun 23, 2017

I got

${10.0}_{4}$ $\text{g}$

of cholesterol to three sig figs (the subscript indicates the first digit of which we are not certain) to reduce the normal freezing point of benzene to ${5.05}^{\circ} \text{C}$.

Given the freezing point depression equation:

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

where:

• ${T}_{f}$ is the freezing point, and $\text{*}$ indicates pure solvent.
• $i$ is the van't Hoff factor, and is $1$ for nonelectrolytes.
• ${K}_{f}$ (not given) is the freezing point depression constant of benzene, the solvent.
• $m$ is the molality of the solution, $\text{mol solute/kg solvent}$.

Since we don't have ${K}_{f}$, we'll have to get it from (Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3):

\mathbf(K_f = (M_"solvent")/("1000 g/kg")(R(T_f^"*")^2)/(DeltabarH_f))

where $M$ is the molar mass in $\text{g/mol}$, $R$ is the universal gas constant, and $\Delta {\overline{H}}_{f}$ is the molar enthalpy of fusion in $\text{J/mol}$.

From NIST, $\Delta {\overline{H}}_{f} = \text{9.87 kJ/mol}$, so the ${K}_{f}$ is:

${K}_{f} = \left(\text{78.11 g/mol")/("1000 g/kg") ("8.314472 J/mol"cdot"K" cdot (5.50 + 273.1"5 K")^2)/("9870 J/mol}\right)$

$= \text{5.11 K"cdot"kg/mol}$

This could be verified here... This allows us to get the molality:

$\textcolor{g r e e n}{m} = \frac{\Delta {T}_{f}}{- i {K}_{f}}$

= (-"0.450 K")/(-(1)("5.11 K"cdot"kg/mol"))

$=$ $\textcolor{g r e e n}{\text{0.0881 mol solute/kg solvent}}$

Since we know that $\text{295 g benzene}$, or $\text{0.295 kg}$, were used, and we know that cholesterol has a molar mass of $\text{386.6 g/mol}$:

$\text{0.0881 mol solute"/cancel"kg solvent" xx 0.295 cancel"kg solvent}$

$=$ $\text{0.0260 mol cholesterol}$

As a result, we have this mass of solute dissolved in benzene:

$0.0260 \cancel{\text{mols cholesterol" xx "386.6 g"/cancel"mols cholesterol}}$

$=$ $\textcolor{b l u e}{\text{10.0 g cholesterol}}$ to three sig figs.