Question

a solution is made by dissolving 0.748 mol of nonelectrolyte solute in 843 g of benzene. calculate the freezing point, (t_f), and boiling point, (t_b), of the solution. constants can be found in the table of colligative constants.
(t_f=)
(t_b=)

Explanation:

Step1: Calculate the molality (m) of the solution

Molality is defined as moles of solute per kilogram of solvent. First, convert the mass of benzene from grams to kilograms: $843\ g=0.843\ kg$. Then, $m=\frac{0.748\ mol}{0.843\ kg}\approx0.887\ m$.

Step2: Determine the freezing - point depression constant ($K_f$) and boiling - point elevation constant ($K_b$) for benzene

For benzene, $K_f = 5.12\ ^{\circ}C/m$ and $K_b=2.53\ ^{\circ}C/m$. The freezing point of pure benzene is $T_{f,0}=5.5\ ^{\circ}C$ and the boiling point of pure benzene is $T_{b,0}=80.1\ ^{\circ}C$. Since the solute is a nonelectrolyte, the van't Hoff factor $i = 1$.

Step3: Calculate the freezing - point depression ($\Delta T_f$)

The formula for freezing - point depression is $\Delta T_f=iK_fm$. Substituting $i = 1$, $K_f = 5.12\ ^{\circ}C/m$, and $m = 0.887\ m$, we get $\Delta T_f=1\times5.12\ ^{\circ}C/m\times0.887\ m\approx4.54\ ^{\circ}C$. Then, $T_f=T_{f,0}-\Delta T_f=5.5\ ^{\circ}C - 4.54\ ^{\circ}C = 0.96\ ^{\circ}C$.

Step4: Calculate the boiling - point elevation ($\Delta T_b$)

The formula for boiling - point elevation is $\Delta T_b=iK_bm$. Substituting $i = 1$, $K_b = 2.53\ ^{\circ}C/m$, and $m = 0.887\ m$, we get $\Delta T_b=1\times2.53\ ^{\circ}C/m\times0.887\ m\approx2.25\ ^{\circ}C$. Then, $T_b=T_{b,0}+\Delta T_b=80.1\ ^{\circ}C+2.25\ ^{\circ}C = 82.35\ ^{\circ}C$.

Answer:

$T_f = 0.96\ ^{\circ}C$
$T_b = 82.35\ ^{\circ}C$