Question

question 22 of 36
a solution is made by dissolving 0.735 mol of nonelectrolyte solute in 855 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.
$t_f =$ $square$ $^circ c$
$t_b =$ $square$ $^circ c$

Explanation:

Step1: Convert solvent mass to kg

$855\ \text{g} = \frac{855}{1000} = 0.855\ \text{kg}$

Step2: Calculate molality of solution

Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.735\ \text{mol}}{0.855\ \text{kg}} \approx 0.8596\ \text{mol/kg}$

Step3: Get colligative constants for benzene

Freezing point of pure benzene $T_f^0 = 5.5^\circ\text{C}$, freezing point depression constant $K_f = 5.12^\circ\text{C}\cdot\text{kg/mol}$; Boiling point of pure benzene $T_b^0 = 80.1^\circ\text{C}$, boiling point elevation constant $K_b = 2.53^\circ\text{C}\cdot\text{kg/mol}$ (standard colligative constants for benzene)

Step4: Calculate freezing point depression

$\Delta T_f = i \cdot K_f \cdot m$. For nonelectrolyte, $i=1$.
$\Delta T_f = 1 \times 5.12^\circ\text{C}\cdot\text{kg/mol} \times 0.8596\ \text{mol/kg} \approx 4.40^\circ\text{C}$

Step5: Calculate solution freezing point

$T_f = T_f^0 - \Delta T_f = 5.5^\circ\text{C} - 4.40^\circ\text{C} = 1.1^\circ\text{C}$

Step6: Calculate boiling point elevation

$\Delta T_b = i \cdot K_b \cdot m$
$\Delta T_b = 1 \times 2.53^\circ\text{C}\cdot\text{kg/mol} \times 0.8596\ \text{mol/kg} \approx 2.17^\circ\text{C}$

Step7: Calculate solution boiling point

$T_b = T_b^0 + \Delta T_b = 80.1^\circ\text{C} + 2.17^\circ\text{C} = 82.3^\circ\text{C}$

Answer:

$T_f = 1.1^\circ\text{C}$
$T_b = 82.3^\circ\text{C}$