Question

question 25 of 36
the boiling point of an aqueous solution is 102.01 °c. what is the freezing point? consult the table of colligative constants.
freezing point: ______ °c

Explanation:

Step1: Calculate boiling point elevation

The boiling point of pure water is $100.00^\circ\text{C}$.
$\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent}) = 102.01^\circ\text{C} - 100.00^\circ\text{C} = 2.01^\circ\text{C}$

Step2: Find molality via boiling point formula

For water, boiling point constant $K_b = 0.512^\circ\text{C}\cdot\text{kg/mol}$. Use $\Delta T_b = K_b \cdot m$:
$m = \frac{\Delta T_b}{K_b} = \frac{2.01^\circ\text{C}}{0.512^\circ\text{C}\cdot\text{kg/mol}} \approx 3.926\text{ mol/kg}$

Step3: Calculate freezing point depression

For water, freezing point constant $K_f = 1.86^\circ\text{C}\cdot\text{kg/mol}$. Use $\Delta T_f = K_f \cdot m$:
$\Delta T_f = 1.86^\circ\text{C}\cdot\text{kg/mol} \times 3.926\text{ mol/kg} \approx 7.30^\circ\text{C}$

Step4: Compute solution freezing point

Freezing point of pure water is $0.00^\circ\text{C}$.
$T_f(\text{solution}) = T_f(\text{pure solvent}) - \Delta T_f = 0.00^\circ\text{C} - 7.30^\circ\text{C}$

Answer:

$-7.30^\circ\text{C}$