**Colligative Properties: Freezing Point Depression and Boiling Point Elevation****Problem Statement:**Assuming 100% dissociation, calculate the freezing point \( (T_f) \) and boiling point \( (T_b) \) of a 2.23 m \( \text{Na}_2\text{SO}_4 \) (aq) solution.Colligative constants can be found in the chempendix.**Solution Structure:**To calculate the freezing point and boiling point:### Freezing Point \( (T_f) \):\[ T_f = \]### Boiling Point \( (T_b) \):\[ T_b = \]---**Explanation:**This problem involves calculating the freezing point depression and boiling point elevation of an aqueous sodium sulfate \( (\text{Na}_2\text{SO}_4) \) solution. The colligative properties in this context depend on the molality of the solute and assume that the solute fully dissociates in solution, meaning every \( \text{Na}_2\text{SO}_4 \) molecule dissociates into two \( \text{Na}^+ \) ions and one \( \text{SO}_4^{2-} \) ion, resulting in three particles per formula unit.Colligative constants (for freezing point depression and boiling point elevation) will be required and can be looked up in a reference table like the chempendix.Formulae required:1. **Freezing Point Depression**: \( \Delta T_f = i \cdot K_f \cdot m \)2. **Boiling Point Elevation**: \( \Delta T_b = i \cdot K_b \cdot m \)Here,- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)- \( K_f \) = freezing point depression constant of the solvent- \( K_b \) = boiling point elevation constant of the solvent- \( m \) = molality of the solution

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Assuming 100%% dissociation, calculate the freezing point (T) and boiling point (T,) of 2.23 m Na, SO, (aq). Colligative constants can be found in the chempendix. Ti =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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