Chapter 12, Problem 12.90SP

Interpretation Introduction

Interpretation:

The freezing point of the given aqueous solution have to determine and should be arranged from lowest to highest freezing point: 0.10 m FeCl₃, 0.30 m glucose ( ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ ), 0.15m CaCl₂.

Concept introduction: The solution of any solvent always have lower freezing point than that of the pure solvent. On cooling of a solution the solvent become separate due to formation of solid and the solute become insoluble in this condition. As the solute is non-volatile thus the vapour pressure over the solution in this condition is the vapour pressure of the solvent only. This vapour pressure always remain lower than that of pure solvent. When a liquid become solid due to decreases of temperature then there remain an equilibrium between the liquid state and solid state of the solvent. The liquid which has lower vapour pressure, need more low temperature for solidification. As the vapour pressure of the solvent in solution is always lower than the pure solvent, thus to solidify the solvent lower temperature is needed. For this condition scientist Raoult’s propose a law-In any solvent if a non-electrolyte, non-volatile solid solute is added then the depression of the freezing point of the solvent is proportional to the molal density of the solute in solution. Thus $\Delta T_f\infty C_m$ , where $\Delta T_f$ = depression of freezing point and $C_m$ is the molal density of the solute.

Given:

The solutes i.e. FeCl₃, C₆H₁₂O₆ and CaCl₂ are completely dissociated in the solution.