CHEM 135 1st Edition Lecture 5 Outline of Last Lecture I. Solutions a. Definitions and componentsb. Types of formationsII. Solubility a. Effect of temperature b. Effect of pressureIII. Colligative PropertiesIV. Vapor pressure and Dynamic Equilibrium V. Raoult’s LawOutline of Current Lecture VI. Review of Raoult’s Law VII. Phase Diagram to indicate Freezing and Boiling pointVIII. Freezing point depression and boiling point elevationCurrent LectureVI. Review of Raoult’s Law- Vapor Pressure of the solution is given by:Psolution=Xsolvent-P solventXsolvent- mole fraction of the solvent (n solvent/n solute + n solvent)- Vapor pressure lowering (P):P= Xsolvent-P solventThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Ex: Calculate the vapor pressure at 25C of a solution containing 55.3 g ethylene glycol (HOCH2CH2CH2OH) and 285.2 g water. The vapor pressure of pure water at 25C is 23.8 torr. Psolution=Xsolvent-P solventGiven: - P solvent= 23.8 torr- Mass of solvent= 285.2 g- Mass of solute= 55.3 gFind: P solutionStrategy: mass moles Xsolvent PsolutionXsolvent= n solventn solute+ n solventXsolvent=15.84 mol0.727+15.84=0.956 molPsolution=Xsolvent-P solvent=(0.956)(23.8)=22.76 torrVII. Phase Diagram to indicate Freezing and Boiling pointVIII. Freezing point depression and boiling point elevation- The freezing point of a solution is lower than the freezing point of the pure solventTf=m-Kf- The boiling point of a solution is higher than the boiling point of the pure solvent. - Tb=m-Kbm= molality of a solution in moles solute per kilogram solventKf, Kb- freezing point depression and boiling point elevation