CHEM 135 1st Edition Lecture 6 Outline of Last Lecture I. Review of Raoult’s Law II. Phase Diagram to indicate Freezing and Boiling pointIII. Freezing point depression and boiling point elevationOutline of Current Lecture IV. Review of freezing point depression and boiling point elevationa. ExamplesV. OsmosisVI. Osmotic Pressurea. ExampleCurrent Lecture Freezing point depression and boiling point elevation- The freezing point of a solution is lower than the freezing point of the pure solventTf=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 constantsEx: Calculate the freezing point and boiling point of a solution containing 10 g of naphthalene (C10H8) in 100 mL benzene (d=0.877 g/cm3). These 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.Equations needed: Mass=density × volumemoles=massmolar massm=moles solutekg solvent∆ Tf=m× KfTf=T °f−∆ TfGiven:Mass of solute=10 g C10H8Volume of solvent= 100 mL benzeneDensity of solvent= 0.877 g/cm3Tf=?Tb=?Strategy: mass moles molality ∆ Tf TfMass= (0.877 g/cm3)(100 mL)= 87.7 g= 0.0877 kgMoles= 10 g C 10 H 8128.1 g/mol=0.78 molesC 10 H 8Molality= .078 mol.0877 kg=.89 m∆ Tf=m× Kf ∆ Tf=(.89 m)(5.12 °Cm) ∆ Tf=4.56 ° CTf=T °f−∆ Tf Tf=(5.5 ° C)−(4.56° C) Tf=0.94 °C*Follow the same steps using the boiling point formula (∆ Tb=m × Kb¿ and (Tb=T °b−∆ Tb¿ to get a final boiling point of 82.35°C .Osmosis is the flow of solvent from a solution of low concentration into a solution of high concentration.Osmotic PressureThe amount of pressure needed to keep osmotic flow from taking place, ie. Osmotic pressure PREVENTS osmosis from occurring∏=MRT∏ (osmotic pressure) is directly proportional to the molarity of the soluteM=molarity (moles/L)R= 0.08206
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