DOC PREVIEW
KU CHEM 135 - Freezing Point and Osmosis
Type Lecture Note
Pages 4

This preview shows page 1 out of 4 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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 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 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


View Full Document

KU CHEM 135 - Freezing Point and Osmosis

Type: Lecture Note
Pages: 4
Download Freezing Point and Osmosis
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Freezing Point and Osmosis and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Freezing Point and Osmosis 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?