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Outline 1/12/07

Outline 1/12/07. First Chem seminar today… Pick up CAPA #2 - in front ChemBoard Assignment = Monday Fill in keypad response unit #’s… Today: Surfactants Colligative Properties Calculating Colligative Effects. Surfactants:.

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Outline 1/12/07

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  1. Outline 1/12/07 • First Chem seminar today… • Pick up CAPA #2 - in front • ChemBoard Assignment = Monday • Fill in keypad response unit #’s… Today: Surfactants Colligative Properties Calculating Colligative Effects

  2. Surfactants: • Anything which alters behavior of solute-solvent mixture….(e.g. soap) Applications? NaCl in cooking Road salt for ice Antifreeze Saline drips Gatorade e.g. Demo of S on water

  3. Effect of solutes on solvent properties... Vapor pressures Boiling points Freezing points Osmotic pressure “Colligative properties”

  4. Calculations of Colligative Property effects • Freezing Point Depression • DT = Kfm • Boiling Point Elevation • DT = Kbm Kf = -1.858°C/m Kb = 0.512°C/m where Kfand Kbare constants and m is concentration of solute in terms of molality (mols/kg solvent)

  5. liquid solid

  6. Calculations of Colligative Property effects • Freezing Point Depression • DT = Kfm • Boiling Point Elevation • DT = Kbm Kf = -1.858°C/m Kb = 0.512°C/m where Kfand Kbare constants and m is concentration of solute in terms of molality (mols/kg solvent)

  7. Example #1: Freezing Point Depression: DT = Kfm • Estimate the freezing point of wine that is 12% by mass ethanol... Kf= -1.858 C kg/mol (for water) Assume 100g of wine, then you have 12g ethanol and 88g H2O... 12g C2H5OH  46 g/mol= 0.261 mol 0.261 molEtOH/88 g H2O = x mol/1000g x = 2.97 m DT = -5.5°C

  8. Example #2: • Compute the molar mass of A, if a solution containing 35.0 g of A dissolved in 135 mL of H2O freezes at -1.75C. • Assume A does not ionize in water. Freezing Point Depression: DT = Kfm Kf= -1.858 C kg/mol (for water) -1.75 = -1.858 m or m = 0.942 mol/kg

  9. Example #2 (cont’d): • m = 0.942mols solute/kg solvent • How many kg of solvent? 135 mL  1 g/mL = 135 g = 0.135 kg 0.942 mol/kg  0.135 kg = 0.127 mol Molar mass: 35.0 g / 0.127 mol = 275 g/mol

  10. Another useful application: • Osmotic Pressure (p) Passage of solute molecules through a semi-permeable membrane Start with something we know: PV = nRT (ideal gases) or P = (n/V)RT Adapt to solutions: (n/V)  (mol/L) = Molarity!

  11. p.107 p = M RT • Osmotic Pressure (p)

  12. p.107

  13. Example: • Compute the molar mass of hemoglobin, if a 25 mL solution containing 0.420 g of hemoglobin has an osmotic pressure of 4.6 torr at 27C. Osmotic pressure: p = M RT M= [m (g) / MM (g/mol)]/V p = m (RT)/ (MM) V = 68300 g/mol

  14. Practice…Worksheet #1 • Remember to keep solute & solvent separate when answering questions!

  15. Henry’s Law • Gases dissolved in a liquid: Concentration = KHp Examples: soda CO2, NOx, SOx and acid rain

  16. For Monday… • Do CAPA set #1 • Post a trial message on Chemboard • Read through Chapter 12 …get ahead! Have a great week-end!

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