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230_Sp10_L15_post - CHEM 230 Sp10 Lecture 15 Chapter 8...

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CHEM 230 Sp10 Lecture 15 Chapter 8: Colligative Properties: 1) Vapor pressure 2) Freezing point depression; boiling point elevation 3) Osmotic pressure Session ID 230
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Demos L13 Phase transition: affect of pressure on bp L14 Solubility (solid in liquid): Like dissolves like Animation of NaCl dissolving in water http://www. chem . iastate .edu/group/Greenbowe/sections/projectfolder/flashfiles/thermoche m/solutionSalt.html L14/L15 Heat of solution/colligative properties: CaCl 2 on ice
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Colligative Properties "Colligative" comes the observation that, in general, the properties are influenced only by how much of something is dissolved in solution, not the nature of the substance Entropy changes with solute dissolving in solvent and creates: Vapor pressure lowering P = x solvent P pure Freezing point depression Δ T f = k f * m Boiling point elevation Δ T b = k b * m Osmotic pressure Π = i RT c
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Accounting Mole fraction x n = moles n moles total Molality m = moles solute kg solvent c Molarity M = moles solute L solution Make a single assumption about a quantity since the solutions are uniform. assume 100 total moles assume 1 kg solvent assume 1 L solution
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An aqueous solution of H 2 O 2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its molarity, molality, and the mole fraction of H 2 O 2 . For density, assume 1 L = 1000 mL 1.11g 1mL *1000 mL = 1110 g solution Use mass % xg solute 1110 g total *100 = 30% x = 333g H 2 O 2 * 1 mol 34 gH 2 O 2 = 9.79 molH 2 O 2 1110 g total- 333g H 2 O 2 = 777g H 2 O * 1 molH 2 O 18 gH 2 O = 43.17 molH 2 O
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Mole fraction x H2O2 = Molality m = Molarity M = molesH 2 O 2 moles total = 9.79 molH 2 O 2 9.79 mol + 43.17 mol = 0.185 molesH 2 O 2 kgsolvent ( H 2 O ) = 9.79 molesH 2 O 2 0.777 kg = 12.6 mol kg molesH 2 O 2 Lsolution = 9.79 molesH 2 O 2 1 L = 9.79 mol L
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