Chapter 12 Lecture 4

Chapter 12 Lecture 4 - b (water) = 0.512 o C/m; Boiling...

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Freezing Point Depression What you already know: Ice Cream churn Rock Salt or white round pellets
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Quantitatively: T f = T f (solution) ─ T f (solvent) = ─ (K f )(m) Kf is molal freezing point depression constant m is molality of solution (mol solute/kg solvent) A 1.45 g sample of a unknown compound is dissolved in 25 mL of benzene. The solution freezes at 4.25 oC. What is the molar mass of the compound? (other stuff: the density of the solution is 0.879 g/mL
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Boiling Point Elevation What you already know: Anti Freeze Adding salt to water for boiling
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Quantitatively: T b = T b (solution) ─ T b (solvent) = ─ (K b )(m) Kb is molal boiling point constant m is molality of solution (mol solute/kg solvent) What is the boiling point of 1.8 M sucrose C 12 H 22 O 11 ? Other stuff you need: d = 1.23 g/mL, K
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Unformatted text preview: b (water) = 0.512 o C/m; Boiling point of water is 100 o C. Osmosis: a semi-permeable membrane that allows only water to pass What you know already: Slug + salt Paramecium + water (pure) Osmosis: The water concentration gradient isotonic NaCl 1M Identical solutions H 2 O NaCl 0.5 M H 2 O hypotonic hypertonic hypotonic hypertonic Hypotonic solutions: solvent leaves Hypertonic solutions: solvent arrives Osmotic Pressure: You have to know how it is measured NaCl 0.5 M H 2 O hypotonic hypertonic NaCl 0.5 M H 2 O hypotonic hypertonic = 0 = 0.7 Water moves from hypotonic to hypertonic solution Or form low osmotic pressure to high osmotic pressure...
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This note was uploaded on 04/16/2008 for the course CHEM 1123 taught by Professor Paul during the Fall '06 term at Arkansas.

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Chapter 12 Lecture 4 - b (water) = 0.512 o C/m; Boiling...

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