BP Elevation & FP Depression (Chapter 7)

BP Elevation & FP Depression (Chapter 7) - Change...

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Calculating Boiling Point Elevation and Freezing Point Depression CHEM 30A Boiling Point Elevation : T b = i K b m BP = BP normal + T b Freezing Point Depression : T f = i K f m FP = FP normal T f Example : What are the BP and FP of a 7.187 m NaCl solution? For water: K b = 0.512 ºC · kg H 2 O / mol particles K f = 1.86 ºC · kg H 2 O / mol particles BP soln = BP normal + T b = 100.00ºC + 7.35ºC = 107.35ºC FP soln = FP normal - T f = 0.00ºC – 26.7ºC = –26.7ºC 1.86 ºC · kg H 2 O mol particles 2 mol particles 1 mol NaCl 7.187 mol NaCl 1 kg H 2 O constant (different for each solvent) molality ( m ) =
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Unformatted text preview: Change in BP (add this number to the normal BP) Change in FP (subtract this number from the normal FP) mol solute kg solvent mol particles mol of solute vant Hoff factor = A. Romero 2008 T b = i K b m = 0.512 C kg H 2 O mol particles = 7.35 C 2 mol particles 1 mol NaCl 7.187 mol NaCl 1 kg H 2 O T f = i K f m = = 26.7 C 100C to 0C 107.35C to 26.7C the range of temperatures over which the solution is a liquid is colligatively extended on both ends Pure Water Water w/ Solute gas BP liquid FP solid...
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This note was uploaded on 12/09/2011 for the course CHEM 1 taught by Professor Staff during the Summer '11 term at Simon Fraser.

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