{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

BP Elevation & FP Depression (Chapter 7)

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ) = Change in BP
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 van’t Hoff factor = A. Romero 2008 T b = i K b m = 0.512 ºC · kg H 2 O mol particles = 7.35 ºC 2 mol particles 1 mol NaCl 7.187 mol NaCl 1 kg H 2 O T f = i K f m = = 26.7 ºC 100ºC to 0ºC 107.35ºC to –26.7ºC 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...
View Full Document

{[ snackBarMessage ]}