CHAPTER 13 (cont.) LECTURE 2

CHAPTER 13 (cont.) LECTURE 2 - CHAPTER 13 (cont.) LECTURE...

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CHAPTER 13  (cont.) LECTURE 2: Boiling-Point Elevation : Non-volatile  solute lowers the vapor pressure. Remember that the normal boiling point of pure     liquid is when the vapor pressure = 1 atm. Therefore, for  solutions,  because the vapor pressure     is lowered, a higher  temperature is required to     reach a vapor pressure of 1 atm for the solution     and we get:  Boiling-Point Elevation = T b (NOTE: THIS IS NOT THE BOILING POINT, IT IS THE CHANGE IN BOILING TEMPERATURE!!). €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€ T b = i K b m Molal boiling-point-elevation constant , K b :     expresses how much  T b  changes with molality,  m. For water, K b  = 0.52  o C/m;  normal b.p. = 100 o C For phenol, K b  = 3.56  o C/m; normal b.p. = 182 o C For benzene, K b  = 2.53  o C/m; normal b.p. = 80.1 o C Freezing-Point Depression : Due to the disruption of the intermolecular forces     in the solvent (solute particles are in the way,     they are attracted to the solvent, etc.) a solution     freezes at a lower  temperature than the pure The decrease in freezing point,  T f ,  is directly     proportional to molality. (K f  is the molal     freezing-point depression constant).
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€€€€€€€€€€€€€€€€€€€€ ∆ T f = i K f m For water, K f  = 1.86  o C/m; normal f.p. = 0 o C For phenol, K f  = 7.40  o C/m; normal f.p. = 43 o C For benzene, K f  = 5.12  o C/m;  normal f.p. = 5.5 o C Problem: What will be the b.p. of a benzene solution                  which freezes at 4.00  o C?  €€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€€    1)  First, find the molality (m) of the solution:
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This note was uploaded on 04/05/2008 for the course CHEM 116 taught by Professor Lalancette during the Spring '08 term at Rutgers.

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CHAPTER 13 (cont.) LECTURE 2 - CHAPTER 13 (cont.) LECTURE...

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