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Unformatted text preview: Chemistry 101 Chapter 12 BOILING POINT ELEVATION
· Normal Boiling Point (BP) of a liquid is the temperature at which the vapor pressure of the liquid equals 1.00 atm I. Effect of a Dissolved Solute on the BP of a Solution What happens if a nonvo lat ile so lute such as ethylene glyco l (C2H6O2) is added to pure water ? Vapor Pressure of Solut ion < Vapor Pressure of Pure Solvent (water) P(A + B) < PA
· It fo llows: 1. The normal BP of solut ion (A+B) is the temperature at which P(A+B) is equal to 1.00 atm 2. This requires a temperature higher than for the pure liquid (water) 3. BP(solut ion) > BP (pure solvent) ® B.P. Elevat ion
DTb = Boiling Point Elevat ion
DTb = BP (solution) - BP (solvent)
· Boiling point elevat ion is a Colligat ive Property Ø depends on the concentration of the Solut ion Ø does not depend on the nature of the Solute DTb = KbCm Cm = Concentration in mo lalit y Kb = Boiling Point Elevat ion Constant (units of °C/m) 18 Chemistry 101 Chapter 12 PHASE DIAGRAM FOR WATER AND AN AQUEOUS SOLUTION OF ETHYLENE GLYCOL 1.00 atm Freezing Curve for Pure Solvent Freezing Curve for Solution Solid Vapor Pressure 0 t ( C) FP of BP of Solvent Solvent FP of BP of Solution Solution DTf DTb 19 Chemistry 101 Chapter 12 Examples: 1. At what temperature would a a 5.00 mo lal so lut ion of ethylene glyco l bo il? (Kb(water) = 0.512 oC/m) DTb = (0.512 oC/m) (5.00 m) = 2.56 oC BP(solution) = 100.00 oC + 2.56 oC = 102.56 oC 2. A so lut ion was made of eugeno l in diethyl ether (ether). If the solut ion was 0.575 m eugenol in ether, what was the Boiling Po int of the solut ion ? BP(diethyl ether) = 34.6 oC Kb(ether) = 2.02 oC/m) DTb = (2.02 oC/m) (0.575 m) = 1.16 oC BP(solut ion) = 34.6 oC + 1.16 oC = 35.8 oC 20 Chemistry 101 Chapter 12 FREEZOMG POINT DEPRESSION
· From the phase diagram, the vapor pressure curve for the solid solut ion is unchanged.
· As the temperature of a solut ion is lowered, the pure solvent freezes out of solut ion. (Sea ice is almost pure water)
· FP of the solut ion shift s toward a lower temperature (F. P. Depression)
DTf = Freezing Po int Depression
DTf = FP(solvent) - FP (solut ion) · Freezing point depression is a Co lligative Property Ø depends on the concentration of the Solut ion Ø does not depend on the nature of the Solute DTf = KfCm Cm = Concentration in mo lalit y Kf = Freezing Point Depressio n Constant (units of °C/m) Practical Applications 1. Use of ethylene glyco l as ant ifreeze in car radiators: Ø lower the FP of coolant Ø elevates the BP of coolant (prevents it from bo iling away) 2. NaCl poured over icy roads lowers the FP of ice below thetemperature of the surrounding air. As a result, the salted ice melts. 3. To obtain Mo lecular Weights of compounds 21 Chemistry 101 Chapter 12 Examples: 1. How many grams o f ethylene glyco l (C2H6O2) must be added to 37.8 g of water to give a Freezing Point of - 0.150 oC ? (Kf = 1.858 °C/m) DTf = FP (solvent) – FP(solut ion) = 0.000 oC – ( - 0.150 oC ) = 0.150 oC DTf = Kf Cm DTf 0.150 oC Cm = = = 0.0807 m K f 1.858 o C/m ? g EG = 37.8 g w ater x 0.0807 mol EG 62.08 g x = 189 g 1000 g w ater 1 mol 0
2. An aqueous solut ion of a mo lecular co mpound freezes at – 0.086 C. What is the mo lalit y o f the 0 solut ion? (Kf = 1.858 C/m) 0
DTf = 0.000 C - (-0.086 C) = 0.086 C DTf = Kf x m DTf 0
0.086 C m = ¾¾¾¾ = ¾¾¾¾¾¾ = 4.6 x 10 2 m Kf 1.858 0 C/m 22 ...
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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.
- Summer '11