front 1 Screen Reader Note: To calculate percent error, find the absolute value of the difference between experimental value and known value. Take this and divide it the known value. After that, multiply by 100 %. End of note. Given the known heat of vaporization (delta H vap) of water is 44.01 kJ/mole, find the percent error using the experimentally determined delta H vap of 43.97 kJ/mole. Express with units of % in the units box.  back 1 Percent error = (43.97 kJ/mole  44.01 kJ/mole x 100%) / 44.01 kJ/mole = 0.09089 % > 0.091 % 
front 2 An unknown liquid has a heat of vaporization of 34.13 kJ/mole. If the normal boiling point is 82, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).  back 2 P1 = 760 torr P2 = ? T1 = 82 degrees C + 273 = 355 K T2 = 25 degrees C + 273 = 298 K delta H = 34.13 kJ/mole x (1000 J / 1 kJ) = 34,130 J/mole R = 8.314 K/J/mole ln (P2 / P1) =  (delta H / R) x (1/T2  1/T1) P2 = P1 e^{[(delta H / R) x (1/T2  1/T1)]} P2 = 760 torr e^{[( 34,130 J/mole / 8.314 K/J/mole) x (1/298 K  1/355 K)]} P2 = 83 torr Ex. In a calculator, input: (760)e^((34130/8.314)(1/2981/355)) 
front 3 An unknown liquid has a heat of vaporization of 7.50 kJ/mole. If the vapor pressure of this liquid at 170 degrees C is 83 torr, what is the normal boiling point of this liquid in degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).  back 3 P1 = 760 torr P2 = 83 torr T1 = ? T2 = 170 degrees C + 273 = 103 K delta H = 7.50 kJ/mole x (1000 J / 1 kJ) = 7,500 J/mole R = 8.314 K/J/mole ln (P2 / P1) =  (delta H / R) x (1/T2  1/T1) ln (83 torr / 760 torr) =  (7,500 J/mole / 8.314 K/J/mole) x (1/103 K  1/T1) T1 = 137.85658472 K T1 = 137.85658472 K  273.15 K = 135.29 T1 = 135 degrees C Ex.

front 4 Screen Reader Note: To calculate percent error, find the absolute value of the difference between experimental value and known value. Take this and divide it the known value. After that, multiply by 100 %. End of note. Given the known heat of vaporization (delta H vap) of water is 44.01 kJ/mole, find the percent error using the experimentally determined delta H vap of 47.04 kJ/mole. Express with units of % in the units box.  back 4 Percent error = (47.04 kJ/mole  44.01 kJ/mole x 100%) / 44.01 kJ/mole = 6.8847989 % > 6.9 % 
front 5 An unknown liquid has a heat of vaporization of 39.15 kJ/mole. If the normal boiling point is 85, what is vapor pressure (in torr) of this liquid at room temperature of 25 degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).  back 5 P1 = 760 torr P2 = ? T1 = 85 degrees C + 273 = 358 K T2 = 25 degrees C + 273 = 298 K delta H = 39.15 kJ/mole x (1000 J / 1 kJ) = 39,150 J/mole R = 8.314 K/J/mole ln (P2 / P1) =  (delta H / R) x (1/T2  1/T1) P2 = P1 e^{[(delta H / R) x (1/T2  1/T1)]} P2 = 760 torr e^{[( 39,150 J/mole / 8.314 K/J/mole) x (1/298 K  1/358 K)]} P2 = 53.784 torr P2 = 54 torr Ex. In a calculator, input: (760)e^((39,150/8.314)(1/2981/358)) 
front 6 An unknown liquid has a heat of vaporization of 5.50 kJ/mole. If the vapor pressure of this liquid at 170 degrees C is 111 torr, what is the normal boiling point of this liquid in degrees C? HINT: Normal boiling point occurs when the vapor pressure of the liquid is the same as atmospheric pressure (1 atm or 760 mm Hg).  back 6 P1 = 760 torr P2 = 111 torr T1 = ? T2 = 170 degrees C + 273 = 103 K delta H = 5.50 kJ/mole x (1000 J / 1 kJ) = 5,500 J/mole R = 8.314 K/J/mole ln (P2 / P1) =  (delta H / R) x (1/T2  1/T1) ln (111 torr / 760 torr) =  (5,500 J/mole / 8.314 K/J/mole) x (1/103 K  1/T1) T1 = 147.04434362 K T1 = 147.04434362 K  273.15 K = 126.10565638 T1 = 126 degrees C^{} 