Principles of Chemistry
Principles of Chemistry CHE 115
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Date Created: 10/11/15
cm 115 Nonelectrolyles XI Solutions of Nonelectrolytes Types 0fSolutes Solutes that exist as dissociated ions in aqueous solutions are called electrolytes Solutes that are present as neutral molecules and not as ions in solutions are called nonelectrolytes A Raoult s Law In solutions that have a volatile solute and a volatile solvent the partial vapor pressure of each component is less than the vapor pressure of the pure component at the same temperature 0 is a molecule of solvent A O is a mOIeCule of solvent A Q is a molecule of solute B O lt P vapor pressure Vapor A vapor p of pure A CIUld gt solution p Pure Solvent A in a Solution of Solvent A and closed container Solute B Assumption The van der Waals forces between the solute solute molecules in the pure solute are identical to the van der Waals force between solvent solvent molecules in the pure solvent and hence the van der Waals forces between the solute and solvent molecules in the solution are the same as the forces in the pure components As solute is added to solvent the number of solvent molecules at the surface of the solution decreases and therefore there will be fewer solvent molecules with suf cient KET to break away and enter the vapor phase There will be fewer solvent molecules in the vapor above the solution than in the vapor above the pure solvent Consequently the partial vapor pressure of the solvent A PA for a solution will be less than the vapor pressure ofpure solvent A P Note PA nARTV A plot of the partial vapor pressure of a component versus the mole fraction of the component illustrates this concepts Concentration expressed as mole fraction cm us mainNW mule fracnun W XA mole fractmn ufB XE Nate XA XE 1000 where nA moles of solvent A and nB moles ofsolute B Partial Vapor Pressures of CCIA and Is He versus XCCII 300 Fem 268 orr 250 7 Wm SUBI r mm X 200 j Pccu a yr Pam al Pfessur u E 6 o o a 0 00 02 04 06 as 10 X ccr 39 39 39 ofcc14 is n 7 39 39 wporr of39V L39 o torr and the partial Vapor pressure ofaeemne is 191 at 50 c Note the partial Vapor pressures are linear functions ofthe mole fractions The equation for a straight line is y Slupe Intercept Uzk y x yx n Xz a For the partial Vapor pressure ofccl4 y Pm X Xm P P P 0 pm MXM 03ka n L XCC2XCC W 1000 CHE 115 Nonelectrolytes Pcci4 P 0001 X001 This equation is the mathematical expression of Raoult s Law for CCl4 Raoult s Law states that the partial vapor pressure of a component is approximately equal to the mole fraction of the component in solution times the vapor pressure of the pure component at the same temperature mmwm Raoult s Law is a good approximation of the partial vapor pressure of each component in solutions where the solute and solvent molecules are similar in size and polarity The total vapor pressure of the solution is given by Dalton s Law of Partial Pressures PT PB PA Substituting Raoult s Law PT PB PA XBPOB XAPOA Example Calculate the total vapor pressure of a benzene C6H6 carbon tetrachloride CCl4 solution for which the mole fraction of CCl4 is 0100 at 50 C The vapor pressures ofpure C6H6 and pure CCl4 at 50 C are 268 and 308 torr PT P0014 l Pc H XCCZAPGCCZA l Xc H ch H Xc H 7 1000 XCCZA 7 1000 0100 7 0900 PT 7 0100308 torr i 0900268 torr 7 2721077 A solution in which the partial vapor pressures of the solute and solvent are given approximately by Raoult s Law is said to be ideal solution B Deviations from Raoult s Law Nonideal Solutions 1 Positive Deviation When the van der Waal s forces between the solute and solvent molecules in the solution are weaker than the forces of attraction between the solute solute molecules in the pure solute and the solvent solvent molecules in the pure solvent then the partial vapor pressure of each component in the solution is greater than the pressure predicted by Raoult s Law Such solutions are said to exhibit a positive deviation and to have heats of solution that are endothermic AHsoln gt 0 Example Acetone carbon disul de solution at 352 C 11 HscCCH3 sCs Acetone Carbon Disulfide Polar Nonpolar cm us Mm Partial Vapor Pressures of CS2 and Acetone versus X65 E E cu E V E zoo E 150 e 100 39gtltcsP cs t so 7 lt s x XMMPML c on 02 on as as 1D Xcsz where PC and PMm are the expermemally determmed Vapor prexmrex and the 2 P251 and mepvmm are the Vapor premium predtcted by Ramlr x Law 2 Negative Deviation n t e van der are Wang L L c A L y w r c L t is Law ch quotquot u Kati C S 4 that are exothermic my lt 0 Example Acetone chloroform solution at 352 C hydmgen bundmg HE C Cl c o H C c I HE C Cl Acetone chloroform Em us mummies Partial Vapor Pressures of HCCI3 and etone versus X Hccll 350 Vie Pfazclme 300 s 250 7 xieetpieetj Pamal Pressure tan c Henry s Law In avery dzlutz mlutzun lg g 005 the partial Vapor pressure ofthe solute is approximately proportional to the mole fraction xsm The proportionality constant k5 called enry s constant is experimentally determined The magnitude ofkmm is dependent on the temperature and the nature ofthe solvent Prom homicide Example Ethyl alcohol mter solution at 20 c Partial Vapor Pressures of Ethyl Alcohol EtOH and Water versus NTo at 20 C 4 ksiceixsica am PM 5 XEKUHP nos Partial Pressure mu x 5 VXuaFua D s on u2 M 05 us 10 X EKOH CHE 115 Nonelectrolytes In a very dilute solution Xsolute s 005 the partial vapor pressure of the solute is given approximately by Henry s Law and the partial vapor pressure of the solvent is given approximately by Raoult s Law PP kXXP solute solute solvent 0 T solute Psolvent solvent D Colligative Properties of Solutions Any property of a solution that is approximately proportional to the number of solute molecules per unit quantity of solvent and independent of the nature of the solute is said to be a colligative property 1 Vapor Pressure Depression Consider a very dilute solution that contains a volatile solvent A and a nonvolatile solute B The vapor pressure of the solution is given approximately by Raoult s Law PA XAPOA Since XA l XB PA l XBPquotA POA XBPquotA Rearranging nB 39 APAP0APAXBP0A POA nBnA7 In a very dilute solution nB lt lt nA and thus nB nA w nA 0 n3 0 APA 7 PA PA P quotA7 where APA is called the vapor pressure depression Example Calculate the depression in the vapor pressure of water when 200 g of sucrose C12H22011 MW 342 gmole are added to 100 g ofwater at 25 C The vapor pressure for pure water at 25 C is 238 torr 7 555mole and n 7 58410 3mole n L mm 342 gmole 7 100g H10 180gmole nHZO 1 nmmz 7 555mole 1000584mole 7 5553mole 7 556mole 555mole 7 nHZO n 3 APHO 7 O 7 lmlammm 7 25010 21077 1 nHZO 1 555mole Em us mummies 2 BoilingPoint Elew on and FreezingPoint Depression ase Diagram lur Pure Salvequot Ph t A and for a Soution of Solvent A and Nonvolatile Solute B mm L mu mummy iv a mu L sun L 500 L u A equlhllmm ADD L minim in minimum W mu s Fressuvei ton son L 200 L 130 L 0 AD 72039 w Get is the boiling pomt ofpureAt is theborlingpomt othe solutlont is the ereezmg pomt ofpureA andt isthe ereezingpomt othe solution prure liquid A is boiling and anonvolatile solute B is added the solution will cease to boil With the addition ofsoluteBthe Vapor pressure ofA is depressed by PA since apoir CL 39 39 is L L 39 39 39 boiling p In order to resume boiling the Wpor pressure ofthe solution must be increased unt39t 39 1 39 L L L u diagram the apoir tL 39 39 39quot 1 39L 39 39 760 torr at temperature 1 The increase in temperature Alb tb 11 required to resume 39 39 39 39 39 he 39 39 Baddedtoa i n 39L t 39 A K39 the andm39 L 39 where Alb magnitude obe39 A r A the solute expressed as mommy CHE 115 Nonelectrolytes molality of the solute m M units M in kilograms of solvent Kg Recall that molarity is de ned as Molarity of the solute solute m units M M liters of solution L The addition of solute B to the solvent A will also result in a lowering of the freezing point by Atf The freezing point tf of the solution is given by the following equation Alf t f If Kfm where Atf is the freezingpoint depression Kf is the freezingpoint depression constant magnitude of Kf is dependent on the nature of the solvent and m is the molality of the solute Example Calculate the boiling point of a solution that contains 200 g of sucrose in 100 g of water Kb for the solvent water is 0512 CKgmole and the freezing point of pure water t b is 10000 C 200g m 7 mole ofsucrose 7 342gmole 7 584104quot Kg of water 0100Kg 1b 7 Kbm i r17 7 0512 C Kg 584102ng 1 10000 C 7 10003 C mole Kg Example A solution that contains 0100 g ofan unknown solute in 100 g of benzene C6H6 has a freezing point of5 10 C Calculate the MW of the unknown solute Kf for benzene is 512 CKgmole and tquotf 551 C for benzene Unknown MW of unknown solute Knowns wt of solute 0100 g wt of solvent C6H6 100 g tf 510 C t f 551 C Kf for C6H6 512 C Kgmole Concepts Atf Kf m m moles of soluteKg of solvent Relationship wt of solute Kf Alf 7 I f If 7 Kfm 7 Kg of solvent Rearranging MW Kf wt of solute 5017 7 AtfKg of solvent MW 512 C Kgmole 0100g 7 123 mole 7 12102 m0le 50W 551 C 510 C00100Kg g g CHE 115 Nonelectrolytes 3 Osmotic Pressure Consider a Utube that contains pure water on the le side of a membrane permeable only to water and an aqueous solution of sucrose on the right side Membrane permeable only to water Water molecules escape from the liquid enter the vapor phase and pass through the membrane Since the vapor pressure of pure water is greater than the vapor pressure of the solution more molecules of water will pass through the membrane from le to right than from right to le Hence there will be a net ow of water from le to right ie from the pure water to the solution The ow from pure solvent to solution or from a dilute solution to a concentrated solution is called osmosis An increase in the external pressure on a liquid will result in an increase in the vapor pressure of the liquid 17dsz vapor vapor liquid dPliquid Where the V with the overline is the volume of one mole of the vapor or of one mole of the liquid dPWOr is the change in vapor pressure and dPhqm01 is the change in the external pressure on the liquid Since 7 781 mL and v 7 18 mL atZS C Vwatzr vapor water 11 gum a large increase in the external pressure dPWmhqmd 434 torr will be required to effect a small increase in the vapor pressure dP 10 torr watemapor The external pressure P2 on the solution can be increased relative to P1 until the vapor pressure of the solution equals the vapor pressure of the pure water and the ow from right to le equals the ow from le to right The diiTerence in pressure P2 P1 required to stop the net ow and keep the solution in equilibrium with the pure water is called the osmotic pressure H The magnitude of H depends on the molarity of the solution