Principles of Chemistry II © Vanden BoutDifferent ways to describe concentrationAll of them are essentiallyAmount of soluteAmount of everything (solvent) Principles of Chemistry II © Vanden BoutMole Fraction!i =moles of itotal molesMolalitym =moles of solutekg of solventMolarityM =moles of soluteL of solution Principles of Chemistry II © Vanden BoutDemo Principles of Chemistry II © Vanden BoutWhat is the key effect of adding the salt to the water?! A.! ! the salt dissolving is endothermic so the temperature drops ! B.! ! the salt dissolving is exothermic so it melts the ice! C.! ! the salt dissolving increases the entropy of the solution! D.! ! the salt dissolving increases the entropy of the solid ice!Principles of Chemistry II © Vanden BoutWhy does the temperature drop?! A.! ! the salt dissolving requires energy (endothermic) ! B.! ! the salt dissolving releases energy (exothermic)! C.! ! the ice melting releases energy (exothermic)! D.! ! the ice melting requires energy (endothermic)! Principles of Chemistry II © Vanden BoutSolutionsThe main effect of making a solution is that the entropy of the solution is higher than the separate solvent and soluteT = 0°C and P = 1 atmGsolid water liquid waterthey have the same free energy at equilibriumsolutionthe solution has a higher entropy and therefore a lower free energyNOW MOST STABLEice will melt to get intothe lower free energy solution Principles of Chemistry II © Vanden BoutThis effect depends on the entropy of the solution which depends on how much "stuff" is dissolved but not what the "stuff" isColligative Propertiesdepend on the concentration of the solutionbut not what is actually dissolved(note: this is approximate as it assumes and ideal solution)The only thing that matters is the number of moles of "stuff" Principles of Chemistry II © Vanden BoutSomethings dissolve into ions making more moles1 M sugar solution = 1 moles of sugar in 1 L solution1 M NaCl solution = 1 moles of Na+ in 1 L solution I mole of Cl- in 1 L solution 2 moles of "stuff"Principles of Chemistry II © Vanden BoutVan't Hoff Numberi =moles of "particles" in solutionmoles of solute dissovledFigure Copyright Houghton Mifflin Company. All rights reserved Principles of Chemistry II © Vanden BoutEffect of making the solutionBoiling Point ElevationFreezing Point DepressionSolution now more stable than vapor. Therefore the boiling point goes upSolution now more stable than solid. Therefore the freezing point goes down Principles of Chemistry II © Vanden BoutBoiling Point Elevation!T = Kbmsolutemolality soluteconstant that depends on solvent!T = iKbmsoluteRemember the number of particles is what matters Principles of Chemistry II © Vanden BoutFreezing Point Depression!T = -Kfmsolutemolality soluteconstant that depends on solvent!T = -iKfmsoluteRemember the number of particles is what mattersPrinciples of Chemistry II © Vanden BoutFigure Copyright Houghton Mifflin Company. All rights reserved Principles of Chemistry II © Vanden BoutWhich would you expect to have the lowest freezing point! A.! ! 2 M sugar solution ! B.! ! 0.5 M NaCl solution! C.! ! 1 M NaCl solution! D.! ! 1 M MgCl2 solution! Principles of Chemistry II © Vanden BoutIf the boiling point is higher,what is the vapor pressure of the solution?! A.! ! higher than the pure solvent ! B.! ! lower than the pure solvent! C.! ! the same as the pure solvent! Principles of Chemistry II © Vanden BoutRaoult's LawPsolvent = XsolventP°vapor pressure of pure solventmole fraction of solvent!Principles of Chemistry II © Vanden BoutOsmosisSolvent can pass through the membrane but the solute can'tSolution is lower in free energy so pure solvent moves to the solution side Principles of Chemistry II © Vanden BoutReverse Osmosis Principles of Chemistry II © Vanden BoutOsmotic Pressure" = MRT " = iMRT Molarity of solution!Osmotic pressure Principles of Chemistry II © Vanden BoutCellshigh conc "stuff"pure waterhigh conc "stuff"pure waterExamplesSolution of 100 g of sugar (sucrose MW 342 g mol-1) in 1 L of water.(100 g)/(342 g mol-1) = 0.292 mol sugar1 L water is approx. 1 kg(1000 g)/(18 g mol-1) = 55.6 molesMole fraction sugar of solutionχsugar = (0.292 mol)/(0.292 + 55.6) = .00522Mole fraction water of solutionχwater = (55.6 mol)/(0.292 + 55.6) = 0.995 (or 1- χsugar)Molalitym = (.292 mol)/(1 kg) = 0.292 mol kg-1MolarityM = (.292 mol)/(1 L) = 0.292 mol L-1Freezing point depressions (given Kf for water is 1.86)ΔT = -iKfm= -(1)(1.86)(.292) = -0.543 °CBoiling point elevation (given Kb for water is 0.51)ΔT = -iKbm= -(1)(0.51)(.292) = +0.15 °COsmotic Pressure (at 25°C)Π = MRT= (1 mol L-1)(0.08206 L atm K-1 mol-1)(298.15 K) = 24.5 atmVapor Pressure (given pure vapor pressure of water at 25°C is 23.76 Torr)PH2O = χH2OP° = (.995)(23.76) = 23.64
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