Chem 104 1st Edition Lecture 17 Outline of Last Lecture I. Continued practice with Raoult’s law.A. Example problems to work with concerning Raoult’s lawII. Vapor Pressure Lowering and its principles and equationIII. Raoult’s law for a volatile soluteIV. Example problem due for extra credit.Outline of Current Lecture I. Ideal vs. Nonideal solutionsII. Vapor pressure of a nonideal solutionIII. Colligative properties related to vapor pressure loweringIV. Freezing point depressionA. Example problems concerning freezing point depression. V. Introduction to OsmosisCurrent LectureI. An ideal solution is made from solute-solvent interactions that are equal to the sum of broken solute-solute and solvent-solvent interactions. Overall, ideal solutions follow Raoult’s Law. Effectively, the solute is diluting the solvent.A nonideal solution is when the solute-solvent interactions are weaker or stronger than the interactions of solute-solute or solvent-solvent solutions. This is shown in the equation below:∆ P=Psolvent−Xsolute∗P °solventII. When everything is perfect, the partial pressure of solute-solvent solution is equal to the vapor pressure. The solute- solute interactions are represented by A and the solvent-solvent interactions are represented by B below. > Represents stronger than and < represents weaker than. If:AB>solute−solventThis means the solution is less than ideal. If:AB<solute−solventThen the solution is higher than ideal.A. Example: The experimentally measured total vapor pressure of the solution is 645 torr. Isthe solution ideal? If not, what can you say about the relative strength of carbon disulfide-acetone interactions?Given: Ptotal(expt)=645 torr, Ptotal(ideal)=443 torrPtotal(expt)=645 torr>Ptotal(ideal)=443 torrThe solution is not ideal and shows positive deviations from Raoult’s Law. So, carbon disulfide-acetone interactions must be weaker than acetone-acetone and carbon disulfide-carbon disulfide interactions.III. Vapor pressure lowering happens at all temperatures. This results in the temperature required to boil the solution has to be higher than the boiling point of the pure solvent. This also means the temperature required to freeze the solution has to be lower than the freezing point of the pure solvent.A. Example: P°H2O boiling point = 100°C, If you added salt to make it salt water its boiling point would lowered. The answer to this is PNaCl boiling point = 10.5°C.This is represented in the equation:∆ Tf=T °f−Tf=kf∗mIV. The freezing point of a solution is lower than the point of the pure solvent. This is represented with the equation:(F Psolvent−F Psolution)=∆ Tf=kf∗mThis is basically the same equation as ∆Tf above.A. Example: What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2?Given: 1.7 m C2H6O2 (aq), kf=1.86 °C/m, FPH2O=0.00Find: Tf °C∆ Tf=kf∗m∆ Tf=(1.7 m)(1.86° Cm)=3.2° CThen calculate the FPH2O:F PH 2 O−F Pso l'n=∆ Tf0.00 °C−F Psol'n=3.2 °CF Psol 'n=−3.2 ° CThis is called colligative because it depends on the numbers not the nature.B. Example: Calculate the molar mass of a compound if a solution of 12.0g dissolved in 80.0g of water freezes at -1.94°C. (kf water = 1.86°C/m)Given: masssolute=12.0g, massH2O=80.0g, FPsol’n = -1.94°C, FPH2O=0.00°CFind: T°f,, °CFirst convert g to kg then solve for FPH2O:80.0 g=0.0800 kgF PH2 O−F Pso l'n=∆ Tf0.00 °C−(−1.94)=1.94 °CNow find the molality:m=∆TfKf , H 2 Om=1.94 °C(1.86°Cm)=1.043 mNow solve for mol:mol=(1.043molkg)(0.0800 kg)=0.08344 molFinally find the molar mass:MM=masssolutemolessoluteMM=12.0 g0.08344 mol=144 g/molV. Osmosis is the flow of solvent from a solution of low concentration to a solution of high concentration. The solutions may be separated by a semi-permeable membrane. Semi-permeable membrane means that a solvent is allowed through it but not a