Chem 104 1st EditionExam # 2 Study Guide Lectures: 14-21 Lecture 14 Solute ConcentrationI. Mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution. This is represented in the equation:mole fractionof A , XA=moles solutetotal moles∈solutionMore simply:ntotal=nA+nB+nC+…+nsolventIn a solution that contains A, B, and C, you want to know the mole fraction of B:XB=nBnA+nB+nCI. Remarks concerning mole fractions:A. X (mole fraction) has no unitB. X ≤ 1C. ∑Xi = 1II. Example 1: What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 ml of solution?Given: 17.2 g C2H6O2, .500 kg H2O, 515 ml solution.Find: MTraore changed the problem to this:1. Solve for molarity of solution2. Calculate molality3. Calculate mole fraction of solute XC2H6O2molarity=moles soluteV (l)solutionThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.First convert the g of C2H6O2 to mol remember that to convert to moles you need to divide by the molar mass:17.2 g(1 mol62 g)=.28 molNext convert the 515 ml of solution to L:515 ml(1 L1000 ml)=.515 LSolve for the molarity:molarity=.28 mol.515 L=.54 MNow solve for molality:molality=¿molessolutemass(kg)solventmolarity=.28 mol.500 kg=.56 mFind the mols of H2O:.500 kg(1000 g1 kg)(1 mol18 g)=27.8 gFinally, solve for the mole fraction:XC2H6O2=nC2H6O2nC2H6O2+nH2O=.28 M.28 M +27.8 mol=0.009III. Example 2: Claculate the molarity of a solution made by dissolving 34.0 g of NH3 in 2.00 × 103 ml of solution (MMNH3 = 17.04 g/mol)Given: 34.0 g NH3, 2.00 × 103 ml solution, MMNH3 = 17.04 g/molFind: MM=n(¿moles solute)V(l)solutionFirst convert the given variables:2.00 ×103ml=2.0 L34.0 g(1 mol17.04 g)=2 molNext input them into the equation:M=2.0 mol2.0 L=1.0 MColligative PropertiesA. Colligative properties are properties whose value depends only on the number of soluteparticles, not on what they are. An example of this is freezing point depression. The vapor pressure changes. The difference in the value of the property between the solution and the pure substance is related to the forces and solute particles occupying solvent molecules positions. In other words, the value depends on the concentration of the solution.Vapor PressureA. The vapor pressure of a solvent above a solution is lower than the vapor of the pure solvent. Adding a nonvolatile solute reduces the rate of vaporization thus decreasing theamount of vapor. Eventually equilibrium will be established.B. Raoult’s Law describes the vapor pressure of a volatile solvent above a solution is equal to its normal vapor pressure multiplied by its mole fraction in the solution. This is represented in the equation:P°solvent∈solution=Xsolvent∗P°Mole fraction ˂ 1, therefore the vapor pressure of the solvent in solution will always be ˂vapor pressure of pure solventC. Example: Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C22H22O11 with 300.0 ml of water.Given: 99.5 g C12H22O11, 300.0 ml H2OFind: PH2OPH 2 O ∈solution= XH 2 O∗P °H 20Convert all variables to moles:99.5 g(1 mol342.30 g)=.2907 mol300.0 ml(1 g1 ml)(1 mol18.02 g)=16.65 molSolve for C12H22O11:16.65.2907+16.65=.9828Solve for PH2O:P° =23.8 torrPH2O=XH2O∗P °H2OPH2O=(0.9828) (23.8 torr)=23.4 torrLecture 16 Raoult’s LawI. Example 1: Calculate the total vapor pressure of a solution made by dissolving 25.0 g of glucose (C6H12O6) in 215 g of water at 50°CGiven: 25.0 g C6H12O6, 215 g H2O, P°= 92.5 torr, MM C6H12O6= 180.2 g, MM H2O = 18.02gFind: PH2OPsolvent∈solution=Xsolvent∗P °First convert g to mol in both H2O and C6H12O6:1mol180.2 g=¿¿25.0 g ¿215 g(1 mol18.02 g)=11.93molNow solve for mole fraction:XH 2O=11.93 mol(0.1387 mol+11.93mol)=0.9885Finally solve for PH2O:PH 2 O=(0.9885) (92.5torr)=91.4 torrVapor Pressure LoweringI. An easy way of viewing vapor pressure lowering is by putting it in terms of what it is equal and not equal to:A. Vapor Pressure Solvent in solution ˂ Vapor pressure of pure solventB. Vapor pressure of solution = amount of solvent in solutionC. Vapor pressure of pure solvent – vapor pressure in solute = vapor pressure lowering.D. It is represented in the equation:∆ P=P °solvent−Psolution= Xsolute∗P° solventNotice that the last part of the equation is Raoult’s law. Another way of viewing Raoult’s law:Xsolute∗P °=(1−Xsolvent)P °solventRaoult’s Law for volatile soluteI. When both the solvent and the solute can evaporate, both molecules will be found in the vapor phase:Ptotal=Psolute+PsolventYou can apply Raoult’s law because their volatile. The solute vapor pressure decreases in the same way the solute decreases the solvents:Psolute=Xsolute∗P °Psolvent= Xsolvent∗P °solventII. This problem needs to be solved and turned into class for extra credit:A. Calculate the component and total vapor pressure of a solution prepared by mixing 3.95 g of CS2 with 2.43 g of C3H6O.Given: 3.95 g CS2, 2.43 g C3H6O, P°CS2=515 torr, P°= 332 torrFind: PCS2, PC3H6O, Ptotal Lecture 17 II. 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 °solventIII. 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