Chem 104 1st Edition Lecture 16 Outline of Last Lecture I. Percent concentration and how to use concentration as a conversion factorOutline of Current Lecture I. Solution Concentration and mole fractionsA. Remarks concerning mole fractionsB. Example of how to solve for mole fractionsII. Colligative propertiesIII. Vapor pressure and Raoult’s lawA. Solving for a vapor pressure by using Raoult’s lawCurrent LectureI. Note that there will be a quiz on Thursday February 19th on how to solve the examples of:A. volume of 10.5% by mass soda (Lecture 13)B. How to prepare 250.0 ml of 19.5% by mass CaCl2 (Lecture 13)C. Finding the molarity, molality, and finally the mole fraction of solute XC2H6O2Solute 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+nCII. Remarks concerning mole fractions:A. X (mole fraction) has no unitB. X ≤ 1C. ∑Xi = 1III. 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)solutionFirst 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=0.009IV. 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