CHEM 1442 1st Edition Lecture 9Outline of Last Lecture I. Raoult’s LawII. Reaction MechanismIII. CatalystsOutline of Current Lecture IV. ExamplesCurrent Lecture* Important Technical Skills1. Be able to use the definitions to calculate the concentration2. Convert the concentration from one unit to anotherExamples1. Mass Percent  Molality Calculate the molality of a 35.4 percent (by mass) aqueous solution of phosphoric acidMolality (m) = moles of solute (H3PO4) / kilograms of solvent35.4% by mass = 35.4 g of H3PO4 / 100 g of solutionMass by solvent = 100 g – 35.4 g = 64.6 g of solventMolality (m) = 5.59 m2. When 0.981 g of an unknown sample is dissolved in 11.23 g of water, the solution freezes at -1.26oC. Calculate the molar mass of this unknown.Molar mass of unknown = mass (g) of unknown (0.981 g) / moles of unknownThese 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.∆T = Kf x mThe solvent water has a melting point (oC) of 0.0 and a molal freezing-point depression constant,Kf (oC/ m) of 1.86.1.26 oC = 1.86 x mm = 0.677 m0.677 moles of solute/ kg of solvent x 0.01123 kg of water = 0.00760 moles of soluteMolar mass = 0.981 g / 0.00760 moles = 129 g/ mol3. Consider the reaction mechanism:Step 1: Cl2 (g)  2 Cl (g)Step 2: Cl (g) + CHCl3 (g)  HCl (g) + CCl3 (g)Step 3: CCl3 (g) + Cl (g)  CCl4 (g)Determine overall reaction:Cl2 (g) + CHCl3 (g)  HCl (g) + CCl4 (g)4. CCl3 (g) + Cl (g)  CCl4 (g)Is this uni-molecular, bi-molecular, or ter-molecular?It is bi-molecular.* Because this is an elementary reaction, it cannot be ter-molecular. 5. Cl2 (g)  2 Cl (g)Determine the rate law of the elementary reaction:Rate = k1 [Cl2]*Because it is an elementary reaction, the reaction order = stoichiometry