CHEM 2200 1nd Edition Lecture 8 Outline of Last Lecture I. ConcentrationII. Collision TheoryIII. The Arrhenius EquationIV. Effects of Catalysts on Reaction RatesOutline of Current Lecture I. In Class QuestionsCurrent Lecture1. Which of these solutions will have the highest boiling point?a. 0.35 m Ethylene Glycolb. 0.20 m KBrc. 0.50 m Glucosed. 0.20 m Na2SO42. a 0.0200 m aqueous solution of an ionic compound, Co(NH3)4Cl3, freezes at -.064 ° C. Kp for water is -1.86a. Calculate the molality from the freezing point depression value.i. ΔTf = -0.0640ii. m = TΔf/Kfpm = 3.44 x 10-2 m 0.0344 m (effective molality)Stated molality is 0.0200 mb. What is the van’t Hoff factor for this compound?i. i = meffective/mstated 0.0344/0.0200 = 1.72 m3. A 0.100 M acetic acid solution has an osmotic pressure of 2.47 atm at 25° C. What is the percent ionization of acetic acid in the solution?a. What is percent ionization? i. Ratio of # of ions ionized/total # of ionsThese 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.ii. 0.100 M is the denominator b. Using R = 0.08206 L atm / mol K, what is the effective molarity of the solution?i. = MRT; M = /RT = π π2.47atm/[(0.08206 L atm / mol K)(298 K)] = 0.101 M (effective molarity) c. Build ICF chartd. Meffective = (0.100 – x) + x + x 0.100 + x (total concentration in solution)x = 0.001 MPercent ionization = 0.001 M/0.100 M = 0.01 x 100% = 1%(weak electrolyte)4. The smell of ripe raspberries is due to 4C5H6O. To find its molecular formula, you dissolve 0.135 g in 25.0g of chloroform (CHCl3). The boiling point of the solution is 61.82° C. What is the molecular formula of the solute?a. CHCl3: BP = 61.70°C; Kbp = +3.63°C/mAnswer:
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