Spring 2006 Laude CH302 Worksheet 3 covering Chapter 8 on Physical Equilibrium 1. Separate these compound into those that are polar (water soluble) and a non-polar( hydrophobic): NaCl, NaCOOCH3, cooking oil, HCl, CH4, CH3OH, CH3CH2OH, gasoline, N2 Water soluble Non-polar solvent soluble NaCl, NaCOOCH3, HCl, CH3OH, CH3CH2OH cooking oil, gasoline, CH4 2. Rank the following substance in increasing order of vapor pressure: NaCl, NH3 solution, Vinegar (CH3COOH), cooking oil, HCl, CH4, He and H2O. NaCl, cooking oil, H2O, Vinegar, HCl, NH3 solution, CH4, He 3. Rank the following solutes in terms of increasing solubility in the solvent: Solvent Solutes Increasing solubility Water NaCl, H2S, CH4, NH3 CH4, H2S, NH3 ,NaCl Hexane NaCOOCH3, cooking oil, CH3OH, N2 N2, NaCOOCH3, CH3OH, cooking oil 4. A liquid is heated at atmospheric pressure. For each of the properties listed, predict whether they would increase or decrease in magnitude.. (a) Viscosity Decrease (b) Density Decrease (c) Surface Tension Decrease (d) Vapor Pressure Increase (e) Tendency to Evaporate Increase 5. Calculate the amount of heat (J) required converting 180 g of water at 10.0°C to steam at 105.0°C. Use constants found in the lecture notes for this calculation. 180 g H2O x (2.26 x 10^3 J/g) = 4.07 x 10^5 J 180 g H2O x (2.03 J/g. °C) x (105.0°C – 100.0°C) 1.8 x 10^3J = 0.018 x 10^5 J Total heat = 6.77 x 10^4 J + 4.07 x 10^5 J + 0.018 x 10^5 J = 4.76 x 10^5 J 6. The molar heat of fusion, ∆Hfus, of Na is 2.6 kJ/mol at its melting point, 97.5°C. How much heat must be absorbed by 5.0g of solid Na at 97.5°C to melt it? 5.0 g Na x (1 mol Na/23 g Na) x (2.6 kJ/1 mol Na) = 0.57 kJ 7. How much heat would be required to convert 234.3 g of solid benzene, C6H6(s), at 5.5 °C into benzene vapor, C6H6(g), at 100.0 °C? Benzene has the following molar heat capacities: C6H6(l) = 136 J/mol °C, and C6H6(g) = 81.6 J/mol °C. The molar heat of fusion for benzene is 9.92 kJ/mol and the molar heat of vaporization for benzene is 30.8 kJ/mol. The melting point of benzene is 5.5 °C; and the boiling point of benzene is 80.1 °C. Benzene's molecular weight is 78.0 g/mol.8. Choose the ion in each pair that would be more strongly hydrated in aqueous solution and justify your answer: (a) Na+ or Rb+ (b) Cl- or Br- (c) Fe3+ or Fe2+ (d) Na+ or Mg2+ Charge density arguments explain the answers. The bigger the charge density, the more hydrated. 9. In Denver, the partial pressure of oxygen is 0.17 atm. What is the molar solubility of oxygen there at 20oC? Henry Law constant for oxygen at 20oC is 0.0013 mol/L atm. S = kHP = (.0013 mol /L atm * .17 atm) = 2.21e-4 mol/L 10. What is the molality of a solution that contains 128 g of CH3OH in 108g of water? 128 g CH3OH x 1 mol CH3OH = 37.0 mol = 37.0 m 0.108 kg H2O 32.0g CH3OH kg H2O 11. (a) How many grams of H2O must be used to dissolve 50.0 g of sucrose, C12H22O11, to prepare a 1.25 m solution of sucrose? 50.0g C12H22O11 x 1 mol C12H22O11 = 0.146 mol C12H22O11 342 g C12H22O11 0.146 mol C12H22O11 = 0.117 kg H2O = 117 g H2O 1.25 mol C12H22O11/ kg H2O (b) Predict the boiling point of this solution; Kb for H2O is 0.512°C/m. ∆Tb = (0.512°C/m)(1.25m) = 0.640°C; BP= 100°C + 0.64°C = 100.64°C (c) Calculate the freezing point of this solution; Kf for H2O is 1.86°C/m. ∆Tf = (1.86°C/m)(1.25m) = 2.32°C Tf(solution) = 0.00°C – 2.32°C = -2.32°C (d) What osmotic pressure would this solution exhibit at 25°C? Its density is 1.34g/mL. 167 g x 1mL = 125mL = .125L 1.34g Msucrose = 0.146 mol = 1.17 mol/L 0.125 L π = MRT = (1.17mol/L)(0.0821 L.atm/mol.K)(298K) = 28.6 atm 12. What are the mole fractions of CH3OH and H2O in the solution described in #1? It contains 128 g of CH3OH and 108 g of H2O. 128g CH3OH x 1 mol CH3OH = 4.00 mol CH3OH 32.0 g CH3OH 108 g H2O x 1 mol H2O = 6.00 mol H2O 18.0 g H2O XCH3OH = 4 mol = 0.400; XH2O = 6 mol = 0.600 (4+6) mol (4+6) mol13. (a) At 40°C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution. Xheptane = 1 mol heptane = 0.200; Xoctane= 1 - Xheptane = 0.800 (1 mol heptane + 4 mol octane) Pheptane = XheptaneP0 heptane = (0.2)(92.0 torr) = 18.4 torr Poctane = XoctaneP0 octane = (0.8)(31.0 torr) = 24.8 torr Ptotal = Pheptane + Poctane = 18.4 torr + 24.8 torr = 43.2 torr (b) Calculate the mole fractions of heptane and octane in the vapor that is in equilibrium with this solution. Xheptane = Pheptane = 18.4 torr = 0.426; Xoctane= Poctane = 24.8 torr = 0.574 Ptotal 43.2 torr Ptotal 43.2 torr 14. When 15.0g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing point of the solution is 7.20°C. The freezing point of pure formic acid is 8.40°C. Solve for Kf for formic acid. 15.0g C2H5OH x 1 mol C2H5OH = 0.435 m 0.750 kg formic acid 46.0 g C2H5OH ∆Tf = (Tf(formic acid)) – (Tf(solution)) = 8.40°C – 7.20°C = 1.20°C Kf = ∆Tf = 1.20°C = 2.76°C/m m 0.435m 15. A 1.20 gram sample of an unknown covalent compound is dissolved in 50.0 g of benzene. The solution freezes at 4.92°C. Calculate the molecular weight of the compound. The freezing point of pure benzene is 5.48°C and Kf is 5.12°C/m. ∆Tf = 5.48°C – 4.92°C = 0.56°C; m = ∆Tf = 0.56°C = 0.11m Kf 5.12°C/m (0.11m)(0.0500kg) = 0.0055 mol solute 1.20 g solute = 0.022 g/mol 0.0055 mol solute 16. 0.500 grams of a sample is dissolved in 30mL of aqueous solution. If this solution has an osmotic pressure of 8.92 torr at 27.0°C, estimate its molecular weight. π = MRT = (n/V)RT; n = πV = (8.92 torr x 1 atm/760 torr)(0.0300L) =0.0000143 mol RT (0.0821 L.atm/mol.K)(300K) 0.500 g/0.0000143 mol = 3.50 x 104 g/mol 17. Rank the following in terms of increasing boiling point elevation when 0.1 moles of each is placed in 1 liter of water. NaCl BaSO4 CaCl2 Urea (a nonionizable, water soluble organic molecule) BaSO4 < urea < NaCl< CaCl2 (remember that BaSO4 is insoluble) 18. Predict the temperature at which water boils if it has a vapor pressure of 355 torr at 80oC. The ∆Hovap of water is 40.7 kJ/mol. Use the Clausius Clapeyron equation and follow my approach in the notes. Make sure you change torr to Pa and you should get a value a little over 373 K depending on how many sig figs you use in your
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