Chapter 11 Review Packet

Chapter 11 Review Packet Name Multiple Choice Portion: 1. Which of the following terms is not a quantitative description of a solution? a. molarity b. molality c. mole fraction d. supersaturation 2. Which of the following molecules would be the most soluble in water? a. CCl 4 b. CH 3 NH 2 c. HI d. CH 4 3. Which of the following molecules would be the most soluble in CH 3 CH 3? a. CH 4 b. KI c. HI d. CH 3 NH 2 4. Oil and water cannot mix together, because: a. The water molecules cannot break the London dispersion forces between oil molecules. b. The hydrogen bonds between the water molecules are stronger than the hydrogen bonds between water and oil molecules. c. The London dispersion forces between the water molecules are not as strong as the ones between the oil molecules. d. The hydrogen bonds between water molecules are hard to break by oil molecules. 5. How will increasing the pressure of a gas affect the solubility of the gas in a solvent? a. increase it b. decrease it c. no effect d. depends on pressure 6. When a substance dissolves in water, heat energy is released if: a. the lattice energy is greater than the hydration energy. b. the hydration energy is positive. c. the hydration energy is greater than the lattice energy. d. the hydration energy is negative.

7. Which statement about hydrogen bonding is true? a. Hydrogen bonding is the intermolecular attractive forces between two hydrogen atoms in solution. b. The hydrogen bonding capabilities of water molecules cause CH 3 CH 2 CH 2 CH 3 to be more soluble in water than CH 3 OH. c. Hydrogen bonding of solvent molecules with a solute will not affect the solubility of the solute. d. Hydrogen bonding interactions between molecules are stronger than the covalent bonds within the molecule. e. Hydrogen bonding arises from the dipole moment created by the unequal sharing of electrons within certain covalent bonds within a molecule. 8. Solid KF has a lattice energy of 804 kj/mol and a heat of solution (in water) of 15 kj/mol. RbF has a lattice energy of 768 kj/mol and a heat of solution (in water) of 24 kj/mol. Which salt forms stronger attractions with water? a. KF, since it has a larger lattice energy. b. RbF, since it has a smaller lattice energy. c. KF, since it has a more negative heat of hydration. d. RbF, since it has a more negative heat of hydration. e. They form equally strong attractions with water, since they both have negative heats of mixing. KF H hyd = H lat + H sol H hyd = ( 804 kj/mol) + ( 15 kj/mol) H hyd = 819 kj/mol 9. The lattice energy of NaI is 686 kj/mol and its heat of solution is 7.6 kj/mol. Calculate the hydration of energy of NaI(s). a. +15.2 b. 678 c. 694 d. +678 e. +694 H hyd = H lat + H sol H hyd = ( 686 kj/mol) + ( 7.6 kj/mol) H hyd = 693.6 kj/mol RbF H hyd = H lat + H sol H hyd = ( 768 kj/mol) + ( 24 kj/mol) H hyd = 792 kj/mol 10. A correct statement of Henry s law is: a. the concentration of a gas in solution is inversely proportional to temperature. b. the concentration of a gas in solution is directly proportional to mole fraction of the solvent. c. the concentration of a gas in solution is independent of pressure. d. the concentration of a gas in solution is inversely proportional to pressure. e. the concentration of a gas in solution is directly proportional to pressure.

11. A solution of two liquids, A and B, shows negative deviation from Raoult s law. This means that: a. the molecules of A interact strongly with other A-type molecules. b. the two liquids have a positive heat of solution. c. molecules of A interact weakly, if at all, with B molecules. d. the molecules of A hinder the strong interaction between B molecules. e. molecules of A interact more strongly with B than A with A or B with B. 12. A solution of CF 3 H in H 2 CO would most likely: a. be ideal. b. show positive deviations from Raoult s law. c. show negative deviations from Raoult s law. d. not be ideal, but the deviations cannot be predicted. e. obey Raoult s law. 13. A liquid-liquid solution is called an ideal solution if: I. it obeys PV=nRT II. it obeys Raoult s law. III. solute-solute, solvent-solvent, and solute-solvent interactions are very similar. IV. solute-solute, solvent-solvent, and solute-solvent interactions are quite different. a. I, II, III b. I, II, IV c. II, III d. II, IV e. I, III, IV 14. Two liquids form a solution and release a quantity of heat. How does the pressure above the solution compare to that predicted by Raoult s law? a. It will be greater. b. It will be less. c. It will be the same. d. It will show positive deviation. e. None of these 15. Osmotic pressure depends on all but which of the following? a. atmospheric pressure b. the molarity of the solution = MRT where c. temperature = osmotic pressure d. the ratio of moles of solute to solution volume M = molarity (which is ratio of moles of solution to solution volume) e. none of these R = universal gas constant T = temperature

16. A 0.2 molar solution of a solute, X, in benzene, displays an osmotic pressure given by the formula = (0.1)RT. Which of the following is most likely to be the case? a. X exists in benzene as X. b. X exists in benzene as X 2. c. X exists in benzene dissociated into two particles. d. This solution strongly deviates from ideal behavior. e. None of these is plausible. If the molarity of the solution is halved, then the particles must be acting as two particles in one. 17. A solution of water and a nonvolatile, non-ionizing compound is placed in a tube with a semi-permeable membrane on one side. The tube is placed in a beaker of pure water. What initial net effect will occur? a. Water will flow from the beaker to the tube. b. Water will flow from the tube to the beaker. c. The compound will pass through the membrane into the solution. d. Nothing will move through the membrane either way. e. Equilibrium is immediately established. 18. To calculate the freezing point of an ideal dilute solution of a single, non-dissociating solute in a solvent, the minimum information one must know is: a. the molality. b. the molality and the freezing point depression. c. the same quantities as in b plus the freezing point of the pure solvent. d. all of the quantities in c plus the molecular weight of the solute. e. all of the quantities in c plus the weight of the solvent. 19. Which of the following solutions is more volatile? a. Solution A (Vapor pressure of Solution A < vapor pressure of water) b. Solution B (Vapor pressure of Solution B = vapor pressure of water) c. Solution C (Vapor pressure of Solution C > vapor pressure of water) The greater the vapor pressure, the more volatile the substance will be. 20. Which of the following compounds will have the highest boiling point? a. MgCl 2 b. NaI The solution with the largest particle concentration will have the c. AlBr 3 highest boiling point. MgCl 2 dissociates into three particles. d. Na 2 S NaI dissociates into two particles. AlBr 3 dissociates into four particles. Na 2 S dissociates into three particles.

Free-Response Portion: 1. The formula and the molecular weight of an unknown hydrocarbon compound are to be determined by elemental analysis and the freezing-point depression method. (a) The hydrocarbon is found to contain 93.46 percent carbon and 6.54 percent hydrogen. Calculate the empirical formula of the unknown hydrocarbon. 1 mol C 93.46 g C = 7.782 mol C = 1.2 C 5 = 6 C 12.01 g C 6.48 mol 1 mol H 6.54 g H = 6.48 mol H = 1 H 5 = 5 H 1.01 g H 6.48 mol Empirical Formula is C 6 H 5 (b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene (molecular weight 147.0) in 25.86 grams of naphthalene (molecular weight 128.2). Calculate the molality of the p-dichlorobenzene solution. 1 mol C 6 H 4 Cl 2 2.53 g C 6 H 4 Cl 2 = 0.0172 mol C 6 H 4 Cl 2 147 g C 6 H 4 Cl 2 molality = 0.0172 mol C 6 H 4 Cl 2 = 0.665 m 0.02586 kg naphthalene (c) The freezing point of pure naphthalene is determined to be 80.2 C. The solution prepared in (b) is found to have an initial freezing point of 75.7 C. Calculate the molal freezing-point depression constant of naphthalene. T f = 80.2 C 75.7 C = 4.5 C T f = K f m 4.5 C = K f 0.665 m 6.8 C/m = K f

(d) A solution of 2.43 grams of the unknown hydrocarbon dissolved in 26.7 grams of naphthalene is found to freeze initially at 76.2 C. Calculate the apparent molecular weight of the unknown hydrocarbon on the basis of the freezing-point depression experiment above. T f = 80.2 C 76.2 C = 4.0 C T f = K f m 4.0 C = (6.8 C/m) m 0.58 m = m molality = moles of solute kg of solvent 0.58 m = x mol unknown 0.0267 kg naphthalene moles of solute = 0.0155 moles Molar Mass = 2.43 of unknown = 157 g/mol 0.0155 moles unknown (e)what is the molecular formula of the unknown hydrocarbon? Empirical Formula from part A is C 6 H 5 (Molar Mass is 77.1 g/mol) 157g/mol = 2 77.1 g/mol Molecular Formula is C 12 H 10

2. A rigid 8.20 L flask contains a mixture of 2.50 moles of H 2, 0.500 mole of O 2, and sufficient Ar so that the partial pressure of Ar in the flask is 2.00 atm. The temperature is 127 o C. a. calculate the total pressure in the flask. For H 2 : PV=nRT (P)(8.20 L) = (2.50 mol)(0.0821 L atm/mol K)(400. K) P = 10.0 atm For O 2 : PV=nRT (P)(8.20 L) = (0.500 mol)(0.0821 L atm/mol K)(400. K) P = 2.00 atm For Ar (given in problem): P = 2.00 atm Total pressure = 10.0 atm + 2.00 atm + 2.00 atm = 14.0 atm b. calculate the mole fraction of H 2 in the flask. PV=nRT (2.00 atm)(8.20 L) = n (0.0821 L atm/mol K)(400. K) 0.499 mol = n (moles of argon = 0.499 mol) H2 = moles of H 2 total moles H2 = 2.50 mol (2.50 mol + 0.500 mol + 0.499 mol) H2 = 0.714

c. Calculate the density (in g L -1 ) of the mixture in the flask. 2.02 g H 2 2.50 mol H 2 = 5.05 g H 2 1 mol H 2 32.0 g O 2 0.500 mol O 2 = 16.00 g O 2 1 mol O 2 40.0 g Ar 0.499 mol Ar = 20.0 g Ar 1 mol Ar Total mass = 5.05 g + 16.00 g + 20.0 g = 41.0 g Density = 41.0 g/8.20 L = 5.00 g/l The mixture in the flask is ignited by a spark, and the reactions represented below occurs until one of the reactants is entirely consumed. 2H 2 (g) + O 2 (g) 2H 2 O(g) d. Give the mole fraction of all species present in the flask at the end of the reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(g) I 2.50 mol 0.500 mol 0 mol C 2(0.500 mol) 0.500 mol +2(0.500 mol) E 1.50 mol 0 mol 1.00 mol New mole fraction will have hydrogen, argon, and water vapor. H2 = moles of H 2 total moles H2 = 1.50 mol (1.50 mol + 0.499 mol + 1.00 mol) H2 = 0.500 Ar = moles of Ar total moles Ar = 0.499 mol (1.50 mol + 0.499 mol + 1.00 mol) Ar = 0.166

H2O = moles of H 2 O total moles H2O = 1.00 mol (1.50 mol + 0.499 mol + 1.00 mol) H2O = 0.334 3. Answer the following questions, which refer to the 100 ml samples of aqueous solutions at 25 C in the stoppered flasks shown below. a. Which solution has the lowest electrical conductivity? Explain. C 2 H 5 OH - The 0.1 M solution with the lowest conductivity will be the one with the fewest number of particles when dissolved. It would be C 2 H 5 OH (ethanol) because it doesn t dissociate. b. Which solution has the lowest freezing point? Explain. MgCl 2 - The freezing-point depression is proportional to the concentration of solute particles. All solutes are at the same concentration, but the van t Hoff factor (i) is largest for MgCl 2. c. Above which solution is the pressure of water vapor greatest? Explain. C 2 H 5 OH - The lowering of vapor pressure of water is directly proportional to the concentration of solute particles in solution. C 2 H 5 OH is the only non-electrolyte, so it will have the fewest solute particles in solution. d. Which solution has the highest ph? Explain. NaF - The fluoride ion, generated upon dissolution of NaF, is a weak base. It is the only solution with ph > 7.

4. Account for each of the following observations about pairs of substances. In your answers, use appropriate principles of chemical bonding and/or intermolecular forces. In each part, your answer must include references to both substances. a. Even though NH 3 and CH 4 have similar molecular masses, NH 3 has a much higher normal boiling point (-33 o C) than CH 4 (-164 o C). NH 3 has hydrogen bonding between molecules and CH 4 has London dispersion forces. The intermolecular forces in NH 3 are stronger than those in CH 4. b. At 25 o C and 1.0 atm, ethane (C 2 H 6 ) is a gas and hexane (C 6 H 14 ) is a liquid. C 2 H 6 and C 6 H 14 both have London dispersion forces. The forces in C 6 H 14 are stronger because the molecule is larger and more polarizable. c. Si melts at a much higher temperature (1,410 o C) than Cl 2 (-101 o C). Si is a network covalent solid with strong covalent bonds between atoms. Cl 2 has discrete molecules with weak London dispersion forces between the molecules. Therefore, more energy is required to break the stronger bonds of Si than the weak intermolecular forces of Cl 2. d. MgO melts at a much higher temperature (2,852 o C) than NaF (993 o C). MgO and NaF are both ionic solids. The +2 and -2 charges in MgO result in a greater attraction between ions than the +1 and -1 charges in NaF (i.e. the lattice energy of MgO is greater than the lattice energy of NaF.) Therefore, more energy is required to break the greater attraction of the MgO.