General Chemistry II CHM202 Unit 1 Practice Test

General Chemistry II CHM202 Unit 1 Practice Test 1. Ion dipole forces always require a. an ion and a water molecule. d. an ion and a polar molecule. b. a cation and a water molecule. e. a polar and a nonpolar molecule. c. an anion and a polar molecule. 2. For a molecule to exhibit dipole dipole interactions, it must a. have a temporary dipole moment. b. have a hydrogen bound to an oxygen, nitrogen, or fluorine. c. have a permanent dipole moment. d. be an ion. e. have 3 or more atoms. 3. Which molecule does not exhibit hydrogen bonding? a. HF d. (CH 3) 3N b. CH 3NH 2 e. CH 3CH 2OH c. CH 3OH 4. Dispersion forces are due to a. permanent dipoles. d. ionic interactions. b. temporary dipoles. e. protons. c. hydrogen bonding. 5. Which of the following nonpolar molecules will have the highest boiling point? a. CO 2 d. C 2H 2 b. C 6H 6 (benzene) e. CF 4 c. C 6F 6 (hexafluorobenzene) 6. Air (consisting mostly of nitrogen (N 2), oxygen (O 2), and argon (Ar)) dissolves in water due to a. dispersion forces. d. hydrogen bonding. b. dipole-induced dipole forces. e. dipole dipole forces. c. ion dipole forces. 7. The Henry s law constant (mol/l atm) for carbon dioxide dissolving in water is 3.5 10 2 mol/(l atm) at 20 C. Calculate the molar concentration of carbon dioxide in a soda can where the air pressure is 2.45 atm. The mole fraction of carbon dioxide in air is 0.0397. a. 1.4 10 3 M d. 8.2 10 2 M b. 3.4 10 3 M e. 8.6 10 2 M c. 2.7 10 2 M 8. Which of the following statements does not correctly characterize the rate of evaporation of a liquid? I. Increases with increasing temperature because molecules have higher kinetic energies. II. Increases with increasing surface area of the liquid because more molecules are located at the surface. III. Increases with stronger intermolecular forces because molecules in the liquid repel each other more. a. I only d. I and II only b. II only e. II and III only c. III only

9. Gasoline is primarily a mixture of hydrocarbons and is sold with an octane rating that is based on a comparison with the combustion properties of isooctane. Gasoline usually contains an isomer of isooctane called tetramethylbutane (C 8H 18), which has an enthalpy of vaporization of 43.3 kj/mol and a boiling point of 106.5 C. Determine the vapor pressure of tetramethylbutane on a very hot summer day when the temperature is 38.0 C. a. 80.0 torr d. 47.9 torr b. 36.7 torr e. 89.3 torr c. 67.8 torr 10. Melting occurs in going from region a. I to II. d. III to II. b. I to III. e. III to I. c. II to III. 11. Which of the following substances has a solid that is less dense than the liquid? a. c. b. d.

12. The phase diagram for carbon dioxide is shown below. What are the phase changes in order as carbon dioxide is cooled from 40 o C to 70 o C, at 200 atm pressure? a. b. c. d. e. supercritical fluid gas liquid solid gas solid supercritical fluid solid supercritical fluid liquid solid gas liquid solid 13. Maple syrup is harder to pour out of a container than water due to its a. boiling point. d. viscosity. b. surface tension. e. higher molecular weight. c. compressibility. 14. Which of the following will require the greatest energy input to separate the ions? a. MgI 2 d. MgBr 2 b. MgF 2 e. NaCl c. MgCl 2 15. Which of the following compounds has the largest melting point? a. NaF d. BeF 2 b. KCl e. NaCl c. RbCl 16. Calculate the lattice energy of sodium fluoride from the following data: Ionization energy of Na: 496 kj/mol Electron affinity of F: 328 kj/mol Energy to vaporize Na: 108 kj/mol F 2 bond energy: 160 kj/mol Energy change for the reaction: a. 931 kj/mol d. 1,011 kj/mol b. 931 kj/mol e. 851 kj/mol c. 1,011 kj/mol

17. Determine the energy change for the reaction from the following data: Lattice energy of LiCl = 861 kj/mol Energy to vaporize Li = 159 kj/mol Ionization energy of Li = 520 kj/mol Cl 2 bond energy: 240 kj/mol Electron affinity of Cl: 349 kj/mol a. 411 kj/mol d. 861 kj/mol b. 528 kj/mol e. 291 kj/mol c. 311 kj/mol 18. Indicate which aqueous solution has the highest vapor pressure. a. 0.1 M KCl d. 0.1 M MgCl 2 b. 0.2 M Na 2CO 3 e. 0.2 M MgCl 2 c. 0.2 M NaCl 19. A solution is prepared by mixing 50.0 g of methanol (CH 3OH) with 50.0 g of ethanol (CH 3CH 2OH). Use the following data to determine the vapor pressure of this solution at 20 C. Both components are volatile. Substance Vapor Pressure at 20 C (torr) Methanol 92 Ethanol 45 a. 69 torr d. 73 torr b. 57 torr e. 83 torr c. 80 torr 20. Calculate the molality of a solution containing 0.355 mol sodium hydrogen carbonate (baking soda, NaHCO 3) and 245 g of water. a. 1.45 m d. 1.12 10 3 m b. 1.12 10 2 m e. 0.0211 m c. 2.11 m 21. Which of the following aqueous solutions will have the highest freezing point? a. 0.03 m potassium perchlorate, KClO 4 b. 0.05 m sodium iodide, NaI c. 0.03 m calcium nitrate, Ca(NO 3) 2 d. 0.02 m lithium phosphate, Li 3PO 4 e. 0.05 m magnesium sulfate, MgSO 4 22. How many atoms are contained in a body-centered-cubic (bcc) unit cell? a. 1 b. 2 c. 4 d. 8 e. 16.

CHM202_Unit_1_Practice_Test Answer Section 1. ANS: D a. An ion and a water molecule exert the ion-dipole force, but there are dipoles other than water. b. Any ion will do, it doesn t have to be a cation. As in a. water is not required. c. Any ion will do, it doesn t have to be a anion. d. This is general enough! e. This combination doesn t exert the force at all. PTS: 1 DIF: Easy REF: 10.3 OBJ: Explain the origin of ion dipole, dipole dipole, and hydrogen-bonding forces. MSC: Remembering 2. ANS: C a. Temporary dipoles exert dispersion forces, not dipole-dipole. b. This is a special dipole interaction. c. Its positive end will be attracted to another atom s negative end, if all the atoms are permanintly polarized. d. Ions are monopoles, not dipoles. e. There are many dipoles with many different combinations of atoms. PTS: 1 DIF: Easy REF: 10.3 OBJ: Identify molecules with dipole dipole interactions, and predict their relative strengths. MSC: Understanding 3. ANS: D a. Hydrogen bonded to fluorine (small, highly electronegative) - hydrogen bonding. b. Hydrogen bonded to nitrogen (small, highly electronegative) - hydrogen bonding. c. Hydrogen bonded to oxygen (small, highly electronegative) - hydrogen bonding. d. All hydrogen atoms are bonded to carbon atoms. Carbon is small, but not highly electronegative - NO HYDROGEN BONDING. e. Hydrogen bonded to oxygen (small, highly electronegative) - hydrogen bonding. PTS: 1 DIF: Easy REF: 10.3 OBJ: Identify the circumstances where hydrogen bonding occurs. MSC: Understanding 4. ANS: B a. Permanent dipoles give rise to dipole-dipole forces. b. Temporary dipoles Dispersion forces ARE due to temporary dipoles. c. Hydrogen bonding is a form of dipole-dipole bonding. d. Ionic interactions - not dispersion. e. Protons - Huh??

Yes, all intermolecular attractions include dispersion forces, but ONLY temporary dipoles cause them. PTS: 1 DIF: Easy REF: 10.2 OBJ: Explain the origin of dispersion forces. MSC: Remembering 5. ANS: C Rational: We are given five compounds and we are asked to determine the compound with the highest boiling point among them. We also know that intermolecular forces are responsible for certain physical properties, including boiling point. Since all of these compounds are symmetrical (determined from Lewis structures), only dispersion forces are present. Dispersion forces increase with increasing molar mass and the size of the molecules, the most massive and largest molecule will give rise to the the largest dispersion forces, and, therefore, highest boiling point, C6F6 (hexafluorobenzene), will have the highest boiling point among these compounds. In this case, we expect the largest molecule, and C 6F 6 is the largest. PTS: 1 DIF: Medium REF: 10.2 OBJ: Predict the relative freezing and boiling points of nonpolar molecules. MSC: Understanding 6. ANS: B We are given the major components of air, N 2, O 2 and Ar. We are asked to determine the intermolecular force(s) responsible for the fact that air dissolves in water. While nitrogen and oxygen molecules form hydrogen bonds with water, argon will not. In order for any of these components of air to form intermolecular bonds with water, though, it is necessary for these symmetrical species to form temporary dipoles in order to interreact with water molecules. Water is strongly polar and will induce dipoles, the only force (including the hydrogen bonds) will be dipole-induced dipole forces.

It makes sense that symmetrical molecules or atoms would have to be induced to be temporary dipoles for this type of bonding. PTS: 1 DIF: Medium REF: 10.5 OBJ: Describe and explain how the solubility of a gas in a liquid depends on the intermolecular forces, the pressure of the gas, and the temperature. MSC: Understanding 7. ANS: B We are given the Henry s Law constant for CO 2, 3.50 10 2 mol/(l atm). at a given temperature and pressure. We are asked to determine the pressure of CO 2 at 2.45 atm when the mole fraction is 0.0397. We will use Henry s Law, C gas = k gas P gas, to compute the concentration at the higher pressure. We will also have to determine the effect of the concentration of CO 2 in air, given the mole fraction. First, determine the partial pressure of CO 2 in air, given the mole fraction: P(CO 2) = X(CO 2) x P(air) = 0.0397 x 2.45atm = 0.097265atm Substitute the known values: C gas = (3.5 10 2 mol/(l atm))( 0.097265atm)=0.003404 mol/l = 3.4 x 10-3 M Since the partial pressure of CO 2 is about a tenth of the overall pressure, we expect the concentration to be smaller as well. The result makes sense. PTS: 1 DIF: Easy REF: 10.5 OBJ: Use Henry s law to relate the solubility of a gas in a solvent to the partial pressure of the gas over the solution. MSC: Applying 8. ANS: C We are given several explanations for the physical process of evaporation. I. Increases with increasing temperature because molecules have higher kinetic energies. II. Increases with increasing surface area of the liquid because more molecules are located at the surface. III. Increases with stronger intermolecular forces because molecules in the liquid repel each other more. Examine each of the reasons in order. to determine if it properly explains evaporation.

I. Increases with increasing temperature because molecules have higher kinetic energies. Faster moving particles will escape more easily. II. Increases with increasing surface area of the liquid because more molecules are located at the surface. This is true. A liquid that is spread out will have more escape routes. III. Increases with stronger intermolecular forces because molecules in the liquid repel each other more. The opposite is true. Intermolecular forces are attractive, not repulsive. III. is NOT correct. Attractive forces will keep molecules from escaping. PTS: 1 DIF: Easy REF: 10.6 OBJ: Explain how and why the rate of evaporation of a pure liquid depends on the temperature, the surface area, and the intermolecular forces. MSC: Understanding 9. ANS: B We are given the ΔH vap (43.3 kj/mol), normal boiling point, T 1, (106.5 C), and the temperature at which to determine vapor pressure, T 2, (38.0 C). The vapor pressure at the normal boiling point, P 1, comes from the definition: temperature at which vapor pressure equals standard atmospheric pressure (760 torr). We have all the information we need to solve the Clausius-Clapeyron equation for P 2. P 1 = 760.0 torr P 2 =? T 1 = 106.5 C = 379.65K T 2 = 38.0 C = 311.15K R = 0.082057 L?9 atm/mol?9 K The important part is to match the temperatures to the pressures (and don t forget Kelvins)!! The rest is algebra and arithmetic. Plug the values into the C-C equation, and solve for the unknown value, P 2. I don t need to do your math for you, but the answer is 36.7 torr. We should expect a lower pressure at a lower temperature, and that s what we find. PTS: 1 DIF: Difficult REF: 10.6 OBJ: Calculate the vapor pressure of a pure liquid. MSC: Applying 10. ANS: A

Region I is the solid phase and Region II is the liquid phase. That s the melting transition. PTS: 1 DIF: Easy REF: 10.7 OBJ: Identify freezing and melting points on a phase diagram. MSC: Remembering 11. ANS: D a. b. and c. all have a boundary between the solid and liquid phases with a positive slope - an indication that the solid is more dense than the liquid. d. has a boundary between solid and liquid that has a negative slope - an indication that the solid form is less dense than the liquid. PTS: 1 DIF: Medium REF: 10.7 OBJ: Explain why the freezing point of water decreases with increasing pressure, unlike most other substances. MSC: Understanding 12. ANS: D We are given a scaled phase diagram with a logarithmic pressure scale. From this we are asked to describe the phase transitions from 40 C to -70 C at a pressure of 200 atm. Using the diagram, we may draw an arrow indicating the transition. We can see that the line begins in the supercritical fluid state (40 C), crosses into and through the liquid phase and ends in the solid phase (-70 C) From a higher to a lower temperature while maintaining a constant pressure should proceed from less to more organized phases.. PTS: 1 DIF: Easy REF: 10.7

OBJ: Use a phase diagram to describe the phase changes that occur as the temperature and pressure change. MSC: Applying 13. ANS: D a. boiling point is the temperature at which the liquid and vapor phases are in equilibrium and has no effect. b. surface tension tends to minimize surface area and has no effect. c. compressibility is a measure of the extent to which a fluid may be compressed and has no effect. d. viscosity - the property of a liquid that measures its resistance to flow. e. higher molecular weight may affect intermolecular forces, but does not necessarily affect flow. PTS: 1 DIF: Easy REF: 10.8 OBJ: Define viscosity of a substance. MSC: Remembering 14. ANS: B We are given five ionic compounds, and are asked a question about lattice energy. Lattice energy is calculated by Coulomb s Law, which is dependent on the magnitude of the charges on the ions and the distance between the charges. This can often be determined in a semiquantitative manner. a. MgI 2 charges are 2+ and 1-, but iodine is a large atom, making the distance between the charges larger. Since the distance is denominator in Coulomb s Law, it inversely affects the lattice energy. Larger distance smaller energy. b. MgF2 fluorine is the smallest atom of the listed ionic compounds and produces the largest lattice energy calculation. c. MgCl 2 charges are 2+ and 1-, but chlorine is a larger atom than fluorine. d. MgBr 2 charges are 2+ and 1-, but bromine is a larger atom than fluorine. e. NaCl charges are 1+ and 1-, smaller lattice energy. Coulomb s Law states that smaller lattices have higher energies agrees with the conclusion. PTS: 1 DIF: Medium REF: 11.1 OBJ: Estimate the relative strengths of ion ion interactions. MSC: Understanding 15. ANS: D We are given five ionic compounds, and are asked a question about melting point, which is determined by lattice energy.

Lattice energy is calculated by Coulomb s Law, which is dependent on the magnitude of the charges on the ions and the distance between the charges. This can often be determined in a semiquantitative manner. a. NaF charges are 1+ and 1-. b. KCl charges are 1+ and 1-. a. RbCl charges are 1+ and 1-. d. BeF2 charges are 2+ and 1-. The numerator will be larger e. NaCl charges are 1+ and 1-, smaller lattice energy. Coulomb s Law states that smaller lattices have higher energies agrees with the conclusion. PTS: 1 DIF: Easy REF: 11.2 OBJ: Rank ionic compounds according to their lattice energies and melting points. MSC: Understanding 16. ANS: B We are given: Ionization energy of Na: 496 kj/mol (ΔH IE) Electron affinity of F: 328 kj/mol (ΔH EA) Energy to vaporize Na: 108 kj/mol (ΔH sub) F 2 bond energy: 160 kj/mol (ΔH BE) Energy change for the reaction: Na(s) + ½ F 2(g) NaF(s) ΔH f = -575kJ and we are asked to determine lattice energy, U Use the Born-Haber Cycle to calculate the lattice energy. ΔH f = ΔH sub + ½ ΔH BE + ΔH IE + ΔH EA + U solving for U; U = ΔH f - ΔH sub - ½ ΔH BE - ΔH IE - ΔH EA U = -575kJ - 108 kj ½(160 kj) ( 328 kj) = -931 kj The lattice energy should be a large negative number.

PTS: 1 DIF: Easy REF: 11.2 OBJ: Calculate lattice energies using the Born Haber cycle. MSC: Applying 17. ANS: A We are given: Lattice energy (U) of LiCl = 861 kj/mol Energy to vaporize (ΔH sub) Li = 159 kj/mol Ionization energy (ΔH IE) of Li = 520 kj/mol Cl 2 bond energy (ΔH BE): 240 kj/mol Electron affinity of Cl (ΔH EA): 349 kj/mol We are asked to determine the energy change for the reaction: Li(s) + ½ Cl 2(g) LiCl(s) ΔH f =? and we are asked to determine lattice energy, U Use the Born-Haber Cycle to calculate the enthalpy of formation. ΔH f = ΔH sub + ½ ΔH BE + ΔH IE + ΔH EA + U ΔHf = 159 kj + ½ (240 kj) + 520 kj + ( 349 kj) + ( 861 kj) = -411kJ The lattice energy should be a exothermic (negative enthalpy). PTS: 1 DIF: Difficult REF: 11.2 OBJ: Calculate lattice energies using the Born Haber cycle. MSC: Applying 18. ANS: A We are given five concentrations of aqueous solutions, and we are asked to determine the highest vapor pressure (a colligative property). Specifically, the smallest value for vapor pressure-lowering. 0.1 M KCl 0.2 M Na 2CO 3 0.2 M NaCl 0.1 M MgCl 2 0.2 M MgCl 2 These are all ionic solutions, so the van t Hoff factor must be considered, and since it is not given, we must assume the theoretical value (complete dissociation).

Colligative properties are dependent upon the number of solute particles only for a given solvent, so the smallest vapor pressure-lowering value thus the highest vapor pressure, since the solvent is water for all solutions will be the lowest concentrations of solute particles. 0.1 M KCl 0.1 M x 2 = 0.2 M 0.2 M Na 2CO 3 0.2 M x 3 = 0.6 M 0.2 M NaCl 0.2 M x 2 = 0.4 M 0.1 M MgCl 2 0.1 M x 3 = 0.3 M 0.2 M MgCl 2 0.2 M x 3 = 0.6 M KCl has only two ions, so the theoretical van t Hoff factor is 2. Coupled with the lowest concentration of solute, we get the lowest concentration and the highest vapor pressure. (MgCl 2 has the same concentration, but a higher van t Hoff factor.) PTS: 1 DIF: Medium REF: 11.3 OBJ: State Raoult s law and successfully use it to relate vapor pressure and mole fraction of solvent. MSC: Understanding 19. ANS: D We are given the vapor pressures of pure methanol and pure ethanol, and the mass of each in a binary solution. Both components are volatile. P methanol = 92 torr (50.0g) P ethanol = 45 torr (50.0g) Raoult s Law predicts the vapor pressure of a solution of volatile liquids as follows: P total = X methanolp methanol + X ethanolp ethanol for each component P solution = X solvent P solvent Where X solvent = mole fraction of the solvent First determine the mole fractions of methanol and ethanol: n methanol = m methanol / Molar mass methanol = 50.0g/32.0g/mol = 1.56 mol n ethanol = m ethanol / Molar mass ethanol = 50.0g/46.0g/mol = 1.09 mol n total = 2.65mol X methanol = n methanol/(n methanol + n ethanol) = 1.56mol/(1.56mol + 1.09mol) = 0.589 X ethanol = n ethanol/(n methanol + n ethanol) = 1.09mol/(1.56mol + 1.09mol) = 0.411

Ptotal = XmethanolPmethanol + XethanolPethanol = (0.589)(92torr) + (1.09)(45torr) = 72.75 torr The vapor pressures is between the two, so the answer makes sense. PTS: 1 DIF: Easy REF: 11.4 OBJ: Calculate the vapor pressure of a solution of volatile substances given the composition of the solution and the vapor pressures of the substances. MSC: Applying 20. ANS: A We are asked to compute molality given the mass of the solute and the solvent, given n solute = 0.355mol m solvent = 245g = 0.245kg The problem is a simple division. m = nsolute/masssolvent = 0.355mol/0.245kg = 1.4489m This is as expected with approximately the same number of moles of NaHCO 3 as the number of kg of water. PTS: 1 DIF: Easy REF: 11.5 OBJ: Calculate molal concentrations, and describe how to prepare a solution of known molality. MSC: Applying 21. ANS: A Given the following solutions and concentrations: 0.03 m potassium perchlorate, KClO 4 0.05 m sodium iodide, NaI 0.03 m calcium nitrate, Ca(NO 3) 2 0.02 m lithium phosphate, Li 3PO 4 0.05 m magnesium sulfate, MgSO 4 we are asked to determine the highest freezing point.

Freezing point depression is a colligative property, so the number of particles of solute determines the change in the freezing point. Since the solutes are all ionic, and no van t Hoff factors are given, we must assume the theoretical value. Multiply each molality by the theoretical van t Hoff factor. The lowest value will have the smallest freezing point depression and the highest freezing point. 0.03 m potassium perchlorate, KClO4 x 2 = 0.06 m 0.05 m sodium iodide, NaI x 2 = 0.10 m 0.03 m calcium nitrate, Ca(NO 3) 2 x 3 = 0.90 m 0.02 m lithium phosphate, Li 3PO 4 x 4 = 0.08 m 0.05 m magnesium sulfate, MgSO 4 x 2 = 0.10 m Lowest concentration of ions (concentration times van t Hoff factor) has the smallest influence on the freezing point. PTS: 1 DIF: Medium REF: 11.5 OBJ: Relate the molal concentration to the freezing point depression or boiling point elevation, correctly using the van t Hoff factor. MSC: Applying 22. ANS: B 1/8 atoms on each of the eight corners and one full atom in the center of the cell = 2 atoms. PTS: 1 DIF: Medium REF: 12.2 OBJ: Determine the number of atoms in a unit cell. MSC: Applying