AP Chemistry: Liquids and Solids Practice Problems Directions: Write your answers to the following questions in the space provided. or problem solving, show all of your work. Make sure that your answers show proper units, notation, and significant digits.. List the three states of matter in order of a. increasing molecular disorder solid, liquid, gas b. increasing intermolecular attractions gas, liquid, solid 2. Which type of intermolecular attractive force operates between a. all molecules London dispersion forces b. polar molecules dipole-dipole c. the hydrogen atom of a polar molecule dipole-dipole and in some cases hydrogen bonding 3. Describe the intermolecular forces that must be overcome to convert each of the following from a liquid to a gas. a. Br 2 Nonpolar covalent molecule; London dispersion forces only b. CH 3 OH Polar covalent molecule with O H bonds; hydrogen bonding, dipole-dipole forces, and London Dispersion forces. c. H 2 S Polar covalent molecule; dipole-dipole and London Dispersion forces. 4. a. What is meant by the term polarizability? Polarizability is the ease with which the charge distribution in a molecule can be distorted to produce a temporary dipole. b. Which of the following atoms would you expect to be the most polarizable: O, S, Se, or Te? Explain. Te is the most polarizable because its valence electrons are farthest from the nucleus and least tightly held. c. Put the following molecules in order of increasing polarizability: GeCl 4, CH 4, SiCl 4, SiH 4, and GeBr 4. CH 4 < SiH 4 < SiCl 4, GeCl 4 < GeBr 4 (Corresponds to increasing molar mass) d. Predict the order of boiling point of the substances in part (c). CH 4 < SiH 4 < SiCl 4, GeCl 4 < GeBr 4 The magnitude of London dispersion forces and thus the boiling points of molecules increase as polarizability increases. 5. Rationalize the differences in boiling points between the members of the following pairs of substances: a. H (20 C) and HCl (-85 C) H has the higher boiling point because hydrogen bonding is stronger than dipole-dipole forces. b. CHCl 3 (6 C) and CHBr 3 (50 C) CHBr 3 has the higher boiling point because it has the higher molar mass, indicating greater polarizability and stronger dispersion forces. c. Br 2 (59 C) and Cl (97 C) Cl has the higher boiling point because the molecules have similar molar masses (hence similar dispersion forces), but Cl is polar giving it dipole-dipole forces that are absent from the nonpolar Br 2.
6. Arrange the following substances in order of increasing boiling point. C 2 H 6 NH 3 Ne AsH 3 lowest Boiling Point Ne < C 2 H 6 < AsH 3 < NH 3 monatomic molecule (small) nonpolar covalent molecule (larger) polar covalent molecule (even larger orces London London London dipole-dipole polar covalent molecule London Hydrogen Bonding 7. How do the viscosity and surface tension of liquids change as intermolecular forces become stronger? Viscosities and surface tensions of liquids both increase as intermolecular forces become stronger. 8. How do the viscosity and surface tension of liquids change as temperature increases? Surface tension and viscosity decrease as temperature and average kinetic energy of molecules increase. 9. Explain the following observations: a. The surface tension of CHBr 3 is greater than that of CHCl 3. CHBr 3 has a higher molar mass, is more polarizable and has stronger dispersion forces, so the surface tension is greater. b. As temperature increases, oil flows faster through a narrow tube. As temperature increases, the viscosity of the oil decreases because the average kinetic energies of the molecules increase. c. Raindrops that collect on a waxed automobile hood take on a nearly spherical shape. Adhesive forces between polar water and nonpolar car wax are weak, so the large surface tension of water draws the liquid into the shape with the smallest surface area, a sphere. 0. Water has a higher capillary action than mercury due to A. higher dipole-dipole forces between the water molecules. B. strong cohesive forces within water. C. very significant induced intermolecular attractions. D. weak adhesive forces in water. E. strong cohesive forces in water which work with strong adhesive forces.; Answer: (E) The strong adhesive forces lead to a creeping effect as water moves up the narrow tubing and the strong cohesive forces attempt to minimize the surface area.. Small drops of water tend to bead up because of A. high capillary action. b. the shape of the meniscus. c. the resistance to increased surface area. d. low London dispersion forces. e. weak covalent bonds. Answer: (C) This is a description of surface tension, which is a result of high dipole-dipole forces between water molecules. These intermolecular forces are also called hydrogen bonds. 2. As you go down the noble gas family on the periodic table, the boiling temperature increases. This trend is due mainly to A. an increase in hydrogen bonding. B. a decrease in dipole-dipole forces. C. the lower atomic masses as you down the family. D. an increase in London dispersion forces. E. an increase in lattice energy. Answer: (D) These very symmetrical atoms are nonpolar; except for the induced dispersion forces, they would all boil at zero Kelvin. As it is, He boils at 4K and Ne at 25K. 2
3. The vapor pressure increases in a predictable order as shown as A. CH 4 < C 2 H 5 OH < C 2 H 5 O C 2 H 5 < Ne. B. Ne < CH 4 < C 2 H 5 O C 2 H 5 < C 2 H 5 OH. C. C 2 H 5 OH < C 2 H 5 O C 2 H 5 < CH 4 < Ne. D. C 2 H 5 O C 2 H 5 < C 2 H 5 OH < CH 4 < Ne E. C 2 H 5 O C 2 H 5 < C 2 H 5 OH < Ne < CH 4 Answer: (C) Examining the M will suggest that only the alcohol (C 2 H 5 OH) has an exposed OH, suggesting strong hydrogen bonding. The other three are essentially controlled by weaker London dispersion forces, which are greater in the larger, more massive compound. Realize that the higher the M, the lower the vapor pressure. 4. Several liquids are compared by adding them to a series of 50 ml graduated cylinders, then dropping a steel ball of uniform size and mass into each. The time required for the ball to reach the bottom of the cylinder is noted. This is s method used to compare the differences in a property of liquids known as A. surface tension. B. buoyancy. C. capillary action. D. viscosity. E. surface contraction Answer: (D) The resistance to flow of any fluid is called viscosity. As you would predict, liquids with high viscosity (e.g., maple syrup) have large intermolecular forces. 5. Plot a vapor pressure curve for Cl 2 O 7 from the following vapor pressures. Determine the boiling point for Cl 2 O 7 under a pressure of 25 torr from the plot: t( C) -24-3 -2 0 29 45 62 79 vp (torr) 5 0 20 40 00 200 400 760 The boiling point of Cl 2 O 7 at 25 torr from the graph is about 34 C. 6. The normal boiling point of ethanol, C 2 H 5 OH, is 78.3 C, and its molar heat of vaporization is 39.3 kj/mol. What is the vapor pressure, in torr, of ethanol at 50.0 C? Use the Clausius-Clapeyron equation. p P T T2 KJ = Hvap R T T 2 4 760 torr 3. 93 0 J / mol P vap, T 8.34 J / K mol 323 K 35.. 3 K 2KJ = 760 torr 760 torr 8. 8. e 3. 25 Pvap, T P 2KJ = T2KJ = = 760 torr P T2 = = 234 torr 3.25 H G KJ K J 3
7. Classify each of the following substances according to the type of solid it forms. a. Gold Solid gold is an atomic solid with metallic properties. b. Carbon dioxide Solid carbon dioxide contains nonpolar carbon dioxide molecules and is a molecular solid c. Lithium fluoride Solid lithium fluoride contains Li + and - ions and is a binary ionic solid. d. Krypton Solid krypton contains krypton atoms that can interact only through London dispersion forces. t is an atomic solid but has the properties characteristic of a molecular solid with nonpolar molecules. 8. Consider the following data concerning four different substances. Compound Conducts Electricity as a Solid Other Properties B 2 H 6 no gas at 25 C SiO 2 no high mp Cs no aqueous solution conducts electricity W yes high mp Label the four substances as either ionic, network, metallic, or molecular solids. B 2 H 6, molecular Cs, ionic SiO 2, network W, metallic 9. The properties of solids vary with their bonding. An example of this is shown by A. ionic solids with electrostatic attractions called ionic bonds, which have high melting temperatures. B. molecular solids with high intermolecular forces which have high melting temperatures. C. ionic solids with highly mobile ions which have high conductance. D. amorphous solids with strong London dispersion forces and high vapor pressure. E. network solids with low melting temperatures due to strong covalent bonds. Answer: (A) onic bonds are unusually strong, requiring high temperatures to melt ionic substances. Table salt, for example, which is Na + Cl -, melts at 804 C. 20. Three simple unit cells are shown below. The dimensions of each unit cell can be described mathematically using the following information. s = length of the edge r = atomic radius a. What is the relationship between s and r for the simple cubic cell? 2r = s b. What is the relationship between s and r for the body-centered cubic cell? 4r = s 3 c. Which of the relationship between s and r for the face-centered cubic cell? 4r = s 2 4
2. Barium has a body-centered cubic structure. f the atomic radius of barium is 222 pm, calculate the density of solid barium. or a body-centered unit cell, the radius of the atom is related to the cube edge length by 4r=s 3 (see previous question). This can also be written as s=4r/ 3. 4r s= = 2. 309 r 3 s= 2. 309( 222 pm) = 53 pm = 5.3 0 8 cm n a body-centered unit cell, there are 2 atoms/unit cell. Calculate the mass. mol Ba 37. 3 g Ba -22 mass = 2 atoms Ba = 4.56 0 g 23 6.02 0 atoms mol Ba Calculate the volume. 3 volume= s = 8 3 22 ( ) ( 53. 0 ) = 35. 0 cm 3 Calculate the density. 22 mass 4. 56 0 g D= = = 3. 38 g / cm 3 Volume 22 3 35. 0 cm 22. Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 g/cm 3. Calculate a value for the atomic radius of nickel. There are 4 Ni atoms in each unit cell. Calculate the mass of a Ni atom. mol Ni 58. 69 g Ni 22 4 Ni atoms = 390. 0 g Ni 23 6. 02 0 atoms mol Ni Calculate the cube edge length. mass Density= volume = mass 22 3. 90 0 g 23 3 = = 5. 70 0 cm volume density 3 6. 84 g / cm volume (edge) 3 edge = 3 3 volume 5.70 0 23 cm 3 8 = = = 3.85 0 cm Use the Pythagorean Theorem to determine the radius. or a face-centered cube: (4r) 2 =s 2 + s 2 = 2s 2 r 8=s r=s/ 8 = 3.85 0-8 cm/ 8=.36 0-8 cm=36 pm 23. Name and describe all the possible phase changes that can occur between different states of matter. Which of these are exothermic, and which are endothermic? Melting Solid to a Liquid Endothermic reezing Liquid to a Solid Exothermic Vaporization Liquid to a Gas Endothermic Condensation Gas to a Liquid Exothermic Sublimation Solid to a Gas Endothermic Deposition Gas to a Solid Exothermic 24. Why does increasing the temperature cause a substance to change in succession from a solid to a liquid to a gas? As the temperature of a substance increases, more particles are able to overcome intermolecular forces and move to a less ordered state, from solid to liquid to gas. 25. Explain how the attractive forces between the particles in a liquid and the equilibrium vapor pressure of that liquid are related. The stronger the attractive forces, the lower the percentage of particles that can escape to the vapor phase and the lower the equilibrium vapor pressure. 26. Why do different substances have different heats of fusion? This has to do with the strength of attraction between the particles. The greater the attraction, the greater the energy required to overcome it, and the higher the heat of fusion. 5
27. n each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: HCl, Ar, or 2 HCl; HCl is polar while Ar and 2 are nonpolar. HCl has dipole forces unlike Ar and 2. b. highest freezing point: H 2 O, NaCl, or H NaCl; onic forces are much stronger than molecular forces. c. lowest vapor pressure at 25 C: Cl 2, Br 2, or 2 2 ; All are nonpolar so the largest molecule ( 2 ) will have the strongest LC forces and the lowest vapor pressure. d. lowest freezing point: N 2, CO, or CO 2 N 2 ; Nonpolar and smallest, so has the weakest intermolecular forces. e. greatest viscosity: H 2 S, H, H 2 O 2 H 2 O 2 ; H O O H structure produces stronger H-bonding interactions than H so has greatest viscosity. f. greatest heat of vaporization: H 2 CO, CH 3 CH 3, CH 4 H 2 CO; H 2 CO is polar so has dipole forces, unlike the other nonpolar covalent compounds. 28. Pressure cookers heat a small amount of water under high constant pressure. What effect does using a pressure cooker have on cooking time? ood in a pressure cooker will take less time to prepare because it will cook at a higher temperature. When the pressure is increased, the boiling point will increase until the vapor pressure of the liquid equal the pressure above the liquid. Once the boiling point is reached, the temperature will not change as long as there is liquid present. Vapor pressure is directly proportional to temperature. 29. Determine the boiling point of water in Breckenridge, Colorado, where the atmospheric pressure is 520 torr. or water, H vap =40.7 kj/mol. Use the Clausius-Clapeyron equation. p P T T 2 KJ = 520 torr 760 torr K J = T = 362 K or 89 C. Hvap R T T 2 4 KJ 4. 07 0 J / mol 8. 3 J / K mol 373 K T KJ 30. The vapor pressure of water at 25 C is 23.8 torr; H vap =40.7 kj/mol. a. Determine the vapor pressure of water at 70 C. H p P T T 2 KJ = K Hvap R T T 2 23.8torr. 4 4 07 0 J / mol G P vap, T2 J = 8. 3 J / K mol 343 K 298 K KJ K J = 206 torr b. Determine the percentage error if your experimental results for vapor pressure at 70 C give a result of 225 torr. % error = accepted value - experimental value accepted value 225 206 00= 00= 8. 4% 225 6
3. How much heat (in kj) must be removed when cooling 55 g of benzene, C 6 H 6, at 20.0 C to solid benzene at 5.48 C? Given for benzene: boiling point at atm = 80. C Melting point at atm = 5.48 C Specific heat (l) =.74 J/ g C Specific heat (g) =.04 J/ g C Heat of fusion = 27 J/g Heat of vaporization = 395 J/g C 6 H 6 (g) 20.0 C C6 H 6 (g) 80. C 2 C6 H 6 (l) 80. C 3 C6 H 6 (l) 5.48 C 4 C6 H 6 (s) 5.48 C. q=mc T=(55 g)(.04 J/ g C)(20.0 C-80. C)=6.43 kj 2. q=mass heat of vaporization = (55 g)(395 J/g) = 6.2 kj 3. q=mc T=(55 g)(.74 J/ g C)(80. C 5.48 C) = 20. kj 4. q=mass heat of fusion = (55 g)(27 J/g) = 9.7 kj ans. 6.43 kj + 6.2 kj + 20. kj + 9.7 kj = 07.4 kj 32. On a phase diagram why does the line that separates the gas and liquid phases end rather than go to infinite pressure and temperature? The liquid-gas line of a phase diagram ends at the critical point, the temperature and pressure beyond which the gas and liquid phases are indistinguishable. 33. Using the phase diagram for CO 2, describe all the phase changes that would occur when CO 2 is heated from -00 C to -0 C at a constant pressure of 6 atm. At 6 atm and -00 C, CO2 is a solid and remains so until about - 60 C, when it melts. At about -45 C, the liquid boils. 34. Your are given the following data for butane, C 4 H 0. Normal melting point -38 C Normal boiling point 0 C Critical temperature 52 C Critical pressure 38 atm Assume that the triple point is slightly lower in temperature than the melting point and that the vapor pressure at the triple point is 3 0-5 torr. a. Sketch a phase diagram for butane. 7
b. Butane at atm and 40 C is compressed to 40 atm. Are two phases present at any time during this process? As butane is compressed from atm (point P) to 40 atm (point R) at 40 C, butane is converted from a gas to a liquid. Both phases are present simultaneously where the vapor pressure curve intersects the vertical isothermal line: T=40 C indicated by point Q. At point Q, both phases are in equilibrium. c. Butane at atm and 200 C is compressed to 40 atm. Are two phases present at any time during this process? Since 200 C is greater than the critical temperature of 52 C, there is no pressure at which two phases exist. At all pressures the liquid phase cannot be distinguished from the gas phase. 35. This question refers to the phase diagram shown. Select the true statement. A. Point E represents both critical temperature and critical pressure. B. Moving from point A to point is sublimation. C. Point C represents both solid and vapor states. D. Point B represents both solid and vapor states. E. Point D represents a point where, no matter how much pressure is exerted, a liquid cannot form. Answer: (B) Be sure you know the details of common pressure vs. temperature phase diagrams, which sections represent each physical state, and what process the movement from one section to another represents. n this case, moving in a line from point A to point represents going from the solid to vapor (gaseous) state. 36. The densities of certain compounds are greater as liquids than as solids. This means that increasing the pressure will result in A. a solid becoming a liquid. B. a liquid becoming a solid. C. lowering of the critical temperature. D. elevation of the freezing temperature. E. lowering of the triple point. Answer: (A) Since the density of the solid is less than that of the liquid, the slope of the solid/liquid line must be negative (like water, unlike carbon dioxide and most other substances). f the pressure in increased, then you move form the solid section of the phase diagram into the liquid section. 37. or each question there are two statements. Decide whether each statement is true or false. Then decide whether Statement is a correct explanation for Statement. Statement a. n an open container, the rate of evaporation of a liquid always equals the rate of condensation. alse b. Water boils at a temperature below 00 C on top of a mountain. True c. The temperature of a substance always increases as heat is added to the substance. alse Statement A dynamic equilibrium exists between a liquid and its vapor in an open container. alse Atmospheric pressure decreases with an increase in altitude. True, correct explanation The average kinetic energy of the particles in a substance increase with an increase in temperature. True d. Solids have a fixed volume. True Particles in a solid cannot move. alse e. Gases are more compressible than liquids. True f. Xenon has a lower boiling point than neon. alse 8 There is more space between particles in a gas than between particles in a liquid. True, correct explanation Dispersion forces between xenon atoms are stronger than those between neon atoms. True
38. Write balanced net ionic equations for each of the following. a. Solid lithium hydride is added to water. LiH + H 2 O Li + + OH - + H 2 b. Excess chlorine gas is passed over hot iron filings. e + Cl 2 ecl 3 (The excess chlorine gas makes e go to its highest oxidation number.) c. Propanol (C 3 H 7 OH) is burned completely in air. 2C 3 H 7 OH + 9O 2 6CO 2 + 8H 2 O d. Hydrogen gas is passed over hot copper() oxide. H 2 + CuO H 2 O + Cu e. Solutions of silver nitrate and lithium bromide are mixed. Ag + + Br - AgBr 9