CHAPTER 13 PROPERTIES OF LIQUIDS SOLUTIONS TO REVIEW QUESTIONS

Transcription

1 EINS v3.qxd 12/22/06 6:06 AM Page 164 CAPTER 13 PROPERTIES OF LIQUIDS SOLUTIONS TO REVIEW QUESTIONS 1. The potential energy is greater in the liquid water than in the ice. The heat necessary to melt the ice increases the potential energy of the liquid, thus allowing the ecules greater freedom of motion. The potential energy of steam (gas) is greater than that of liquid water. 2. At 0 C, all three substances, 2 S, 2 Se, and 2 Te, are gases, because they all have boiling points below 0 C. 3. Liquids are made of particles that are close together, they are not compressible and they have definite volume. Solids also exhibit similar properties. 4. Liquids take the shape of the container they are in. Gases also exhibit this property. 5. The pressure of the atmosphere must be 1.00 atmosphere, otherwise the water would be boiling at some other temperature. 6. O + 7. Melting point, boiling point, heat of fusion, heat of vaporization, density, and crystal structure in the solid state are some of the physical properties of water that would be very different, if the ecules were linear and nonpolar instead of bent and highly polar. For example, the boiling point, melting point, heat of fusion and heat of vaporization would be lower because linear ecules have no dipole moment and the attraction among ecules would be much less. 8. Prefixes preceding the word hydrate are used in naming hydrates, indicating the number of ecules of water present in the formulas. The prefixes used are: mono = 1 hexa = 6 di = 2 hepta = 7 tri = 3 octa = 8 tetra = 4 nona = 9 penta = 5 deca = The distillation setup in Figure would be satisfactory for separating salt and water, but not for separating ethyl alcohol and water. In the first case, the water is easily vaporized, the salt is not, so the water boils off and condenses to a pure liquid. In the second case, both substances are easily vaporized, so both would vaporize (though not to and equal degree) and the condensed liquid would contain both substances

2 EINS v3.qxd 12/22/06 6:06 AM Page The thermometer would be at about 70 C. The liquid is boiling, which means its vapor pressure equals the confining pressure. From Table 13.1, we find that ethyl alcohol has a vapor pressure of 543 torr at 70 C. 11. The water in both containers would have the same vapor pressure, for it is a function of the temperature of the liquid. 12. Vapor pressure is the pressure exerted by a vapor when it is in equilibrium with its liquid. The liquid ecules on the surface and the gas ecules above the liquid are the important players in vapor pressure. One liter of water will evaporate at the same rate as 100 ml of water if they are at the same temperature and have the same surface area. The liquid and vapor in both cases will come to equilibrium in the same amount of time and they will have the same vapor pressure. Vapor pressure is a characteristic of the type of liquid, not the number of ecules of a liquid. 13. In Figure 13.1, it would be case (b) in which the atmosphere would reach saturation. The vapor pressure of water is the same in both (a) and (b), but since (a) is an open container the vapor escapes into the atmosphere and doesn t reach saturation. 14. If ethyl ether and ethyl alcohol were both placed in a closed container, (a) both substances would be present in the vapor, for both are volatile liquids; (b) ethyl ether would have more ecules in the vapor because it has a higher vapor pressure at a given temperature. 15. The vapor pressure observed in (c) would remain unchanged. The presence of more water in (b) does not change the magnitude of the vapor pressure of the water. The temperature of the water determines the magnitude of the vapor pressure. 16. At 30 torr, would boil at approximately 29 C, ethyl alcohol at 14 C, and ethyl ether at some temperature below 0 C. 17. (a) At a pressure of 500 torr, water boils at 88 C. (b) The normal boiling point of ethyl alcohol is 78 C. (c) At a pressure of 0.50 atm (380 torr), ethyl ether boils at 16 C. 18. Based on Figure 13.7: (a) Line BC is horizontal because the temperature remains constant during the entire process of melting. The energy input is absorbed in changing from the solid to the liquid state. (b) During BC, both solid and liquid phases are present. (c) The line DE represents the change from liquid water to steam (vapor) at the boiling temperature of water

3 EINS v3.qxd 12/22/06 6:06 AM Page Physical properties of water: (a) melting point, 0 C (b) boiling point, 100 C (at 1 atm pressure) (c) colorless (d) odorless (e) tasteless (f) heat of fusion, 335 J> g (80 cal> g) (g) heat of vaporization, 2.26 kj> g (540 cal> g) (h) (i) density = 1.0 g>ml (at 4 C) specific heat = J> 20. For water, to have its maximum density, the temperature must be 4 C, and the pressure sufficient to keep it liquid. d = 1.0 g>ml 21. Apply heat to an ice-water mixture, the heat energy is absorbed to melt the ice (heat of fusion), rather than warm the water, so the temperature remains constant until all the ice has melted. 22. Ice at 0 C contains less heat energy than water at 0 C. eat must be added to convert ice to water, so the water will contain that much additional heat energy. 23. Ice floats in water because it is less dense than water. The density of ice at 0 C is g> ml. Liquid water, however, has a density of 1.00 g> ml. Ice will sink in ethyl alcohol, which has a density of g> ml. 24. The heat of vaporization of water would be lower if water ecules were linear instead of bent. If linear, the ecules of water would be nonpolar. The relatively high heat of vaporization of water is a result of the ecule being highly polar and having strong dipole-dipole and hydrogen bonding attraction for other water ecules. 25. Ethyl alcohol exhibits hydrogen bonding; ethyl ether does not. This is indicated by the high heat of vaporization of ethyl alcohol, even though its ar mass is much less than the ar mass of ethyl ether. 26. A linear water ecule, being nonpolar, would exhibit less hydrogen bonding than the highly polar, bent, water ecule. The polar ecule has a greater interecular attractive force than a nonpolar ecule. 27. Water, at 80 C, will have fewer hydrogen bonds than water at 40 C. At the higher temperature, the ecules of water are moving faster than at the lower temperature. This results in less hydrogen bonding at the higher temperature NC 2 C 2 N 2 has two polar N 2 groups. It should, therefore, show more hydrogen bonding and a higher boiling point (117 C) versus 49 C for C 3 C 2 C 2 N

4 EINS v3.qxd 12/22/06 6:06 AM Page Rubbing alcohol feels cold when applied to the skin, because the evaporation of the alcohol absorbs heat from the skin. The alcohol has a fairly high vapor pressure (low boiling point) and evaporates quite rapidly. This produces the cooling effect. 30. (a) Order of increasing rate of evaporation: mercury, acetic acid, water, toluene, benzene, carbon tetrachloride, methyl alcohol, bromine. (b) ighest boiling point is mercury. Lowest boiling point is bromine. 31. Water boils when its vapor pressure equals the prevailing atmospheric pressure over the water. In order for water to boil at 50 C, the pressure over the water would need to be reduced to a point equal to the vapor pressure of the water (92.5 torr). 32. In a pressure cooker, the temperature at which the water boils increases above its normal boiling point, because the water vapor (steam) formed by boiling cannot escape. This results in an increased pressure over water and, consequently, an increased boiling temperature. 33. Vapor pressure varies with temperature. The temperature at which the vapor pressure of a liquid equals the prevailing pressure is the boiling point of the liquid. 34. As temperature increases, ecular velocities increase. At higher ecular velocities, it becomes easier for ecules to break away from the attractive forces in the liquid. 35. Water has a relatively high boiling point because there is a high attraction between ecules due to hydrogen bonding. 36. Ammonia would have a higher vapor pressure than SO 2 at -40 C because it has a lower boiling point ( N 3 is more volatile than SO 2 ). 37. As the temperature of a liquid increases, the kinetic energy of the ecules as well as the vapor pressure of the liquid increases. When the vapor pressure of the liquid equals the external pressure, boiling begins with many of the ecules having enough energy to escape from the liquid. Bubbles of vapor are formed throughout the liquid and these bubbles rise to the surface, escaping as boiling continues. 38. F has a higher boiling point than Cl because of the strong hydrogen bonding in F (F is the most electronegative element). Neither F 2 nor Cl 2 will have hydrogen bonding, so the compound, F 2, with the lower ar mass, has the lower boiling point. 39. The boiling liquid remains at constant temperature because the added heat energy is being used to convert the liquid to a gas, i.e., to supply the heat of vaporization for the liquid at its boiling point

5 EINS v3.qxd 12/22/06 6:06 AM Page C, the boiling point of ethyl ether. (See Table 13.2) 41. If the lake is in an area where the temperature is below freezing for part of the year, the expected temperature would be 4 C at the bottom of the lake. This is because the surface water would cool to 4 C (maximum density) and sink. 42. The formation of hydrogen and oxygen from water is an endothermic reaction, due to the following evidence: (a) Energy must continually be provided to the system for the reaction to proceed. The reaction will cease when the energy source is removed. (b) The reverse reaction, burning hydrogen in oxygen, releases energy as heat

6 EINS v3.qxd 12/22/06 6:06 AM Page 169 CAPTER 13 SOLUTIONS TO EXERCISES 1. Acid anhydride: [ClO 4, Cl 7 ] 2. Acid anhydride: [ 2 SO 3, SO 2 ] 3. Basic anhydrides: [LiO, Li ] [ 2 CO 3, CO 2 ] [ 2 SO 4, SO 3 ] [NaO, Na ] [ 3 PO 4, P 5 ] [NO 3, N 5 ] [Mg(O) 2, MgO] 4. Basic anhydrides: [KO, K ] [Ba(O) 2, BaO] [Ca(O) 2, CaO] 5. (a) (b) (c) (d) (e) (f) 6. (a) (b) (c) (d) (e) (f) Ba(O) " 2 BaO + 2 C 3 O + 3O 2 2 CO Rb RbO + 2 SnCl 2 # 2 " SnCl2 + 2 NO 3 + NaO NaNO 3 + CO CO 3 Li + 2 LiO 2 KO " K + Ba + 2 Ba(O) Cl 2 + Cl + ClO SO SO 4 2 SO KO K 2 SO (a) barium bromide dihydrate (b) aluminum chloride hexahydrate (c) iron(iii) phosphate tetrahydrate 8. (a) magnesium ammonium phosphate hexahydrate (b) iron(ii) sulfate heptahydrate (c) tin(iv) chloride pentahydrate 9. Deionized water is water from which the ions have been removed. (a) ard water contains dissolved calcium and magnesium salts. (b) Soft water is free of ions that cause hardness ( Ca 2+ and Mg 2+ ) but it may contain other ions such as Na + and K

7 EINS v3.qxd 12/22/06 6:06 AM Page Deionized water is water from which the ions have been removed. (a) Distilled water has been vaporized by boiling and recondensed. It is free of nonvolatile impurities, but may still contain any volatile impurities that were initially present in the water. (b) Natural waters are generally not pure, but contain dissolved minerals and suspended matter, and can even contain harmful bacteria. 11. ydrogen bonding occurs in ecules containing hydrogen which is bonded to either fluorine, oxygen, or nitrogen. (a) C 2 6 will not hydrogen bond; hydrogen is not bonded to either fluorine, oxygen, or nitrogen. (b) N 3 will hydrogen bond; hydrogen is bonded to nitrogen. (c) will hydrogen bond; hydrogen is bonded to oxygen. (d) I will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (e) C 2 5 O will hydrogen bond; one of the hydrogen is bonded to oxygen. 12. ydrogen bonding occurs in ecules containing hydrogen which is bonded to either fluorine, oxygen, or nitrogen. (a) F will hydrogen bond; hydrogen is bonded to fluorine. (b) C 4 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. (c) 2 will hydrogen bond; hydrogen is bonded to oxygen. (d) C 3 O will hydrogen bond; one of the hydrogens is bonded to oxygen. (e) 2 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or nitrogen. 13. N 3 N N O O C 2 5 O C C O O C C

8 EINS v3.qxd 12/22/06 6:06 AM Page F F F 2 O O O C 3 O O C O O C 15. Water spreading out on a glass plate is an example of adhesive forces. The water ecules have stronger attractive forces toward the glass surface than they do to other water ecules. 16. Water forming beaded droplets is an example of cohesive forces. The water ecules have stronger attractive forces for other water ecules that they do for the surface (100. g CoCl 2 # 6 )a g b = CoCl 2 # 6 (100. g FeI 2 # 4 )a g b = FeI 2 # 4 (100. g CoCl 2 # 6 )a g b 6 1 CoCl 2 # 6 = O (100. g FeI 2 # 4 )a g b 4 1 FeI 2 # 4 = O 21. Assume 1 of the compound which contains seven es of water. g g2 % = g MgSO 4 # = a b11002 = 51.17% g 22. Assume 1 of the hydrate which contains 18 of water. % = g g Al 2 (SO 4 ) 3 # g2 = a b11002 = 48.67% g

9 EINS v3.qxd 12/22/06 6:06 AM Page Assume 100. g of the compound g2 = 14.2 g g2 = 85.8 g Pb(C 2 3 O 2 ) 2 (14.2 g )a g b = O (85.8 g Pb(C 2 3 O 2 ) 2 )a g b = Pb(C 2 3 O 2 ) 2 In the formula for the hydrate, there is one e of Pb(C 2 3 O 2 ) 2, so divide each of the es calculated by Pb(C 2 3 O 2 ) 2 = 1 Pb(C O 2 ) = Therefore, the formula is Pb(C 2 3 O 2 ) 2 # g hydrate g FePO 4 = 8.1 g driven off (8.1 g 2O)a g b = O (16.9 g FePO 4)a g b = FePO 4 The formula is FePO 4 # = = (a) Energy (E a ) to heat the water from 20. C 100. C E a = (m)(specific heat)( T) = (120. g)a J b(80. C) = 4.0 * 104 J (b) Energy (E to convert water to steam: heat of vaporization = 2.26 * 10 3 b ) J>g E b = (m)(heat of vaporization) = (120. g)(2.26 * 10 3 J>g) = 2.71 * 10 5 J E total = E a + E b = (4.0 * 10 4 J) + (2.71 * 10 5 J) = 3.11 * 10 5 J 26. (a) Energy (E a ) to cool the water from 24 C 0 C E a = (m)(specific heat)( T) = (126 g)a J b(24 C) = 1.3 * 104 J (b) Energy (E b ) to convert water to ice E b = (m)(heat of fusion) = (126 g)(335 J>g) = 4.22 * 10 4 J E total = E a + E b = (1.3 * 10 4 J) + (4.22 * 10 4 J) = 5.5 * 10 4 J

10 EINS v3.qxd 12/22/06 6:06 AM Page Energy released in cooling the water: 25 C to 0 C E = (m)(specific heat)( T) = (300. g)a J b(25 C) = 3.1 * 104 J Energy required to melt the ice E = (m)(heat of fusion) = (100. g)(335 J>g) = 3.35 * 10 4 J Less energy is released in cooling the water than is required to melt the ice. Ice will remain and the water will be at 0 C. 28. Energy to heat the water = energy to condense the steam (300. g)a J b(100. C - 25 C) = (m)(2259 J>g) m = 42 g (grams of steam required to heat the water to 100. C) 42 g of steam are required to heat 300. g of water to 100. C. Since only 35 g of steam are added to the system, the final temperature will be less than 100. C. Not sufficient steam. 29. Energy lost by warm water = energy gained by the ice x = final temperature mass( ) = (1.5 L )a (1500 g)a J (4.707 * 10 5 J) - (6276x J> C) = 2.51 * 10 4 J x J> C 4.46 * 10 5 J = x 10 3 x J/ C x = 68 C 1000 ml ba 1.0 g L ml b = 1500 g b(75 C - x) = (75 g)a335 J g J b + (75 g)a b(x - 0 C) 30. E = (m)(heat of fusion) (500. g)(335 J>g) = 167,000 J needed to melt the ice 9560 J 6 167,500 J Since 167,500 J are required to melt all the ice, and only 9560 J are available, the system will be at 0 C. It will be a mixture of ice and water. 31. (a) 2 Na NaO + 2 (1.00 g Na)a g b 2 2O 2 Na a g b = g

11 EINS v3.qxd 12/22/06 6:06 AM Page 174 (b) (c) 32. (a) (b) (c) MgO + Mg(O) 2 (1.00 g MgO)a g b 1 2O 1 MgO a g b = g N NO 3 (1.00 g N 5 )a g b 1 2O a g b = g 1 N 5 2 K KO + 2 (1.00 K) 2 2O 2 K a g b = 18.0 g Ca + 2 Ca(O) (1.00 Ca) 2 2O 1 Ca a g b = 36.0 g SO SO 4 (1.00 SO 3 ) 1 2O a g b = 18.0 g 1 SO Water forms droplets because of surface tension, or the desire for a droplet of water to minimize its ratio of surface area to volume. The ecules of water inside the drop are attracted to other water ecules all around them, but on the surface of the droplet the water ecules feel an inward attraction only. This inward attraction of surface water ecules for internal water ecules is what holds the droplets together and minimizes their surface area. 34. Steam ecules will cause a more severe burn. Steam ecules contain more energy at 100 C than water ecules at 100 C due to the energy absorbed during the vaporization stage (heat of vaporization). 35. The alcohol has a higher vapor pressure than water and thus evaporates faster than water. When the alcohol evaporates it absorbs energy from the water, cooling the water. Eventually the water will lose enough energy to change from a liquid to a solid (freeze). 36. When one leaves the swimming pool, water starts to evaporate from the skin of the body. Part of the energy needed for evaporation is absorbed from the skin, resulting in the cool feeling

12 EINS v3.qxd 12/22/06 6:06 AM Page (a) From 0 C to 40. C solid X warms until at 40. C it begins to melt. The temperature remains at 40.0 C until all of X is melted. After that, liquid X will warm steadily to 65 C where it will boil and remain at 65 C until all of the liquid becomes vapor. Beyond 65 C, the vapor will warm steadily until 100 C. T ( C) 20 0 Joules (b) Joules needed (0 C to 40 C) = (60. g)(3.5 J>)(40. C) = 8400 J Joules needed at 40 C = (60. g)(80. J>g) = 4800 J Joules needed (40 C to 65 C) = (60. g)(3.5 J>)(25 C) = 5300 J Joules needed at 65 C = (60. g)(190 J>g) = 11,000 J Joules needed (65 C to 100 C) Total Joules needed = (60. g)(3.5 J>)(35 C) = 7400 L 37,000 J (each step rounded to two significant figures) 38. During phase changes (ice melting to liquid water or liquid water evaporating to steam), all the heat energy is used to cause the phase change. Once the phase change is complete the heat energy is once again used to increase the temperature of the substance. 39. As the temperature of a liquid increases, the ecules gain kinetic energy thereby increasing their escaping tendency (vapor pressure). 40. Since boiling occurs when vapor pressure equals atmospheric pressure, the graph in Figure 13.4 indicates that water will boil at about 75 C at 270 torr pressure. 41. CuSO 4 (anhydrous) is gray white. When exposed to moisture, it turns bright blue forming CuSO 4 # 5. The color change is an indicator of moisture in the environment. 42. MgSO 4 # 7 Na 2 PO 4 # Soap can soften hard water by forming a precipitate with, and thus removing, the calcium and magnesium ions. This precipitate is a greasy scum and is very objectionable, so it is a poor way to soften water

13 EINS v3.qxd 12/22/06 6:06 AM Page Chlorine is commonly used to destroy bacteria in water. Ozone and ultraviolet radiation are also used in some places. 45. Ozone, O When organic pollutants in water are oxidized by dissolved oxygen, there may not be sufficient dissolved oxygen to sustain marine life, such as fish. Most marine life forms depend on dissolved oxygen for cellular respiration. 47. Liquids that are stored in ceramic containers should never be drunk, for they are likely to have dissolved some of the lead from the ceramic. If the ceramic is glazed, the liquid is less apt to dissolve lead from the ceramic. 48. Na 2 zeolite(s) + Mg 2+ (aq) Mg zeolite(s) + 2 Na + (aq) 49. Softening of hard water using sodium carbonate: CaCl 2 (aq) + Na 2 CO 3 (aq) CaCO 3 (s) + 2 NaCl(aq) 50. eating Curve for 2 S Temperature ( C) Liquid S gas (phase change) liquid Solid S liquid (phase change) solid gas 100 eat added 51. (a) Melt ice: E a = (m)(heat of fusion) = (225 g)(335 J>g) = 7.54 * 10 4 J (b) (c) Warm the water: E b = (m)(specific heat)( T) = (225 g)a J b(100. C) = 9.41 * 104 J Vaporize the water: E c = (m)(heat of vaporization) = (225 g)(2259 J>g) = 5.08 * 10 5 J E total = E a + E b + E c = 6.78 * 10 5 J

14 EINS v3.qxd 12/22/06 6:06 AM Page The heat of vaporization of water is 2.26 kj> g Energy liberated when steam at C condenses to water at C Energy liberated in cooling water from C to 30.0 C Total energy liberated = 2.04 * 10 6 J * 10 5 J = 2.30 * 10 6 J 55. Energy to warm the ice from C to 0 C 56. (2.26 kj>g)a g b = 40.7 kj> cal E = (m)(specific heat)( T) = (250. g)a b( C) = 3.1 * 10 3 cal (3.1 kcal) (50.0 steam)a g ba (50.0 )a g (100. g)a 2.01 J b(10.0 C) = 2010 J Energy to melt the ice at 0 C (100. g)(335 J>g) = 33,500 J Energy to heat the water from 0 C to 20.0 C (100. g)a J b(20.0 C) = 8370 J E total = 2010 J + 33,500 J J = 4.39 * 10 4 J = 43.9 kj 2 (l) 2 2 (g) + O 2 (g) The conversion is L O 2 O 2 g (25.0 L O 2 )a L ba2 2O ba g b = 40.2 g 1 O The conversion is day ecules day O day 2.26 kj g ba J b(100.0 C C) = 2.64 * 105 J ecules hr * 1023 ecules a ba 1000 J b = 2.04 * 10 6 J kj ecules min day ba 1 hr 24 hr 60 min = ecules s ba1 min 60 s b 6.97 * ecules>s

15 EINS v3.qxd 12/22/06 6:06 AM Page Liquid water has a density of 1.00 g> ml. 59. d = m V 1.00 e of water vapor at STP has a volume of 22.4 L (gas) 2 2 (g) + O 2 (g) 2 (g) (a) (80.0 ml 2 ) 1 ml O 2 = 40.0 ml O react with 80.0 ml of 2 ml Since 60.0 ml of O 2 are available, some oxygen remains unreacted. (b) 60.0 ml ml = 20.0 ml O 2 unreacted. 60. Energy absorbed by the student when steam at 100. C changes to water at 100. C 2.26 kj (1.5 g steam)a b = 3.4 kj (3.4 * 10 3 J) g Energy absorbed when water cools from 100. C to 20.0 C E = (m)(specific heat)( t) E = (1.5 g)a J b(100. C C) = 5.0 * 102 J E total = 3.4 * 10 3 J * 10 2 J = 3.9 * 10 3 J 61. (a) won t hydrogen bond no hydrogen in the ecule (b) will hydrogen bond C 3 O O C 3 ƒ (c) (d) (e) V = m d = g 1.00 g>ml = 18.0 ml won t hydrogen bond; no hydrogen covalently bonded to a strongly electronegative element will hydrogen bond O ƒ ƒ won t hydrogen bond; hydrogen must be attached to N, O, or F 62. During the fusion (melting) of a substance the temperature remains constant so a temperature factor is not needed. 63. Energy needed to heat Cu to its melting point: E = (50.0 g)(0.385 J>)(1083 C C) = 2.04 * 10 4 J Energy needed to melt the Cu (volume of 1 e)

16 EINS v3.qxd 12/22/06 6:06 AM Page 179 E = (50.0 g)(134 J>g) = 6.70 * 10 3 J E total = 2.04 * 10 4 J * 10 3 J = 2.71 * 10 4 J 64. A lake freezes from the top down because the density of water as a solid is lower than that of the liquid, causing the ice to float on the top of the liquid. Since the liquid remains below the solid, marine, and aquatic life continues. 65. eat lost by warm water = heat gained by ice m = grams of ice to lower temperature of water to 0.0 C. (120. g)a J b(45 C C) = (m)(335 J>g) 68 g = m (grams of ice melted) 68 g of ice melted. Therefore, 150. g - 68 g = 82 g ice remain. 66. Mass solution - mass = mass 2 SO 4 (122 ml)(1.26 g>ml) - (100. ml)(1.00 g>ml) = 54 g 2 SO 4 added