![11.1 Solution Composition ,[object Object],[object Object],[object Object],[object Object],[object Object]](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Ch11 z5e solutions
- 1. Ch11 Solutions pp
- 9. Figure 11.1 The Steps in the Dissolving Process pp
- 12. Table 11.3 p.521 pp
- 17. Soap P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
- 20. Skeletal structure of Soap _ P O - CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O - O -
- 25. Figure 23.29 Soap Micelles
- 35. 20 40 60 80 100
- 37. Figure 11.7 The Solubilities of Several Gases in Water
- 58. 1 atm Vapor Pressure of solution Vapor Pressure of pure water
- 59. 1 atm Freezing and boiling points of water
- 60. 1 atm Freezing and boiling points of solution
- 61. 1 atm T f T b
- 62. Figure 11.14 p. 536 The Development of Osmotic Pressure pp AP requires you to diagram the change when adding solute to a solvent The effect of adding a solute is to extend the range of a solvent.
- 64. Figure 11.16 Osmotic Pressure. Net transfer of solvent molecules until hydrostatic pressure equalizes the solvent flow in both directions
- 65. Figure 11.18 Osmosis at Equilibrium
- 67. Figure 11.17 Osmosis When pressure exceeds this then reverse osmosis occurs.
Notas del editor
- Section 11.1 Solution Compositions Z5e 514
- Z5e 514 Section 11.1 Solution Composition
- Z5e 514.
- SE 11.1 Z5e 514 Molarity = 0.215 M Mass % = 0.990% Mole fraction = 0.00389 Molality - 0.217 m
- SE 11.1 Z5e 514 Molarity = 0.101 L Mass % = 0.990% Mole fraction = 0.00389 Molality - 0.217 m
- Section 11.2 The Energies of Solution Formation 5e 518
- Fig. 11.1 Z5e 519
- Rf. Fig. 11-2 Z5e 519
- Rf. Z553 #35.
- Rf. Z553 #35.
- Rf. Z553 #35.
- Section 11.3 Factors Affecting Solubility Z5e 522
- Bottom diagram shows skeletal structure of soap with nonpolar and polar ends
- Z5e 1133 Fig. 23.29.
- Z5e 523.
- Rf. Fig. 11.5 Z5e 523
- Z5e 524 SE 11.4
- Z5e 524 SE 11.4
- Z5e 524 SE 11.4
- Z5e 525.
- Rf. Fig. 11.6 Z5e 525 Note: most become more soluble with higher temperature, but note cerium sulfate
- Rf. Z5e 526 Warmer water on lakes blocks oxygen getting to older water below and the warmer water contains less oxygen.
- Fig. 11.7 Z5e 527
- Section 11.4 The Vapor Pressures of Solutions Z5e 538
- Z5e 529
- This is because the dissolved nonvolatile solute decreases the number of solvent molecules at the surface, and so proportionately lowers the escaping tendency of the solvent molecules. Rf. Z5e 528 and Fig. 11.9
- Note: all substances are molecules (so no colligative property problem) Use mole fraction 77 b.
- Note: all substances are molecules (so no colligative property problem) Use mole fraction 77 b.
- Note: all substances are molecules (so no colligative property problem) Use mole fraction 77 b.
- Z5e 532
- Z5e 554 # 51. Use density to get number of moles. Then find of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get See solutions guide p. 267
- Z5e 554 # 51. Use density to get number of moles. Then find of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get See solutions guide p. 267
- Use density to get number of moles. Then find of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get See solutions guide p. 267
- Z5e 554 # 51. Use density to get number of moles. Then find of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get See solutions guide p. 267
- Use density to get number of moles. Then find of liquid pentane & hexane. Then use Raoult’s law to get P in torr of both partial and total pressure. b. Since partial pressure of gas is proportional to number of moles of gas present use mol pentane in vapor/total mol vapor to get See solutions guide p. 267
- Z5e 532
- Z5e 532
- Z5e 532 SE 11.6
- Z5e 532
- Section 11.5 Bp elevation & Fp depression Stays liquid longer because intermolecular forces between solute/solvent help hold the liquid together.
- Stays liquid longer (before freezing) because IMF of solute/solution prevents forming orderly crystal
- Fig. 11.14 Z5e 536 AP question will require you to diagram the change of addition of a solute to a solvent. The effect of adding a solute is to extend the range of a solvent.
- Section 11.6 Osmotic Pressure Z5e 540
- Z5e 540 fig. 11.16
- Z5e 540 Fig. 11.17
- Z5e 546 Section 11.7 Colligative Properties of Electrolyte Solutions