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13 Properties of Solutions Solutions to Exercises volume percent concentration. Volume percent is a volume ratio that is valid for any volume unit. For convenience, assume 30 mL C₂H₆O₂, and 70 mL The density of C₂H₆O₂ is 1.1088 g/mL. The density of H₂O is 0.997 g/mL at 25°C. Find molality and then for water. 30 mL = = 33 mL X 62.1gC₂H₆O₂ 1 mol C₂H₆O₂ 1 mL H₂O 1000 kg = 7.6752 = m In this 30% solution, ethylene glycol is the solute (present in lesser amount) and water is the solvent (present in greater amount). It is the freezing point of water, 0.0°C, that is depressed by the nonvolatile solute ethylene glycol. If pure ethylene glycol is used in the radiator, it freezes at its regular (not depressed) freezing point, -11.5°C. The freezing point of the solution, -14°C, is lower than the freezing point of pure ethylene glycol. 13.103 (a) 0.100 m is 0.300 m in particles. is the solvent. = = -1.86(0.300) = -0.558; = 0.558 = -0.558°C = -0.6°C (b) (nonelectrolyte) = -1.86(0.100) = -0.186; = 0.0 0.186 = -0.186°C = -0.2°C (measured) = i X (nonelectrolyte) From Table 13.4, i for 0.100 m = 2.32 13.104 (a) m = mol solute CS₂ = 0.250 CS₂ 1.261 1 mL CS₂ CS₂ 1000 1kg = 0.4956 = 0.496 m 0.4956 (b) m = Kb = 2.34°C/m = = 0.33 m m = kg unknown CS₂ ; m X kg CS₂ = unknown unknown ; MM 50.0 CS₂ 1 mL = = CS₂ MM = 0.333 5.39 m 0.06305 unknown CS₂ = = 2.6 10² g/mol 13.105 M = RT П = 57.1 298 torr K 760 torr mol K atm = 3.072 10⁻³ = 3.07 X 10⁻³ M 389