Lesson_14_-_Example_Solutions-2009

# Lesson_14_-_Example_Solutions-2009 - Data M 2 = 245.3 g/mol...

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Lesson 14 Example 14-1 When 39.8 g of a non-dissociating, non-volatile sugar is dissolved in 200 g of water, the boiling point of water is raised by 0.30 o C. Estimate the molar mass of the sugar. Data: m 2 = 39.8 g m 1 = 200 g of water T b = 0.3 o C K b = 0.512 K.kg/mol T b = K b m kg mol 5859 . 0 mol / kg K 512 . 0 K 3 . 0 K T m b b = = = ( 29 kg / g 1000 / m M / m m 1 2 2 = 1

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Lesson 14 mol / g 65 . 339 kg / mol 5859 . 0 g 8 . 39 g 200 kg / g 1000 m m m kg / g 1000 M 2 1 2 = = = 2
Lesson 14 Example 14-2 Lanthanum (III) trichloride (LaCl 3 ) dissociates into ions as it dissolves in water: LaCl 3 ( s ) La 3+ ( aq ) + 3 Cl - ( aq ) Suppose 0.2453 g of LaCl 3 is dissolved in 100 g of H 2 O. What is the boiling point of the solution at atmospheric pressure, assuming no association among ions and that the solution behaves ideally?

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Unformatted text preview: Data: M 2 = 245.3 g/mol m 2 = 0.2453 g m 1 = 100 g M 1 = 18 g/mol K b = 0.512 K.kg/mol 3 Lesson 14 3 2 LaCl mol 001 . g 3 . 245 mol 1 g 2453 . n = = kg mol 01 . O H kg 1 . LaCl mol 001 . m 2 3 = = Since 4 ions are formed for each LaCl 3 dissociated, i = 4 ∆ T b = i K b m = K 0205 . kg mol 01 . mol kg K 512 . 4 = Therefore, the boiling point of the solution is T b = 100 + 0.0205 = 100.0205 o C 4 Lesson 14 5...
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Lesson_14_-_Example_Solutions-2009 - Data M 2 = 245.3 g/mol...

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