Lesson_12_-_Example_Solutions-2009

Lesson_12_-_Example_Solutions-2009 - Lesson 12 Example 12-1...

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Lesson 12 Example 12-1 What are the molality and molarity of a solution of ethanol (C 2 H 5 OH) in water if the molar fraction of ethanol is 0.05? Assume that the density of the solution is 0.997 g/ml Basis: 100 moles of solution ( n H2O + n etOH ) Data: C 2 H 5 OH/H 2 O solution, M H2O = 18 g/mol M etOH = 46 g/mol x etOH = 0.05 ρ = 0.997 g/ml Equations: O 2 H etOH etOH etOH n n n x + = solution L 0 . 1 n molarity etOH = 1
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Lesson 12 O H kg 0 . 1 n molality 2 etOH = n etOH = x etOH ( n et + n H2O ) = 0.05(100) = 5 moles n H2O = 100 – 5 = 95 moles ( 29 g 1710 mol g 18 mol 95 m O 2 H = = =1.71 kg kg mol 92 . 2 O H kg 71 . 1 etOH moles 5 molality 2 = = ( 29 g 1940 mol g 46 mol 5 g 1710 m m m etOH 0 2 H solution = + = + = ( 29 ml 8 . 1945 g 997 . 0 ml g 1940 m V solution solution = = ρ = =1.9458 L 2
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Lesson 12 L mol 57 . 2 solution L 9458 . 1 etOH moles 5 molarity = = = 2.57 M 3
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Lesson 12 Example 12-2
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Lesson_12_-_Example_Solutions-2009 - Lesson 12 Example 12-1...

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