Lesson_13_-_Example_Solutions-2009doc

Lesson_13_-_Example_Solutions-2009doc - Lesson 13 Example...

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Lesson 13 Example 13-1 Ethanol (C 2 H 5 OH) and methanol (CH 3 OH) form a mainly ideal solution. At 20 o C, the vapour pressure of ethanol is 44.5 mmHg and that of methanol is 88.7 mmHg. a) Calculate the partial pressures and the total vapour pressure of this solution and the mole fraction of ethanol in the vapour if n ethanol = 1.3 mol and n methanol = 1.25 mol. b) Assume that you can condense the vapour phase alone. Calculate the vapour pressure of the condensed vapour phase and the mole fraction of ethanol in the vapour. a) 5098 . 0 25 . 1 3 . 1 3 . 1 n n n x meOH etOH etOH etOH = + = + = x meOH = 1 – x etOH = 1 – 0.5098 = 0.4902 Raoult’s law: P etOH = x etOH P o etOH = (0.5098)(44.5 torr) = 22.7 torr P meOH = x meOH P o meOH = (0.4902)(88.7 torr) = 43.5 torr P = P etOH + P meOH = 22.7 torr + 43.5 torr = 66.2 torr 1
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Lesson 13 Using Dalton’s law of partial pressures: P etOH = y etOH P 34 . 0 torr 2 . 66 torr 7 . 22 P P y etOH etOH = = = y meOH = 1 – y etOH = 1 – 0.34 = 0.66 The vapour phase is richer in the more volatile (higher P vap ) compound.
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