Lesson_11_-_Example_Solutions-2009

Lesson_11_-_Example_Solutions-2009 - = = = Therefore, 0.051...

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Lesson 11 Example 11-1 The vapour pressure of benzene (C 6 H 6 ) at 25 o C is 94.7 mmHg. After 1.00 g of benzene is injected into a 10 L container held at 25 o C, what is the partial pressure of benzene in the bulb and how many grams remain as liquid? Data: T = 25 o C = 298 K P vap = 94.7 mmHg = 0.1246 atm m = 1.0 g V = 10 L M = 78 g/mol There are 2 options: 1) All benzene vapourizes; 2) Part of benzene vapourizes 1
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Lesson 11 Let’s initially assume option 2) is the right one: V gas = V = 10 L (even if no benzene vapourizes, the volume of 1.0 g of benzene is negligible compared with 10 L) 2 C 6 H 6 (g) C 6 H 6 (g) C 6 H 6 (l) P by ideal gas P vap (25 o C)
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Lesson 11 ( 29 ( 29 mol 051 . 0 K 298 1 atm L 082 . 0 K mol L 10 atm 1246 . 0 RT V P RT PV n vap =
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Unformatted text preview: = = = Therefore, 0.051 moles of C 6 H 6 must evaporate so that the pressure in the gas phase equals the vapour pressure of C 6 H 6 at 25 o C. ( 29 g 98 . 3 mol g 78 mol 051 . nM m = = = Since only 1.0 g of C 6 H 6 was injected in the bulb, option 2) cannot be right. Therefore, option 1) must be the correct one. ( 29 ( 29 torr 8 . 23 atm 0313 . L 10 1 K 298 K mol atm L 082 . g 78 mol g . 1 MV mRT V nRT P = = = = = 3...
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This note was uploaded on 09/23/2010 for the course CHEM 103 taught by Professor Chem during the Fall '09 term at University of West Georgia.

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Lesson_11_-_Example_Solutions-2009 - = = = Therefore, 0.051...

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