159_spring_08_chapt_13_secs_3_4_a

# 159_spring_08_chapt_13_secs_3_4_a - Typical Exam Problem 8...

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Ch 13 21 Typical Exam Problem 8 At 60 C and at a pressure of O 2 of 1.00 atm, the solubility of O 2 (g) in water is 0.00285 g O 2 /100 g H 2 O. What mass of O 2 can dissolve in 250 g of water under 1.00 atom of air pressure at 60 C? The mole percent of O 2 in air is 21%.

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Ch 13 22 Solution to Typical Exam Problem 8 S gas = kH X Pgas Sgas/Pgas = K H = (0.00285 g O 2 /100g)(1/1.00 atm) •K H = 2.85 x 10-5 g O2 /g H 2 O atm = 2.85 x 10-5 g O 2 /g H 2 O atm •S gas = K H P gas = 2.85 x 10-5 g O 2 /g H 2 O atm (0.21) (1.00 atm) = 6.0 x 10-6 g O 2 /g H 2 O 6.0 x 10-6 g O 2 / g H 2 O (250 g H 2 O) = 1.5 x 10 -3 g O 2
Ch 13 23 Typical Exam Problem 10 A concentration of 8.9μg of lead per 355 g of solution is equivalent to how many ppb?

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Ch 13 24 Solution to Typical Exam Problem 10 Mass Fraction = 8.9 μg Pb / 355 g solution ( 1g/ 10 6 μg Mass Fraction = 2.5 x 10 -8 g Pb/ g solution ppb = 2.5 x 10 -8 g Pb / g solution x 10 9 = 25 ppb 25 ppb means 25 g Pb 10 9 g of solution The same way that a 25 % solution of anything means: 25 g of anything 100 g of solution
Ch 13 25 Colligative Properties of Volatile Nonelectrolyte Solutions – p. 413 (used in some homework problems) Question - What is the effect on vapor pressure when the solute is volatile, that is, when the vapor pressure consists of solute and solvent molecules? Just as a nonvolatile solute lowers the vapor pressure of the solvent by making the mole fraction of the solvent less than 1, the presence of each volatile component lowers the vapor pressure of the other by making each mole fraction less than 1.

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## 159_spring_08_chapt_13_secs_3_4_a - Typical Exam Problem 8...

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