CH116 Homework 2

# CH116 Homework 2 - The system is at equilibrium. 28. K = 23...

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CH116R General Chemistry 2 Recitation Chapter 11 Colligative Properties 62. molality = 13.4 m n solu = 201 mol V solu = 11.2 L; molality = 13.4 m T b - T b ° = 6.8 ° C T b = 106.8 ° C 64. molality = 0.56 molar mass = 210 g/mol (molar mass/empirical formula mass) = 2 molecular formula = C 14 H 8 O 2 66. molarity = 4.0 × 10 - 5 M n solu = 4.0 × 10 - 5 mol molar mass = 2.5 × 10 5 g/mol 70. a. pure water, b. CaCl 2 solution, c. CaCl 2 solution, d. pure water, e. CaCl 2 solution 72. i = 2 π = 4.8 atm P > 4.8 atm 76. molarity of solute molecules = 5.11 × 10 - 2 M molar mass of solute = 97.8 g/mol Additional Exercises 82. π tree - π ground 2 atm h fluid 30 m Challenge Problem 92. a. 0.25, b. –0.562 ° C 98. a. 601.2 mg NaCl, 29.9 mg KCl, 20.0 mg CaCl 2 2H 2 O, and 309.5 mg NaC 3 H 5 O 3 ; b. 6.59 atm to 7.30 atm Marathon Problem 103. Eu(NO 3 ) 3 6H 2 O Chapter 13 The Equilibrium Constant 20. a. K p = 5.3 × 10 - 3 , b. K p = 2.9 × 10 - 5 , c. K p = 190 24. K = 4.08 × 10 8 ; Q = 4.08 × 10 8 = K
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Unformatted text preview: The system is at equilibrium. 28. K = 23 30. Reaction d has n = 0 on the gases so reaction d has K p = K . 32. P H2 = 21.3 torr K p = 4.07 Equilibrium Calculations 36. a. Q = 220 > K so [H 2 O] will decrease. b. Q = 2.2 = K so [H 2 O] will the same. c. Q = 0.4 < K so [H 2 O] will increase. d. Q = 2.2 = K so [H 2 O] will the same. e. [H 2 O] = 0.55 M at equilibrium. f. Water is a solute in the reaction. 48. a. At equilibrium, [H 2 O] = 5.0 10-2 M , [Cl 2 O] = 1.8 10-2 M , and [HOCl] = 9.2 10-3 M . b. At equilibrium, [H 2 O] = [Cl 2 O] = 0.22 M and [HOCl] = 0.07 M . 56. P NH2 = P HCl = 2.2 atm K p = 4.8 Additional Exercise 66. a. P NO = 1 10-16 atm [NO] = 4 10-18 mol/L = 2 10 3 molecules/cm 3 , b. In nature, the reaction is not at equilibrium....
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## This note was uploaded on 04/08/2008 for the course CHEM 116 taught by Professor Staff during the Fall '07 term at Stevens.

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