hw3solution_pdf

# P total p a p b 89 torr 3 6 torr 93 torr you might

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P total = P A + P B + · · · = 89 torr + 3 . 6 torr = 93 torr You might want to check to see that you get the same answer no matter what value

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casey (rmc2555) – Homework 3 – holcombe – (51395) 4 you assume as the equal masses of the two components. As an additional challenge, can you solve this problem WITHOUT assuming a definite number of grams, by setting the mass of each component equal to an algebraic variable? 010 4.0points The vapor pressure of a volatile component in a solution decreases as its mole fraction decreases. This is known as 1. LeChatelier’s Principle. 2. Bragg’s Law. 3. Henry’s Law. 4. Raoult’s Law. correct Explanation: According to Raoult’s Law, the vapor pres- sure of the volatile component of a solution equals its mole fraction in the solution times the vapor pressure of the pure solvent. P = X P 0 011 4.0points A container holds a mixture of acetone and water at 40 C. If χ acet , the mole fraction of acetone, is 0.1, what is the total vapor pressure? (The equilibrium vapor pressures for pure acetone and pure water at 40 C are roughly 400 torr and 50 torr respectively.) 1. 85 torr correct 2. 365 torr 3. 50 torr 4. 400 torr 5. 40 torr Explanation: P = P H 2 O + P acet = χ H 2 O P H 2 O + χ acet P acet = (1 - χ acet ) P H 2 O + χ acet P acet = (0 . 9)(50) + (0 . 1)(400) = 85 012 4.0points Pure water remains in the liquid phase from 273 K up to 373 K, a range of 100 K. Over what range of temperatures would a solu- tion consisting of 58 . 44 g of NaCl dissolved in 0 . 5 kg of H 2 O remain in the liquid phase? Use K b = 0 . 512 K · m 1 and K f = 1 . 86 K · m 1 . 1. 102 K 2. 94 . 61 K 3. 109 . 5 K correct 4. 9 . 488 K 5. 107 . 4 K Explanation: n salt = 58 . 44 g 58 . 44 = 1 moles m eff = i · n salt m H 2 O = 2 · 1 moles 0 . 5 kg = 4 mol · kg 1 range = 100 + m eff · ( K b + K f ) = 109 . 5 K 013 4.0points Estimate the boiling point elevation of a sat- urated solution of LiF at 100 C, assuming it undergoes complete dissociation. The solubil- ity of LiF is 230 mg / 100 g of water at 100 C, and the boiling point constant of water is 0 . 51 K · kg / mol. Correct answer: 0 . 0904416 C. Explanation: s = 230 mg 100 g H 2 O = 0 . 23 g 0 . 1 kg H 2 O k b = 0 . 51 K · kg / mol MW = 25 . 9394 g / mol
casey (rmc2555) – Homework 3 – holcombe – (51395) 5 LiF dissociates to give 2 ions in solution, so i = 2 and Δ T b = i k b m = 2 (0 . 51 K · kg / mol) × parenleftbigg 0 . 23 g 25 . 9394 g / mol parenrightbigg 0 . 1 kg = 0 . 0904416 K = 0 . 0904416 C . 014 4.0points How many grams of antifreeze (C 2 H 4 (OH) 2 ) would be required per 500 g of water to pre- vent the water from freezing at a temperature of - 16 C ? Correct answer: 267 . 011 g. Explanation: solute is C 2 H 4 (OH) 2 (antifreeze) m solvent = 0 . 5 kg H 2 O FP solution = - 16 C m solute = ? FP solution = FP solvent + Δ t f Δ t f = FP solution - FP solvent = - 16 C - 0 . 000 C = - 16 C Δ t f = K f m m = Δ t f K f = - 16 C - 1 . 86 C /m = 8 . 60215 m m = n solute m solvent (kg) n solute = ( m ) ( m solvent ) = (8 . 60215 m )(0 . 5 kg) = 4 . 30108 mol (4 . 30108 mol solute) parenleftbigg 62 . 08 g solute mol solute parenrightbigg = 267 . 011 g C 2 H 4 (OH) 2 015 4.0points Given that the freezing point depression con- stant for water is 1.86 K m 1 , what is the freezing point of a solution that con- tains 0.5 moles KNO 3 and 1 mole of sucrose (C 12 H 22 O 11 ) in 500 g of water?

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• Fall '07
• Holcombe
• Chemistry, Amount of substance, Freezing-point depression

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