hw3solution_pdf

# 1 372 c 2 372 c 3 186 c 4 279 c 5 558 c 6 744 c

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1. -37.2 C 2. -3.72 C 3. -1.86 C 4. +2.79 C 5. -5.58 C 6. -7.44 C correct 7. -2.79 C Explanation: The van’t Hoff factor for KNO 3 is 2. So 0.5 moles is effectively 1 mole of ions. There is another mole of sucrose. This is 2 total moles in 500 g making a 4 molal solution of all solutes. Δ T f = - mK f = - (4)(1 . 86) = - 7 . 44 016 4.0points An aqueous solution freezes at - 2 . 94 C. What is the molality of the solution? Assume that there is no dissociation of the solute and that k f is 1 . 42 K · kg / mol. Correct answer: 2 . 07042 mol / kg. Explanation: k f = 1 . 42 K · kg / mol Δ T f = - 2 . 94 C = 2 . 94 K Because the freezing point of water is 0 C, the freezing point of the solution equals the freezing point of depression. Δ T f = k f m m = Δ T f k f = 2 . 94 K 1 . 42 K · kg / mol = 2 . 07042 mol / kg 017 4.0points What is the osmotic pressure of a solution

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casey (rmc2555) – Homework 3 – holcombe – (51395) 6 that contains 4.56 × 10 3 moles of lactose in 100 mL of solution at 25 C? 1. 848 torr correct 2. 536 torr 3. 71 torr 4. 1053 torr 5. 113 torr Explanation: n = 4 . 56 × 10 3 mol V = 100 mL T = 25 C + 273 . 15 = 298 . 15 K M = parenleftbigg 456 × 10 3 mol 100 mL parenrightbiggparenleftbigg 1000 mL L parenrightbigg = 4 . 56 × 10 2 mol L Osmotic pressure π = M R T = ( 4 . 56 × 10 2 mol / L ) × parenleftbigg 62 . 36 L · torr K · mol parenrightbigg (298 . 15 K) = 847 . 8 torr 018 4.0points What would be the osmotic pressure exerted by Na + on a semi-permeable membrane sep- arating two chambers with Na + concentra- tions of 0 . 1 M and 0 . 6 M. (Note: at room temperature the product of the gas con- stant and temperature is roughly equal to 25 L · atm · mol 1 .) 1. 0 . 015 atm 2. 0 . 25 atm 3. 12 . 5 atm correct 4. 0 . 125 atm 5. 0 . 0025 atm 6. 15 atm Explanation: Π = M · R · T Note that the molarity term used in the os- motic pressure actually refer to the difference in the molarities of two chambers separated by a semi-permeable membrane. Π = 0 . 5 M · 25 L · atm · mol 1 = 12 . 5 atm 019 4.0points Water from a local stream is added to one side of the U-tube shown below. Pure water is placed in the tube on the other side of the semipermeable membrane. With the left side open to barometric pressure of 1.0 atm and 1.15 atm applied to the right side, the two liquids do not move. 1.00 atm 1.15 atm B A In which half of the U-tube is the pure water located? 1. Not enough information is given. 2. B 3. A correct Explanation: 020 4.0points According to Raoult’s Law, a mixture of cyclopentane (C 5 H 10 ) and cyclohexane (C 6 H 12 ) would show 1. no deviation. correct 2. positive deviation. 3. negative deviation. Explanation:
casey (rmc2555) – Homework 3 – holcombe – (51395) 7 Raoult’s Law applies to the vapor pressure of the mixture, so a positive deviation means that the vapor pressure is higher than ex- pected for an ideal solution and a negative deviation a lower vapor pressure. Negative deviation will occur when the interactions between the different molecules are some- what stronger than the interactions between molecules of the same kind.

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

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