Exam 1-solutions

# Exam 1-solutions - Version 262 – Exam 1 – sutcliffe...

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Unformatted text preview: Version 262 – Exam 1 – sutcliffe – (51630) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Use the density of water = 1g/ml. How many grams of methanol (CH 3 OH) are needed to lower the freezing point of 400 mL of water by 2.5 ◦ C? ( K f water = 1 . 86 ◦ C/molality) 1. 12.8 2. 9.71 3. 18.3 4. 21.8 5. 17.2 correct Explanation: K f H 2 O = 1 . 86 ◦ C /m V water = 400 L FP depression = 2.5 ◦ C Δ T f = K f · m m = Δ T f K f = 2 . 5 ◦ C 1 . 86 ◦ C /m = 1 . 344 m m H 2 O = 400 mL × 1 . 00 g mL × 10 − 3 kg g = 0 . 4 kg H 2 O m = n CH 3 OH kg water = g CH 3 OH MW CH 3 OH kg water g CH 3 OH = m (kg water) (MW CH 3 OH ) = parenleftbigg 1 . 344 mol CH 3 OH kg water parenrightbigg × (0 . 400 kg water) × parenleftbigg 32 . 0 g CH 3 OH mol CH 3 OH parenrightbigg = 17 . 2 g CH 3 OH 002 10.0 points Note: Assume ideal behaviour. And read the next sentence carefully... What is the change in vapor pressure from that of pure water for an aqueous solution at 40 ◦ C in which the mole fraction of fructose is 0 . 1? The vapor pressure of pure water at 40 ◦ C is 55 . 34 torr. 1. 4 . 427 torr 2. 3 . 874 torr 3. 6 . 087 torr 4. 5 . 534 torr correct 5. 4 . 981 torr 6. 6 . 641 torr Explanation: P pure solvent = 55 . 34 torr T = 40 χ fructose = 0 . 1 χ water = 1- . 1 = 0 . 9 P = χ solvent P pure solvent = 0 . 9(55 . 34 torr) = 49 . 806 torr . The change in vapor pressure is Δ P = 55 . 34 torr- 49 . 806 torr = 5 . 534 torr . 003 10.0 points Consider the reaction NO(g) + 1 2 O 2 (g) → NO 2 (g) Find the equilibrium constant for the re- action at 298 K if Δ H ◦ =- 56 . 52 kJ and Δ S ◦ =- 72 . 60 J · K − 1 at 298 K. 1. 1 . 30 × 10 6 correct 2. 8 . 08 × 10 9 3. 7 . 67 × 10 − 7 4. 3 . 23 × 10 4 Version 262 – Exam 1 – sutcliffe – (51630) 2 5. 1 . 65 × 10 − 4 Explanation: Δ G ◦ = Δ H ◦- T Δ S ◦ =- 56520- (298)(- 72 . 60) =- 34885 . 2 J K = e − Δ G/ ( R T ) = e 34885 . 2 / (8 . 314 · 298) = 1 . 3033 × 10 6 004 10.0 points An unknown liquid has a vapor pressure of 88 mmHg at 45 ◦ C and a heat of vaporization of 32 kJ / mol, what is its vapor pressure at 55 ◦ C? 1. 127 mmHg correct 2. 89 . 4 mmHg 3. 156 mmHg 4. 61 mmHg Explanation: Use the Clausius-Clapeyron Equation. Here, the only thing we don’t know is P 2 . ln parenleftbigg P 2 P 1 parenrightbigg = Δ H vap R parenleftbigg 1 T 1- 1 T 2 parenrightbigg ln parenleftbigg P 2 88 parenrightbigg = 32 , 000 8 . 31 parenleftbigg 1 318- 1 328 parenrightbigg ln P 2 = 32 , 000 8 . 31 parenleftbigg 1 318- 1 328 parenrightbigg + ln88 P 2 = e 4 . 84 = 127 mmHg 005 10.0 points What is the boiling point elevation of a solu- tion of Na 2 SO 4 (142.1 g/mol, complete disso- ciation) made by dissolving 10.0 g of Na 2 SO 4 into 250 g water ( K b = 0.512 ◦ C/ m )?...
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Exam 1-solutions - Version 262 – Exam 1 – sutcliffe...

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