Exam 1-solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 )?...
View Full Document

Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online