Exam 1-solutions-1 - Version 002 Exam 1 laude (51635) 1...

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 002 Exam 1 laude (51635) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Constants R = 8 . 314 J K- 1 mol- 1 Definitions Q [C] c [D] d [A] a [B] b G H- TS K w [H + ][OH- ] pH - log [H + ] pOH - log [OH- ] Equations G = H- T S ln P 2 P 1 = H vap R 1 T 1- 1 T 2 = MRT T b = iK b m T f = iK f m P i = P i i P = i P i q = mc T q = m H G =- RT ln K K = e- G RT ln K 2 K 1 = H R 1 T 1- 1 T 2 Q = K [at equil.] [H + ] = ( K a C a ) 1 / 2 [OH- ] = ( K b C b ) 1 / 2 [H + ] = K a ( C a /C b ) pH = p K a + log( C b /C a ) [OH- ] = K b ( C b /C a ) pOH = p K b + log( C a /C b ) 001 6.0 points Which of the solutions below will have the greater boiling point and what will it be? K b . 5 C kg / mol for water. I) 180 g of glucose (C 6 H 12 O 6 ) dissolved in 1 kg water II) 110 g of CaCl 2 dissolved in 2 kg of water 1. Solution II with a boiling point of 101 . 5 C 2. Solution I with a boiling point of 101 C 3. Solution I with a boiling point of 100 . 5 C 4. Solution II with a boiling point of 100 . 75 C correct Explanation: First calculate the molality of each solution, which is mol solute per kg solvent. To do this, you will need to convert from mass to mols, using molar mass. Remember also that i = 3 for CaCl 2 and i = 1 for glucose. Using T = imK b you will find that T = 0 . 5 C for the glucose solution and 0.75 for the salt solution, which means the boiling point of the salt solution will be 100.75 C. 002 6.0 points Carbon Dioxide 3 1 4 2 A B 200 250 300 350 400 10 10 1 10 2 10 3 10 4 Pressure,bar Temperature, K A sample of carbon dioxide is stored at 10 4 bar and 250 K. This sample is then decompressed to 10 bar at constant temper- ature. Then, at constant pressure it is heated to 400 K. Next, it is compressed at con- Version 002 Exam 1 laude (51635) 2 stant temperature to 200 bar. According to the phase diagram, how many phase transi- tions has the carbon dioxide gone through, and what is its final state? 1. 2, gas 2. 2, supercritical fluid correct 3. 2, liquid 4. 3, supercritical fluid 5. 3, gas Explanation: Navigate through the diagram through each process. In the first decompression, the solid is decompressed and becomes a liquid and then a gas (2 phase transitions). Upon heating, it becomes a supercritical fluid (no phase transitions). 003 6.0 points What would be the pH of a solution pre- pared from 2 L of 0 . 5 M HClO and 1 L of . 316 M NaClO? The K a of chlorous acid is 3 . 16 10- 8 . 1. 7 correct 2. 8 3. 7 . 3 4. 7 . 5 Explanation: Because this is a buffer, it is acceptable to use moles in place of concentrations [H + ] = K a ( C a /C b ) = ( 3 . 16 10- 8 ) (1 M / . 316 M) = 1 10- 7 pH = 7 004 6.0 points Rank the following compounds from most soluble to least soluble. Assume that all bonds except the OH are ionic. (You can estimateexcept the OH are ionic....
View Full Document

This note was uploaded on 08/24/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

Page1 / 10

Exam 1-solutions-1 - Version 002 Exam 1 laude (51635) 1...

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