HW3_Solutions

# HW3_Solutions - estrada(gse77 – Homework 3 – Sutcliffe...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: estrada (gse77) – Homework 3 – Sutcliffe – (52410) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Read the last third of the questions care- fully; you may have to take into account the van’t Hoff factor, i. 001 10.0 points 67 . 7 g sodium nitrate is dissolved in water to make 754 g of solution. What is the percent sodium nitrate in the solution? 1. . 0897878% 2. 8 . 97878% correct 3. 100% 4. 9 . 86449% Explanation: m NaNO 3 = 67 . 7 g m soln = 754 g percent = mass solute mass solution × 100 % = 67 . 7 g NaNO 3 754 g solution × 100 % = 8 . 97878% 002 10.0 points How much CH 2 O is needed to prepare 445 mL of a 2.65 M solution of CH 2 O? 1. 17.6 g 2. 45.6 g 3. 0.0393 g 4. 79.6 g 5. Not enough information is given. 6. 35.4 g correct 7. 179 g Explanation: M = 2 . 65 M V = 445 mL = 0 . 445 L ? g CH 2 O = (0 . 445 L) × 2 . 65 mol CH 2 O L soln × 30 g CH 2 O 1 mol = 35 . 3775 g 003 (part 1 of 2) 10.0 points A student investigating the properties of so- lutions containing carbonate ions prepared a solution containing 5 . 54 g Na 2 CO 3 in a flask of volume 250 mL. Some of the solution was transferred to a buret. What volume of so- lution should be dispensed from the buret to provide 6 . 221 mmol Na 2 CO 3 ? Correct answer: 29 . 7547 mL. Explanation: m Na 2 CO 3 = 5 . 54 g V = 250 mL = 0 . 25 L n Na 2 CO 3 = 6 . 221 mmol = 0 . 006221 mol FW Na 2 CO 3 = 2 (22 . 9958 g / mol) + 12 . 011 g / mol + 3 (15 . 9994 g / mol) = 105 . 99 g / mol M Na 2 CO 3 = 5 . 54 g (105 . 99 g / mol) (0 . 25 L) = 0 . 209076 M Na 2 O 3 V = (0 . 006221 mol Na 2 CO 3 ) × parenleftbigg 1 L Na 2 CO 3 . 209076 mol Na 2 CO 3 parenrightbigg = 0 . 0297547 L = 29 . 7547 mL 004 (part 2 of 2) 10.0 points What volume of solution should be dispensed from the buret to provide 8 . 359 mmol CO 2 − 3 ? Correct answer: 39 . 9806 mL. Explanation: n CO 2- 3 = 8 . 359 mmol estrada (gse77) – Homework 3 – Sutcliffe – (52410) 2 V = 0 . 008359 mol CO 2 − 3 parenleftBigg 1 mol Na 2 CO 3 1 mol CO 2 − 3 parenrightBigg × parenleftbigg 1 L Na 2 CO 3 . 209076 mol Na 2 CO 3 parenrightbigg = 0 . 0399806 L = 39 . 9806 mL 005 10.0 points Do not forget the moLaLity is moles of solute per kg of SOLVENT!! Calculate the molality of perchloric acid in 9.2 M HClO 4 (aq). The density of this solution is 1.54 g/mL. 1. 18 m 2. 31 m 3. 21 m 4. 26 m 5. 15 m correct Explanation: 006 (part 1 of 2) 10.0 points Calculate the molality of KOH in a solution prepared from 12 . 93 g of KOH and 66 g of water. Correct answer: 3 . 49165 m . Explanation: m KOH = 12 . 93 g m H 2 O = 66 g MW K = 39 . 1 g / mol MW O = 16 g / mol MW H = 1 . 0079 g / mol FW KOH = 39 . 1 g / mol + 16 g / mol + 1 . 0079 g / mol = 56 . 1079 g / mol ....
View Full Document

## This note was uploaded on 03/03/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

### Page1 / 7

HW3_Solutions - estrada(gse77 – Homework 3 – Sutcliffe...

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

View Full Document
Ask a homework question - tutors are online