fak chem hw 2

# fak chem hw 2 - To(st22362 Homework 2 Fakhreddine(52420...

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1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 1.0 points What additional in±ormation, i± any, would enable you to calculate the molality o± a 7.35 molar solution o± a nonelectrolyte solid dis- solved in water? 1. Only the density o± the solution would be needed. 2. Only the density o± water would be needed. 3. None is needed. 4. Both the density o± the solution and the molecular weight o± the solute would be needed. correct 5. Only the molecular weight o± the solute would be needed. Explanation: molarity = mol solute L solution molality = mol solute kg solvent The density o± the solution can be used to convert volume (1 L) o± solution into mass o± solution. Then the molecular weight o± the solute (given or calculated ±rom the ±ormula) can be used to convert the number o± moles solute in 1 L solution into mass o± solute in grams. The mass o± the solvent is the di²erence between the mass o± the solution and the mass o± the solute (both o± which have been calculated). Substitute the values into the molality ±ormula and calculate. 002 1.0 points The mole ±raction o± a certain nonelectrolyte compound in a solution containing only that substance and water is 0.100. The molecular weight o± water is 18.0 g/mol. What addi- tional in±ormation is needed to determine the molality o± the solution? 1. The density o± the solute. 2. The molecular weight o± the compound. 3. The density o± the solution. 4. The mole ±raction o± water in the solu- tion. 5. No additional in±ormation; the molality can be calculated ±rom the in±ormation given. correct Explanation: Here we can assume that we have 1 mol total. (In ±act, we can choose any number o± moles, but the math is easier i± you choose 1 mol.) I± the mole ±raction o± the substance is 0.100, you can then assume that you have 0.100 mol o± the substance, and the remaining 0.900 mol is H 2 O. The molality o± a solution is determined by the ±ollowing ±ormula: m = mol solute kg solvent We’ve already assumed that we have 0.100 mol o± solute, and we can determine the kg o± H 2 O in the usual way: 0 . 900 mol H 2 O p 18 . 0 g 1 mol Pp 1 kg 1000 g P = 0.0162 kg, and we can calculate the molality o± this solu- tion: m = 0 . 100 mol 0 . 0162 kg H 2 O = 6.17 m So, it is possible to determine the molality o± this solution without any additional in±orma- tion. 003 1.0 points A solution is 40.0% silver nitrate (AgNO 3 ) by mass. The density o± this solution is 1.48 grams/mL. The ±ormula weight o± AgNO 3 is 170 grams/mol. Calculate the molality o± AgNO 3 in this solution. Correct answer: 3

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## This note was uploaded on 04/16/2010 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas.

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fak chem hw 2 - To(st22362 Homework 2 Fakhreddine(52420...

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