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chem hw2

# chem hw2 - rose(kmr2427 Homework 2 fakhreddine(51615 This...

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rose (kmr2427) – Homework 2 – fakhreddine – (51615) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0points The mole fraction of a certain nonelectrolyte compound in a solution containing only that substance and water is 0.100. The molecular weight of water is 18.0 g/mol. What addi- tional information is needed to determine the molality of the solution? 1. The mole fraction of water in the solu- tion. 2. The density of the solution. 3. No additional information; the molality can be calculated from the information given. correct 4. The density of the solute. 5. The molecular weight of the compound. Explanation: Here we can assume that we have 1 mol total. (In fact, we can choose any number of moles, but the math is easier if you choose 1 mol.) If the mole fraction of the substance is 0.100, you can then assume that you have 0.100 mol of the substance, and the remaining 0.900 mol is H 2 O. The molality of a solution is determined by the following formula: m = mol solute kg solvent We’ve already assumed that we have 0.100 mol of solute, and we can determine the kg of H 2 O in the usual way: 0 . 900 mol H 2 O parenleftbigg 18 . 0 g 1 mol parenrightbiggparenleftbigg 1 kg 1000 g parenrightbigg = 0.0162 kg, and we can calculate the molality of this solu- tion: m = 0 . 100 mol 0 . 0162 kg H 2 O = 6.17 m So, it is possible to determine the molality of this solution without any additional informa- tion. 002 1.0points What additional information, if any, would enable you to calculate the molality of a 7.35 molar solution of a nonelectrolyte solid dis- solved in water? 1. None is needed. 2. Only the density of the solution would be needed. 3. Both the density of the solution and the molecular weight of the solute would be needed. correct 4. Only the density of water would be needed. 5. Only the molecular weight of the solute would be needed. Explanation: molarity = mol solute L solution molality = mol solute kg solvent The density of the solution can be used to convert volume (1 L) of solution into mass of solution. Then the molecular weight of the solute (given or calculated from the formula) can be used to convert the number of moles solute in 1 L solution into mass of solute in grams. The mass of the solvent is the difference between the mass of the solution and the mass of the solute (both of which have been calculated). Substitute the values into the molality formula and calculate. 003 1.0points A solution is 40.0% silver nitrate (AgNO 3 ) by mass. The density of this solution is 1.48 grams/mL. The formula weight of AgNO 3 is 170 grams/mol. Calculate the molality of AgNO 3 in this solution. Correct answer: 3 . 92157 m .

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rose (kmr2427) – Homework 2 – fakhreddine – (51615) 2 Explanation: density = 1.48 g/mL FW AgNO 3 = 170 g/mol % AgNO 3 = 40% by mass molality = n AgNO 3 kg water In 100 g of 40.0% AgNO 3 (aq) there are 40.0 g AgNO 3 and 60.0 g water.
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chem hw2 - rose(kmr2427 Homework 2 fakhreddine(51615 This...

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