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Homework 06

Homework 06 - Su Yung Homework 6 Due 11:00 pm Inst...

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Su, Yung – Homework 6 – Due: Oct 29 2007, 11:00 pm – Inst: FAKHREDDINE 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Each question is worth 2 points. 001 (part 1 of 1) 2 points A one molar solution is one having 1. one mole of solute per one mole of sol- vent. 2. one mole of solute per one liter of solute. 3. one mole of solute per one kilogram of solvent. 4. one mole of solute per one kilogram of solution. 5. one mole of solute per liter of solution. correct Explanation: Molarity is moles of solute per liter of solu- tion. A one molar solution would have 1 mole of solute per liter of solution. 002 (part 1 of 1) 2 points What volume of a solution that is 40.0% NaNO 3 by mass contains 0.15 mole of NaNO 3 ? Density of the solution = 1.32 g/mL. 1. 24.1 mL correct 2. 9.60 mL 3. 38.2 mL 4. 3.86 mL 5. 42.0 mL Explanation: n NaNO 3 = 0 . 15 mol density soln = 1 . 32 g / mL The NaNO 3 solution is 40.0% NaNO 3 by mass. We can write this as the ratio 40 . 0 g NaNO 3 100 g soln The density of the solution is 1.32 g/mL. This can be more fully written as 1 . 32 g soln 1 mL soln We want the volume of the solution that contains 0.15 mole NaNO 3 . We convert from mole to grams NaNO 3 : ? g NaNO 3 = 0 . 15 mol NaNO 3 × 85 g NaNO 3 1 mol NaNO 3 = 12 . 75 g NaNO 3 We use the percent by mass to convert from grams solution to milliliters NaNO 3 solution. Then we use the solution density to convert from g solution to mL solution: ? mL soln = 12 . 75 g NaNO 3 × 100 g soln 40 . 0 g NaNO 3 × 1 mL soln 1 . 32 g soln = 24 . 1 mL soln 003 (part 1 of 1) 2 points What is the molarity of a solution prepared by dissolving 10.5 grams of NaF in enough water to produce 750 mL of solution? 1. 0.333 M correct 2. 7.9 M 3. 0.014 M 4. 3.0 M Explanation: m NaF = 10.5 g V = 750 mL = 0.75 L Molarity is moles solute per liter of solution. We use the molar mass of NaF to convert from grams to moles NaF: ? mol NaF = 10 . 5 g NaF × 1 mol NaF 42 g NaF = 0 . 25 mol NaF

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Su, Yung – Homework 6 – Due: Oct 29 2007, 11:00 pm – Inst: FAKHREDDINE 2 To calculate the molarity of the NaF solution, we divide the moles of NaF in the solution by the total volume of the solution in liters: ? M NaF = 0 . 25 mol NaF 0 . 750 L solution = 0 . 333 M NaF 004 (part 1 of 1) 2 points What mass of hydrated copper(II) sulfate (CuSO 4 · 5H 2 O) is needed to prepare one liter of a 0.6675 molar solution of CuSO 4 ? 1. 0.675 g 2. 166.7 g correct 3. none of these 4. 107.8 g 5. 10.8 g Explanation: V = 1 L M = 0.6675 M The formula weight of CuSO 4 · 5H 2 O is 249.62 g/mol. Each liter of CuSO 4 · 5H 2 O con- tains one mole of CuSO 4 · 5H 2 O, so we need 0.6675 moles of CuSO 4 · 5H 2 O, or 0 . 6675 mol × 249 . 62 g / mol = 166 . 7 g CuSO 4 · 5H 2 O . 005 (part 1 of 1) 2 points Consider an aqueous solution that contains 38 % (by mass) of a hypothetical solute Z. The formula weight of the solute Z is 247 . 3.
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Homework 06 - Su Yung Homework 6 Due 11:00 pm Inst...

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