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Unformatted text preview: Version 209 – hw01 – McCord – (53580) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Here are some questions to get you thinking quantitatively again. Review of concentration calculations and then a little thermo and then solutions. 001 10.0 points How much NaNO 3 is needed to prepare 225 mL of a 1.55 M solution of NaNO 3 ? 1. 4.10 g 2. 12.3 g 3. 0.132 g 4. 29.6 g correct 5. 0.244 g Explanation: V = 225 mL M = 1 . 55 M ? g NaNO 3 = 225 mL × 1 L soln 1000 mL × 1 . 55 mol NaNO 3 1 L soln × 85 g NaNO 3 1 mol NaNO 3 = 29 . 6 g NaNO 3 002 10.0 points What is the final concentration of NaOH when 335 mL of 0.75 M NaOH are mixed with 165 mL of 0.15 M NaOH? 1. 0.18 M 2. 0.55 M correct 3. 1.67 M 4. 0.45 M 5. 0.82 M Explanation: V 1 = 335 mL M 1 = 0.75 M V 2 = 165 mL M 2 = 0.15 M Molarity is moles solute per liter of solution. Two solutions are being mixed together in this problem. We find the moles of NaOH in each of the individual solutions: ? mol NaOH = 335 mL soln × 1 L soln 1000 mL soln × . 75 mol NaOH 1 L soln = 0 . 2512 mol NaOH ? mol NaOH = 165 mL soln × 1 L soln 1000 mL soln × . 15 mol NaOH 1 L soln = 0 . 0248 mol NaOH The total moles of NaOH in the new solu tion will be the sum of the moles in the two individual solutions: ? mol NaOH = 0 . 2512 mol + 0 . 0248 mol = 0 . 276 mol NaOH The total volume of the new solution will be the combined volume of the individual solu tions: ? mL new soln = 335 mL + 165 mL = 500 mL soln To calculate the molarity of NaOH in the new solution, we divide the total moles of NaOH in the new solution by the total volume of the new solution:...
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This note was uploaded on 03/23/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
 Spring '07
 Holcombe

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