Exam 3-solutions - Version 020 Exam 3 fakhreddine(51010 This print-out should have 32 questions Multiple-choice questions may continue on the next

Exam 3-solutions - Version 020 Exam 3 fakhreddine(51010...

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Version 020 – Exam 3 – fakhreddine – (51010) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In which of these compounds would you find ONLY dispersion forces existing between the molecules? I. CCl 4 ; II. NH 3 ; III. HBr; IV. CO 2 . 1. I and IV only correct 2. I and III only 3. II only 4. II and IV only 5. I and II only 6. II and III only 7. I only 8. IV only 9. III and IV only 10. III only Explanation: A nonpolar covalent molecule would have only dispersion forces with another nonpolar covalent molecule. 002 10.0 points Consider a pure sample of each of these com- pounds RbCl FeCl 3 FeN SrCl 2 Order from weakest to strongest intermolecu- lar forces. 1. RbCl < FeN < SrCl 2 < FeCl 3 2. RbCl < SrCl 2 < FeCl 3 < FeN correct 3. FeCl 3 < SrCl 2 < RbCl < FeN 4. FeN < RbCl < SrCl 2 < FeCl 3 5. FeN < SrCl 2 < RbCl < FeCl 3 Explanation: 003 10.0 points What mass of (NH 4 ) 2 SO 4 would be required to prepare 500 mL of a 7.50% solution of am- monium sulphate ((NH 4 ) 2 SO 4 )? Its density is 1.04 g/mL. 1. 41.6 g 2. 39.0 g correct 3. 36.0 g 4. 28.2 g 5. 43.8 g Explanation: V = 500 mL density soln = 1 . 04 g / mL The (NH 4 ) 2 SO 4 is 7.50% (NH 4 ) 2 SO 4 by mass. We can write this as the ratio 7 . 50 g (NH 4 ) 2 SO 4 100 g soln . The density of the solution is 1.04 g/mL. This can be more fully written as 1 . 04 g soln 1 mL soln . We want to prepare 500 mL of the solution. We use the solution density to convert from mL solution to g solution: ? g soln = 500 mL soln × 1 . 04 g soln 1 mL soln = 520 g soln Now we can use the percent by mass to con- vert from grams to grams (NH 4 ) 2 SO 4 : ? g (NH 4 ) 2 SO 4 = 520 g soln × 7 . 50 g (NH 4 ) 2 SO 4 100 g soln = 39 g (NH 4 ) 2 SO 4 004 10.0 points
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Version 020 – Exam 3 – fakhreddine – (51010) 2 At constant temperature if the volume of a gas sample is tripled the pressure will 1. remain the same. 2. be increased by a factor of 6. 3. be one third the original. correct 4. be cut in half. 5. be tripled. Explanation: 3 V 1 = V 2 Boyle’s Law relates the volume and pressure of a sample of gas: P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 (3 V 1 ) P 1 = 3 P 2 P 2 = 1 3 P 1 005 10.0 points What volume of 0 . 135 M K 3 PO 4 is required to react with 87 mL of 0 . 401 M MgCl 2 according to the equation 2 K 3 PO 4 + 3 MgCl 2 Mg 3 (PO 4 ) 2 + 6 KCl 1. 54.3243 2. 114.484 3. 32.9169 4. 44.8035 5. 19.5065 6. 172.368 7. 29.8988 8. 146.092 9. 31.9531 10. 31.4894 Correct answer: 172 . 368 mL. Explanation: [K 3 PO 4 ] = 0 . 135 M V = 87 mL [MgCl 2 ] = 0 . 401 M First, we determine the moles of MgCl 2 : ? mol MgCl 2 = 87 mL soln × 1 L soln 1000 mL soln × 0 . 401 mol MgCl 2 1 L soln = 0 . 034887 mol MgCl 2 Using the mole-to-mole ratio from the bal- anced chemical equation we calculate the moles of K 3 PO 4 needed to react with this amount of MgCl 2 : ? mol K 3 PO 4 = 0 . 087 mol MgCl 2 × 2 mol K 3 PO 4 3 mol MgCl 2 = 2 . 3258 × 10 5 mol K 3 PO 4 Now we use the molarity of the K 3 PO 4 solu- tion to convert from moles of K 3 PO 4 to the volume of K 3 PO 4 : ? mL soln = 2 . 3258 × 10 5 mol K 3 PO 4 × 1 L soln 0 . 135 mol K 3 PO 4 × 1000 mL soln 1 L soln = 172 . 368 mL soln It would require 172 . 368 mL of the K 3 PO 4 solution to provide enough moles of K 3 PO 4 to react all of the MgCl 2 .
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  • Fall '08
  • wandelt
  • Chemistry, Mole, Intermolecular force, Van der Waals force

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