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Exam 3-solutions

# Exam 3-solutions - Version 241 – Exam 3 – sutcliffe...

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Unformatted text preview: Version 241 – Exam 3 – sutcliffe – (52380) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How much 0.100 M H 2 SO 4 are needed to make 25.0 mL of 0.00500 M solution? 1. 500 mL 2. 1.25 mL correct 3. 0.00200 mL 4. 0.800 mL Explanation: M 1 = 0.100 M M 2 = 0.00500 M V 2 = 25 . 0 mL A portion of the 0 . 1 M H 2 SO 4 solution will be diluted with water to form 25 mL of the . 005 M H 2 SO 4 solution. All of the moles of H 2 SO 4 in the new dilute solution must come from the more concentrated solution. (There is no H 2 SO 4 in the water!) We use the desired volume and molarity of the dilute solution to determine the number of moles of H 2 SO 4 needed to make this solution: ? mol H 2 SO 4 = 25 mL soln × 1 L soln 1000 mL soln × . 005 mol H 2 SO 4 1 L soln = 0 . 000125 mol H 2 SO 4 We need enough of the 0 . 1 M solution to provide 0 . 000125 mol H 2 SO 4 . We use the molarity to convert from moles to volume of the solution: ? mL soln = 0 . 000125 mol H 2 SO 4 × 1 L soln . 1 mol H 2 SO 4 × 1000 mL soln 1 L soln = 1 . 25 mL soln 1 . 25 mL of the 0 . 1 M solution diluted to 25 mL with water would give us the desired solution. 002 10.0 points The density of the vapor of allicin, a compo- nent of garlic, is 1 . 14 g · L − 1 at 125 ◦ C and 175 Torr. What is the molar mass of allicin? 1. 50 . 8 g · mol − 1 2. 162 g · mol − 1 correct 3. 273 g · mol − 1 4. 869 g · mol − 1 5. 21 . 6 g · mol − 1 Explanation: T = 125 ◦ C + 273.15 K = 398 . 15 K P = (175 Torr) 1 atm 760 Torr = 0 . 230263 atm ρ = 1 . 14 g / L The ideal gas law is P V = nRT n V = P RT with unit of measure mol/L on each side. Multiplying each by molar mass (MM) gives n V · MM = P RT · MM = ρ , with units of g/L. MM = ρ RT P = (1 . 14 g / L) ( . 08206 L · atm mol · K ) . 230263 atm × (398 . 15 K) = 161 . 755 g / mol 003 10.0 points Consider the following reaction: 2 FeS + 3 N 2 → 2 FeN + 2 SN 2 This reaction has a 65.0% yield. Version 241 – Exam 3 – sutcliffe – (52380) 2 How much N 2 is consumed if 13.6 g of FeN are produced? 1. 0.200 mol 2. 31.4 mol 3. 0.190 mol 4. 0.449 mol correct 5. 412 mol 6. 0.150 mol 7. 0.899 mol Explanation: % yield = 65% m FeN = 13 . 6 g 13.6 g FeN is the actual yield. We need to first calculate the theoretical yield. % yield = actual yield theoretical yield theoretical yield = actual yield % yield ? mol N 2 = 13 . 6 g FeN × 100 65 × 1 mol FeN 69 . 86 g FeN × 3 mol N 2 2 mol FeN = 0 . 449 mol N 2 004 10.0 points Which of the following statements about dis- persion forces is NOT correct? Dispersion forces 1. are also called London forces....
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Exam 3-solutions - Version 241 – Exam 3 – sutcliffe...

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