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1. While performing the Dumas experiment, if the outside of the flask were not completely dry (Part I), how would this affect the calculated molar mass of the unknown volatile liquid? Would it be higher or lower? Explain.

1. While performing the Dumas experiment, if the outside of the flask were not completely dry (Part I), how would this affect the calculated molar mass of the unknown volatile liquid? Would it be higher or lower? Explain.

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter10: Solutions
Section: Chapter Questions
Problem 89QAP
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3:12 & K Post-laboratory Qu... POST-LABORATORY QUESTIONS Due after Laboratory Experiment 1. While performing the Dumas experiment, if the outside of the flask were not completely dry (Part I), how would this affect the calculated molar mass of the unknown volatile liquid? Would it be higher or lower? Explain. 2. While determining the pressure during the Dumas experiment, if the barometric pressure reported were lower than the actual vapor pressure (Part 1), how would this error affect the calculated molar mass of the unknown volatile liquid? Explain. 3. While determining the volume during this experiment (Part II), if the total volume of the flask were incorrectly reported as 125 mL, how would this error affect the calculated molar mass of the unknown volatile liquid? Explain. 4. Calculate the number of moles of Argon gas contained in a 4.00 dm³ container at 620. kPa and 25.0°C. If the gas is Helium instead of Argon, will the answer change? Explain 5. In the Dumas method for determining the molar mass of an unknown volatile liquid, a sample of a volatile liquid that boils below 100°C is vaporized in a boiling water bat and the mass of the vapor required to fill the flask is determined. The following data was collected: 6. Flask Volume Mass of Flask + Cap Mass of Flask + Condensed Liquid Mass of Condensed Liquid Temperature of Water Bath (°C) Barometric Pressure (atm). 100.0 100.0 +79.6 mL = 279.6 ml 143.85 g 144.95 g 1.10 g 84.1 °C 0.985 atm a. What is the molar mass of the unknown? b. What will its temperature be when the pressure is 1.20 atm and the volume is 250.0 mL? HE mical analysis of an unknown compound shows that it contains 64.9% C, 13.5% and 21.6% 0 by mass. At 120°C a 50 mmHg, 2.00 L of the gaseous compound weighs 4.60 g.
Transcribed Image Text:3:12 & K Post-laboratory Qu... POST-LABORATORY QUESTIONS Due after Laboratory Experiment 1. While performing the Dumas experiment, if the outside of the flask were not completely dry (Part I), how would this affect the calculated molar mass of the unknown volatile liquid? Would it be higher or lower? Explain. 2. While determining the pressure during the Dumas experiment, if the barometric pressure reported were lower than the actual vapor pressure (Part 1), how would this error affect the calculated molar mass of the unknown volatile liquid? Explain. 3. While determining the volume during this experiment (Part II), if the total volume of the flask were incorrectly reported as 125 mL, how would this error affect the calculated molar mass of the unknown volatile liquid? Explain. 4. Calculate the number of moles of Argon gas contained in a 4.00 dm³ container at 620. kPa and 25.0°C. If the gas is Helium instead of Argon, will the answer change? Explain 5. In the Dumas method for determining the molar mass of an unknown volatile liquid, a sample of a volatile liquid that boils below 100°C is vaporized in a boiling water bat and the mass of the vapor required to fill the flask is determined. The following data was collected: 6. Flask Volume Mass of Flask + Cap Mass of Flask + Condensed Liquid Mass of Condensed Liquid Temperature of Water Bath (°C) Barometric Pressure (atm). 100.0 100.0 +79.6 mL = 279.6 ml 143.85 g 144.95 g 1.10 g 84.1 °C 0.985 atm a. What is the molar mass of the unknown? b. What will its temperature be when the pressure is 1.20 atm and the volume is 250.0 mL? HE mical analysis of an unknown compound shows that it contains 64.9% C, 13.5% and 21.6% 0 by mass. At 120°C a 50 mmHg, 2.00 L of the gaseous compound weighs 4.60 g.
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