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2. According to the average molar mass your calculated, which one of the following liquids is possibly your unknown sample (show the calculations for how you selected your answer). 1 a. Water H₂O b. Acetone C3H6O c. Cyclohexane C6H12 d. Ethanol C₂H5OH

2. According to the average molar mass your calculated, which one of the following liquids is possibly your unknown sample (show the calculations for how you selected your answer). 1 a. Water H₂O b. Acetone C3H6O c. Cyclohexane C6H12 d. Ethanol C₂H5OH

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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R = 0.0821 L.atm/mol.°K Sample ID: A Mass of the empty Erlenmeyer flask + rubber band + foil Pressure Temperature (°C) Temperature (°K) Mass of flask + vapor + rubber band + foil Mass of vapor (grams) Volume of Flask Calculated number of moles (n) Molar Mass (g/mol) Average Molar Mass (g/mol) n= / 107.2189 I atm 95°℃ 368.15⁰K 108.209 0.9829 MM- = pv 1* 0.33 RT 0.0821 368.15 Trial 1 330ml=0.33L a. Water H₂O b. Acetone C3H6O c. Cyclohexane C6H12 d. Ethanol C₂H5OH n = :0.010 met 0.982 0.010 = 98.29/mol n= Trial 2 107.2189 I atm 95.1°C 368.25 k 108.2969 1.0789 330ml=0.33L pv RT MM = 103 g/mol 1*0.33 0.0821368.25 .:0.010 mol m 1.078 n 0.010 = 2. According to the average molar mass your calculated, which one of the following liquids is possibly your unknown sample (show the calculations for how you selected your answer). =107.89/mol
Transcribed Image Text:R = 0.0821 L.atm/mol.°K Sample ID: A Mass of the empty Erlenmeyer flask + rubber band + foil Pressure Temperature (°C) Temperature (°K) Mass of flask + vapor + rubber band + foil Mass of vapor (grams) Volume of Flask Calculated number of moles (n) Molar Mass (g/mol) Average Molar Mass (g/mol) n= / 107.2189 I atm 95°℃ 368.15⁰K 108.209 0.9829 MM- = pv 1* 0.33 RT 0.0821 368.15 Trial 1 330ml=0.33L a. Water H₂O b. Acetone C3H6O c. Cyclohexane C6H12 d. Ethanol C₂H5OH n = :0.010 met 0.982 0.010 = 98.29/mol n= Trial 2 107.2189 I atm 95.1°C 368.25 k 108.2969 1.0789 330ml=0.33L pv RT MM = 103 g/mol 1*0.33 0.0821368.25 .:0.010 mol m 1.078 n 0.010 = 2. According to the average molar mass your calculated, which one of the following liquids is possibly your unknown sample (show the calculations for how you selected your answer). =107.89/mol
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