A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 120 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 32 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)
A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 120 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 32 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter9: Gases
Section: Chapter Questions
Problem 83E: Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass,...
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A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 120 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 32 s for 1.0 L of O2 gas to effuse.
Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)
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