An effusion container is filled with 2 L of an unknown gas. It takes 98 s for th

Question

An effusion container is filled with 2 L of an unknown gas. It takes 98 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 97 s for 2.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas.
An effusion container is filled with 2 L of an unknown gas. It takes 98 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 97 s for 2.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas.

Explanation / Answer

rate of unknown effusion = 2000 / 98 mL / sec = 20.4 mL / sec
rate of O2 = 2000 / 97 = 20.61 mL / sec

using Graham's law we see

(Rate X / Rate O2) ^2= MM O2 / MM X

(20.4 / 20.61) ^2 = 32 / MM X

MM X = 32 / 0.979 = 32.66 g/ mol

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