An effusion container is filled with 2 L of an unknown gas. It takes 98 s for th
ID: 790335 • Letter: A
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