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Topic: Effusion( Grahams law)  (Read 4787 times)

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Immortal

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Effusion( Grahams law)
« on: March 16, 2006, 10:55:00 PM »
 :-\ the question is
"a gas of an unknown molar mass was allowed to effuse through a small openingn under constant-pressure conditions. It required 105 s for 1.0L of teh gase to effuse. Under identical conditions it required 31 s for 1.0L of O2 gas to effuse. Calculate the molar mass of the unknown gas."

i calculated it but i think i did it wrong.. the rate given are in a weird form..
i had the problem setup as follows sqrt Mx ( molar mass unknown/ sqrt 32= 1L/31s/ 1L/105s. the molar mass i found was 4.377 is this correcT?

Offline Mitch

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Re:Effusion( Grahams law)
« Reply #1 on: March 16, 2006, 10:59:16 PM »
How could a gas with a lower MW than oxygen effuse slower than Oxygen? Don't loose your chemical intuition when dealing with these kinds of problems
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Immortal

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Re:Effusion( Grahams law)
« Reply #2 on: March 16, 2006, 11:04:32 PM »
yea the concept doesn match my numbers, that why i was wondering how i would do this.. a ta mentioned "The time it takes to effuse (for any given volume of gas) is inversly proportional to u (rms speed)." but i dunno how i would setup the prob.

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