Graham's Law question.
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Graham's Law question.

[From: ] [author: ] [Date: 11-12-13] [Hit: ]
MM2=[83.8^1/2 * (1/87.3)/(1/131.MM2=189.......
A sample of Kr gas escapes through a tiny hole in 87.3 sec. Under the same conditions, the same amount of unknown gas effuses in 131.3 sec. What is the molar mass of the unknown gas?

When I tried this question, I got the Molar mass of the unknown gas as 189.56 g/mol. I'm not sure if it is right. Please tell me quick. I have university exam tomorrow!!

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r1/ r2 = √MM2 / √MM1 solve for MM2
MM2 =[ √MM1*nr1/r2]^2
MM1 Kr =83.8g
t1=87.3, t2=131.3 , r1=1/t1, r2=1/t2

MM2=[83.8^1/2 * (1/87.3)/(1/131.3)]^2
MM2=189.6 g/m
1
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