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Ideal Gas Law/Molar Mass

  • Thread starter Canuck156
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  • #1
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Hi, this is the question I'm having trouble with:

A gas mixture is known to contain equal numbers of moles of two gases. The mixture has a density of 1.47g/L at 1.00 atm and 298K. In a diffusion experiment, one of the gases was found to diffuse 1.25 times faster than the other under the same conditions. What are the respective molar mases of the two gases?

I've found that [tex]1.25=\sqrt{\frac{M_B}{M_A}}[/tex] Therefore [tex]1.5625=\frac{M_B}{M_A}[/tex], where [tex]M_B[/tex] is the molar mass of gas B. However, from there I have no idea where to go next. I think I'm supposed to use the ideal gas law, but I'm not sure how to apply it when there's a mixture of gases. Any help is appreciated! Thanks!
 

Answers and Replies

  • #2
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Nobody knows how to do this? Come on, surely someone must be able to help? :smile:
 
  • #3
lightgrav
Homework Helper
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does the ideal gas law depend on
what those 10^23 objects are?
(an ideal gas = atoms don't interact)

so find (n_moles/V)(m/mole).
 
  • #4
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Oh... OK, I get it now. Thanks lightgrav. I didn't realise that that was one of the properties of an ideal gas... Maybe I shouldn't be doing University Chem without having done High School Chem first... :rolleyes:
 

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