Ideal Gas Law/Molar Mass

  • Thread starter Canuck156
  • Start date
  • #1
Canuck156
32
0
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
Canuck156
32
0
Nobody knows how to do this? Come on, surely someone must be able to help? :smile:
 
  • #3
lightgrav
Homework Helper
1,248
30
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
Canuck156
32
0
Oh... OK, I get it now. Thanks lightgrav. I didn't realize 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:
 

Suggested for: Ideal Gas Law/Molar Mass

  • Last Post
Replies
8
Views
899
  • Last Post
Replies
4
Views
321
  • Last Post
Replies
2
Views
484
Replies
10
Views
230
Replies
8
Views
915
  • Last Post
Replies
4
Views
833
  • Last Post
Replies
1
Views
632
Replies
10
Views
648
Top