I would do it this way although it may be the long way around.
You know
(rate1/rate2)= sqrt(M2/M1) where M2 and M1 are the molar masses.
Let's assume a rate1 for N2O4 to be 1 L/sec. Use PV=nRT to calculate volume. Plug that volume into the rate1/rate2 equation to find rate2. Since the time is the same, use PV = nRT to calculate n for M2. I can't tell if you calculated it and 0.0226 is your answer or if it is the answer in the book. At any rate, I get that answer also.
Effusion/Diffusion Chemistry Help!!
If 0.0129 mol of dinitrogen tetroxide effuses through a pinhole in a certain amount of time, how much NO would effuse in that same amount of time under the same conditions? (0.0226)
3 answers
I cannot figure it out this way at all, I have been trying for a while with no success. If there is any other figures that you could add it would be appreciated.
I was trying to do it this way: Rate1 / Rate2 = square root of (Mass2 / Mass 1) but I am not coming up with .0226? I do not know where to plug the .0129 into??
I was trying to do it this way: Rate1 / Rate2 = square root of (Mass2 / Mass 1) but I am not coming up with .0226? I do not know where to plug the .0129 into??
sqrt of 92.011/30.006=1.751 x .0129 = .0226 mol of NO