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)
sqrt of 92.011/30.006=1.751 x .0129 = .0226 mol of NO
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.
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??
To find out how much NO would effuse in the same amount of time as dinitrogen tetroxide (N2O4) under the same conditions, we need to use Graham's law of effusion/diffusion. Graham's law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.
First, we need to find the molar mass of dinitrogen tetroxide (N2O4) and nitrogen monoxide (NO). The atomic masses of nitrogen (N) and oxygen (O) are 14.01 g/mol and 16.00 g/mol, respectively.
Molar mass of N2O4 = 2(N) + 4(O) = 2(14.01 g/mol) + 4(16.00 g/mol) = 92.02 g/mol
Next, we can calculate the rate of effusion of dinitrogen tetroxide (N2O4) using the given amount of 0.0129 mol and the molar mass of N2O4.
Rate of effusion of N2O4 = (0.0129 mol) / (92.02 g/mol) = 0.000140 mol/g
Now, using Graham's law, we can find the rate of effusion of nitrogen monoxide (NO) by multiplying the rate of effusion of N2O4 by the square root of the ratio of their molar masses.
Rate of effusion of NO = (0.000140 mol/g) * √(molar mass of N2O4 / molar mass of NO)
To calculate the rate of effusion of NO, we need the molar mass of NO, which is calculated as follows:
Molar mass of NO = (N) + (O) = (14.01 g/mol) + (16.00 g/mol) = 30.01 g/mol
We can now substitute the values into the formula to find the rate of effusion of NO:
Rate of effusion of NO = (0.000140 mol/g) * √(92.02 g/mol / 30.01 g/mol)
Calculating this, we get:
Rate of effusion of NO ≈ 0.000431 mol/g
Finally, to find the amount of NO that would effuse in the same amount of time as N2O4 (0.0226), we multiply the rate of effusion of NO by the molar mass of NO:
Amount of NO = (Rate of effusion of NO) * (molar mass of NO) * (time)
Amount of NO = (0.000431 mol/g) * (30.01 g/mol) * (0.0226)
Calculating this, we get:
Amount of NO ≈ 0.0293 mol
Therefore, approximately 0.0293 mol of NO would effuse in the same amount of time as 0.0129 mol of N2O4 under the given conditions.