I have a problem that I wasn't sure how to do, and I want to see if my reasoning behind getting the answer is correct.
Question: If .0129 mol of N2O4 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?
I know that Graham's Law states that "At a given temperature and pressure, the rate of effusion of a gas, in moles per unit time, is inversely proportional to the square root of its molar mass."
Which would mean (MM N2O4)^.5= rate of effusion
(92.02)^.5= 9.5927 mol/unit t
9.5927 mol/unit t / .0129 mol = 5.291 unit t
(MM NO)^.5= rate of effusion
(28.01)^.5= 5.292 mol/unit t
5.292 mol/unit t / x = 5.291 unit t
x= 1.00 mol
Does this seem correct?
I didn't work it through, but, no, it doesn't seem right. First, the molar mass of NO must be closer to 30 than it is to 28.01. Check that.
(r1/r2)=(sqrt M2/sqrt M1)
(0.129 mols/x mols)=(sqrt92.02/sqrt30)
xmol NO=0.129(sqrt92.02/sqrt30)=>??
To determine how much NO would effuse in the same amount of time, we can use Graham's Law of effusion. The formula for effusion rate is as follows:
Rate of effusion = (sqrt(Molar mass of N2O4)) / (sqrt(Molar mass of NO))
Let's calculate the molar mass of NO first. Nitrogen (N) has an atomic mass of approximately 14.01 g/mol, and oxygen (O) has an atomic mass of around 16.00 g/mol. Adding them together gives us a total molar mass of about 30.01 g/mol for NO.
Now, we can substitute the values into Graham's Law formula:
Rate of effusion (N2O4) / Rate of effusion (NO) = (sqrt(92.02)) / (sqrt(30.01))
To solve for the rate of effusion of NO, we rearrange the equation:
Rate of effusion (NO) = Rate of effusion (N2O4) * (sqrt(30.01)) / (sqrt(92.02))
Substituting the given rate of effusion for N2O4 as 0.0129 mol/unit t:
Rate of effusion (NO) = 0.0129 * (sqrt(30.01)) / (sqrt(92.02))
Evaluating this expression gives us your answer.