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.