can anyone explain to me how to find the factorial of avogadro number ? what is its value ?
It's such a large number that you might be able to justify using Stirling's approximation.
Google for Stirling's approximation.
I can tell you that the number of base 10 digits is > 6.02x10^23. You can't calculate this. Recently I have been coding my own big number c++ library. I often test it on 1,000,000!. It takes 5 minutes, and produces somewhere around 5 million digits. I doubt you are asked to calculate the exact value of 6.02E23!. If you want an approximation use Stirling approx. I use it in my e-calculator to find how many terms I need to sum to get the the desired accuracy.
ln(n!) = n(ln(n) - 1) Note this is only a good approximation.
This therefore means that n! = en(ln(n) - 1)) (approximately)
assumes usual metric value of
6.022141×10^23 mol^(-1) (reciprocal moles)
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