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Q. Given that the abundances of isotopes $^{54}Fe, ^{56}Fe$ and $^{57}Fe$ are $5\%, 90\%$ and $5\%$, respectively, the atomic mass of Fe is

Structure of Atom

Solution:

Average atomic mass of $Fe$
$=\frac{\left(54\times5\right)+\left(56\times90\right)+\left(57\times5\right)}{100}=55.95$