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WBJEEGiven, that the abundances of isotopes, \({ }_{54} \mathrm{Fe},{ }_{56} \mathrm{~F}\) and \({...
Given, that the abundances of isotopes, \({ }_{54} \mathrm{Fe},{ }_{56} \mathrm{~F}\) and \({ }_{57} \mathrm{Fe}\) are \(5 \%\), \(90 \%\) and \(5 \%\), respectively, the atomic mass of \(\mathrm{Fe}\) is
Solution:
Average atomic weight
\(\)
\frac{54 \times 5+56 \times 90+57 \times 5}{100}=55.95
\(\)
\(\)
\frac{54 \times 5+56 \times 90+57 \times 5}{100}=55.95
\(\)
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