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WBJEEGiven, that the abundances of isotopes, 54Fe,56 F and \({...
Given, that the abundances of isotopes, 54Fe,56 F and 57Fe are 5%, 90% and 5%, respectively, the atomic mass of 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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