Given that the abundance of isotopes 54Fe,56Fe, and 57Fe is 5%, 90% and 5%, respectively. The atomic mass of Fe is (IIT-JEE, 2009)
A
55.85
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B
55.95
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C
55.75
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D
55.05
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Solution
The correct option is B 55.95 The atomic mass of iron is the average of all the masses. So I take the percentage abundance of each multiplied by its molecular weight, and divide it by 100. Which is: =(5×54)+(90×56)+(5×57)100=55.95