Chemistry, asked by abhishek104, 1 year ago

what will be the answer??

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Answered by prashant42

Percentage of Fe = 0.33% molecular weight of haemoglobin Weight of Fe present = (0.33/100) × 67200 = 221.8one Fe weighs 55.8 g Therefore, 221.8/55.8 = 4Hence, 4 atoms of iron are present in one molecule of haemoglobin.

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