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Haemoglobin contains $0.33 \%$ of irons by weight. The molecular weight of haemoglobion is approximately $67200$. The number of iron atoms (at wt of $Fe=56$) present in one molecule of haemolgobin is

$(a)\;6 \\ (b)\;1 \\(c)\;4 \\(d)\;2 $
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Molecular weight of haemoglobin $=67200(Given)$
$\%$ of Fe present in the sample $=0.33 \%$
$\therefore$ Amount of $Fe$ present in haemoglobin $=\large\frac{0.33 }{100}$$ \times 67200$
$\qquad= 222\;amu$
Thus the number of atoms present $=\large\frac{222}{56}$
$\qquad= 3.9 =4 \;amc$
Hence c is the correct answer.
answered Apr 1, 2014 by meena.p

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