Hemoglobin contains $0.34 \%$ of iron by mass. The number of Fe atoms in $3.3 \mathrm{~g}$ of hemoglobin is
(Given: Atomic mass of Fe is $56 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$.)
Solution
According to the question,
<br/><br/>
$100 \mathrm{~g}$ of hemoglobin contains $0.34 \mathrm{~g}$ of iron<br/><br/> $3.3 \mathrm{~g}$ of hemoglobin contains $\frac{0.34}{100} \times 3.3 \mathrm{~g}$ of iron <br/><br/>moles of $$\mathrm{Fe}=\frac{0.34 \times 3.3}{100 \times 56}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$$
<br/><br/>
$N=\frac{0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$
<br/><br/>
$=1.21 \times 10^{20}$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
This question is part of PrepWiser's free JEE Main question bank. 163 more solved questions on Some Basic Concepts of Chemistry are available — start with the harder ones if your accuracy is >70%.