JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 1)
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 $$56 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$$.)
$$1.21 \times 10^{5}$$
$$12.0 \times 10^{16}$$
$$1.21 \times 10^{20}$$
$$3.4 \times 10^{22}$$
Explanation
According to the question,
$$100 \mathrm{~g}$$ of hemoglobin contains $$0.34 \mathrm{~g}$$ of iron
$$3.3 \mathrm{~g}$$ of hemoglobin contains $$\frac{0.34}{100} \times 3.3 \mathrm{~g}$$ of iron
moles of $$\mathrm{Fe}=\frac{0.34 \times 3.3}{100 \times 56}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$$
$N=\frac{0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$
$$ =1.21 \times 10^{20} $$
$$100 \mathrm{~g}$$ of hemoglobin contains $$0.34 \mathrm{~g}$$ of iron
$$3.3 \mathrm{~g}$$ of hemoglobin contains $$\frac{0.34}{100} \times 3.3 \mathrm{~g}$$ of iron
moles of $$\mathrm{Fe}=\frac{0.34 \times 3.3}{100 \times 56}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$$
$N=\frac{0.34 \times 3.3 \times 6.022 \times 10^{23}}{100 \times 56}$
$$ =1.21 \times 10^{20} $$
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