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Hemoglobin contains \mathrm{ 0.34 \%} of iron by mass. The number of \mathrm{ {Fe}} atoms in \mathrm{3.3 \mathrm{~g}} of hemoglobin is

\mathrm{\text { (Given: Atomic mass of } \mathrm{Fe} \text { is } 56 \mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1} \text {.) }}

Option: 1

1.21 \times 10^{5}


Option: 2

12.0 \times 10^{16}


Option: 3

1.21 \times 10^{20}


Option: 4

3.4 \times 10^{22}


Answers (1)

best_answer

Hemoglobin contains 0.34% of iron by mass,

So, in 100 g Hemoglobin 34 g is iron ,

then in 3.3 g Hemoglobin \rightarrow 3.3\times \frac{0.34}{100}\text{g} of iron

\Rightarrow 3.3 g Hemoglobin \rightarrow 1.12 \times 10^{-2}\text{g of iron}

Now, \mathrm{mole \; of\; iron=\frac{1.12\times10^{-2}}{56}mole=0.01986\times10^{-2}mole}

We know, \mathrm{1\; mole = 6.022\times 10^{23}atoms}

then,

\begin{aligned} 0.019 \times 10^{-2} \text { mole } &=0.015 \times 10^{-2} \times 6.022 \times 10^{23} \text { atoms } \\ &=1.20 \times 10^{20} \text { atom } \end{aligned}

Hence, Option (3) is correct.

Posted by

manish

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