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Some Basic Concepts of Chemistry question

2022 · 26 Jul · Shift 2 · Q1
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Some Basic Concepts of Chemistry question

2022 · 26 Jul · Shift 2 · Q1

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
Hemoglobin contains 0.34%0.34 \%0.34% of iron by mass. The number of Fe atoms in 3.3 g3.3 \mathrm{~g}3.3 g of hemoglobin is (Given: Atomic mass of Fe is 56 u,NA=6.022×1023 mol−156 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}56u,NA​=6.022×1023 mol−1.)
  1. A
    1.21×1051.21 \times 10^{5}1.21×105
  2. B
    12.0×101612.0 \times 10^{16}12.0×1016
  3. C
    1.21×10201.21 \times 10^{20}1.21×1020
  4. D
    3.4×10223.4 \times 10^{22}3.4×1022
View written solutionFree

Correct answer: C

  1. Mass of iron in hemoglobin

Given that hemoglobin contains 0.34%0.34\%0.34% iron by mass, the mass of Fe in 3.3 g3.3\,\text{g}3.3g hemoglobin is

Mass of Fe=0.34100×3.3=0.0034×3.3=0.01122 g\text{Mass of Fe} = \frac{0.34}{100} \times 3.3 = 0.0034 \times 3.3 = 0.01122\,\text{g}Mass of Fe=1000.34​×3.3=0.0034×3.3=0.01122g
  1. Moles of Fe

Using molar mass of iron =56 g mol−1=56\,\text{g mol}^{-1}=56g mol−1,

n(Fe)=0.0112256=2.0036×10−4 moln(\text{Fe}) = \frac{0.01122}{56} = 2.0036 \times 10^{-4}\,\text{mol}n(Fe)=560.01122​=2.0036×10−4mol
  1. Number of Fe atoms
Number of Fe atoms=n×NA\text{Number of Fe atoms} = n \times N_ANumber of Fe atoms=n×NA​ =2.0036×10−4×6.022×1023= 2.0036 \times 10^{-4} \times 6.022 \times 10^{23}=2.0036×10−4×6.022×1023 ≈1.21×1020\approx 1.21 \times 10^{20}≈1.21×1020
  1. Match with options

Thus, the number of Fe atoms is

1.21×1020\boxed{1.21 \times 10^{20}}1.21×1020​

So the correct option is C.

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