Question

49. Number of Fe atoms in 22.4 g Haemoglobin if it
contains 0.25% Fe (Atomic mass of Fe = 56) is
(1) 6.02 x 1021
(2) 6.02 x 1020
(3) 3.01 x 1020 (4) 3.01 x 1021​

Answer 1

Answer:

Option (2) $6.02\times 10^{20}$

Explanation:

Haemoglobin contains 0.25% Fe

Thus 22.4 gram of haemoglobin will contain Fe

$=22.4\times\frac{0.25}{100}$ g

$=22.4\times\frac{1}{400}$

$=0.056$ g

Atomic mass of Fe = 56

Therefore, moles of Fe in 22.4 g of haemoglobin

$=\frac{\text{Weight in grams}}{\text{Atomic Weight}}$

$=\frac{\text{0.056}}{\text{56}}$

$=10^{-3}$ mole

1 mole of Fe contains = $6.023\times 10^{23}$ atoms

∴ $10^{-3}$ mole of Fe will contain

$=6.023\times 10^{23}\times 10^{-3}$ atoms

$=6.02\times 10^{20}$ atoms

Hope this helps.