Haemoglobin contains 0.33 % of iron by weight. The molecular weight of haemoglobin is approximately 67200 g. The number of iron atoms (at. weight of Fe is 56 ) present in one molecule of haemoglobin are

Question

Haemoglobin contains 0.33 % of iron by weight. The molecular weight of haemoglobin is approximately 67200 g. The number of iron atoms (at. weight of Fe is 56 ) present in one molecule of haemoglobin are (A) 1 (B) 6 (C) 4 (D) 2

Tardigrade Admin · Accepted Answer

∵ 0.33 % of iron by weight means 100 g of haemoglobin has 0.33 g of iron 100 g of haemoglobin contains iron =0.33 g ∴ 67200 g of haemoglobin contains iron =(0.33 × 67200/100) g =221.76 g text of Fe Number of Fe-atoms =(221.76/56) =3.96 ≈ 4

Q. Haemoglobin contains $0.33%$ of iron by weight. The molecular weight of haemoglobin is approximately $67200 g$ . The number of iron atoms (at. weight of $F e$ is $56$ ) present in one molecule of haemoglobin are

1

6

4

2

$∵ 0.33%$ of iron by weight means $100 g$ of haemoglobin has $0.33 g$ of iron
$100 g$ of haemoglobin contains iron $= 0.33 g$
$∴ 67200 g$ of haemoglobin contains iron
$= \frac{0.33 \times 67200}{100} g = 221.76 g of F e$
Number of Fe-atoms $= \frac{221.76}{56}$
$= 3.96 \approx 4$

Q. Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200g. The number of iron atoms (at. weight of Fe is 56 ) present in one molecule of haemoglobin are

1

6

4

2

∵0.33% of iron by weight means 100g of haemoglobin has 0.33g of iron 100g of haemoglobin contains iron =0.33g ∴67200g of haemoglobin contains iron =1000.33×67200​g=221.76g of Fe Number of Fe-atoms =56221.76​ =3.96≈4

Q. Haemoglobin contains $0.33%$ of iron by weight. The molecular weight of haemoglobin is approximately $67200 g$ . The number of iron atoms (at. weight of $F e$ is $56$ ) present in one molecule of haemoglobin are

$∵ 0.33%$ of iron by weight means $100 g$ of haemoglobin has $0.33 g$ of iron
$100 g$ of haemoglobin contains iron $= 0.33 g$
$∴ 67200 g$ of haemoglobin contains iron
$= \frac{0.33 \times 67200}{100} g = 221.76 g of F e$
Number of Fe-atoms $= \frac{221.76}{56}$
$= 3.96 \approx 4$