Haemoglobin contains 0.33% of iron by weight. The molecular mas of haemoglobin is about 67200. the number of iron atoms (atomic mass of Fe=56) present in molecule of haemoglobin is:

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Solution

The correct option is A 4
Percentage of $F e = 0.33$ % molecular weight of haemoglobin
Weight of $F e$ present = $\frac{0.33}{100} \times 67200$ = 221.8
One $F e$ weighs 55.8 g
Therefore, $\frac{221.8}{55.8}$ $= 4$

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

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. The number of iron atoms (Atomic weight of

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Haemoglobin contains 0.33% of iron by weight. The molecular mas of haemoglobin is about 67200. the number of iron atoms (atomic mass of Fe=56) present in molecule of haemoglobin is:

The correct option is A 4Percentage of Fe=0.33 % molecular weight of haemoglobinWeight of Fe present = 0.33100×67200 = 221.8One Fe weighs 55.8 gTherefore, 221.855.8 =4

The correct option is A 4
Percentage of $F e = 0.33$ % molecular weight of haemoglobin
Weight of $F e$ present = $\frac{0.33}{100} \times 67200$ = 221.8
One $F e$ weighs 55.8 g
Therefore, $\frac{221.8}{55.8}$ $= 4$