Haemoglobin contains $0.33$ % of $F e$ by weight. If $1$ molecule of haemoglobin contains $2 F e$ atoms, the molecular weight of haemoglobin will be nearly:

[Atomic weight of $F e = 56$ ]

67000

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34000

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17000

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20000

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Solution

The correct option is A $34000$
$% F e = \frac{Weight of Fe}{Molecular weight of compound} \times 100$

Weight of Fe $2 \times 56 = 112 g m$

$\Rightarrow \frac{112}{x} \times 100 = 0.33$

$∴ x = 34000 =$ Molecular weight of Hemoglobin.

Hence, option B is correct.

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Q. 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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Haemoglobin contains 0.33% of Fe by weight. If 1 molecule of haemoglobin contains 2Fe atoms, the molecular weight of haemoglobin will be nearly:[Atomic weight of Fe=56]

The correct option is A 34000 %Fe=Weight of FeMolecular weight of compound×100Weight of Fe 2×56=112 gm⇒112x×100=0.33∴x=34000= Molecular weight of Hemoglobin.Hence, option B is correct.

Haemoglobin contains $0.33$ % of $F e$ by weight. If $1$ molecule of haemoglobin contains $2 F e$ atoms, the molecular weight of haemoglobin will be nearly:

[Atomic weight of $F e = 56$ ]

The correct option is A $34000$
$% F e = \frac{Weight of Fe}{Molecular weight of compound} \times 100$

Weight of Fe $2 \times 56 = 112 g m$

$\Rightarrow \frac{112}{x} \times 100 = 0.33$

$∴ x = 34000 =$ Molecular weight of Hemoglobin.

Hence, option B is correct.