Question

Haemoglobin contains $0.33\%$ of $Fe$ by weight. If $1$ molecule of Haemoglobin contains two $Fe$ atoms, the molecular weight of Haemoglobin will be (at wt.of $Fe=56$) nearly:

• A. $67000$
• B. $34000$
• C. $17000$
• D. $20000$

Answer 1

The correct option is B $34000$

Given,

Haemoglobin contains $0.33\%$ of Fe by its molecular weight.

Let "x" be the molecular mass of haemoglobin.

The weight of Fe present =$\frac{0.33\%\times x}{100}$

One molecule of haemoglobin contains two atoms. (given)It is known that the number of atoms present can be calculated by dividing the molecular weight of the compound by the atomic weight of the respective atom.

i.e)$\Rightarrow 2=\frac{\frac{0.33\%\times x}{100}}{56g}$

$\Rightarrow x=\frac{2\times 56g\times 100}{0.33}$

$\Rightarrow x=33,939g/mol$

i.e)$33,939g/mol\sim 34,000g/mol$

Thus, the molecular weight of haemoglobin will be nearly $34,000$ (Option-B)