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Haemoglobin $(C_{2952}H_{9664}N_{812}O_{832}S_8Fe_4)$ is oxygen carrier. If a person has $5L$ blood and each mL of blood has $5\times 10^9$ R.B.C and each R.B.C has $2.8\times 10^8$ haemoglobin molecules, then how many grams of haemoglobin is present in the body of that person?

$\begin{array}{1 1}(a)\;756.8g&(b)\;75.68g\\(c)\;1531.7g&(d)\;153.1g\end{array}$

1 Answer

Molecular mass of Haemoglobin $(C_{2952}H_{9664}N_{812}O_{832}S_8Fe_4)$ =65248 g
Moles of Hb=$\large\frac{2.8\times 10^8\times 5\times 10^9\times 5\times10^3}{6.023\times 10^{23}}$
$ = 11.6\times 10^{-3}Mol$
$\therefore$ mass of Hb=mol $\times$ mol.mass
$= 11.6\times 10^{-3}\times 65248$
$= 756.8g$
Hence (a) is the correct answer.
answered Oct 25, 2013 by sreemathi.v
edited Mar 18, 2014 by mosymeow_1

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