Question

Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. In the blood of an adult human, there are approximately 2.64 × 1013 red blood cells with a total of 2.90 g of iron. On the average, how many iron atoms are present in each red blood cell? (molar mass Fe = 55.85 g/mol)

Answer 1

Answer : The number of iron atoms present in each red blood cell are, $1.077\times 10^9$

Explanation :

First we have to calculate the moles of iron.

$\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles$

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains $6.022\times 10^{23}$ number of iron atoms

So, 0.0519 mole of iron contains $0.0519\times 6.022\times 10^{23}=3.125\times 10^{22}$ number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = $\frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}$

Number of iron atoms are present in each red blood cell = $\frac{3.125\times 10^{22}}{2.90\times 10^{13}}$

Number of iron atoms are present in each red blood cell = $1.077\times 10^9$

Therefore, the number of iron atoms present in each red blood cell are, $1.077\times 10^9$