Answer:
The percentage of volume taken by iron(II) in the blood cell is 0.0019%.
Explanation:
Radius of iron(II) ions ,r= 75.1 pm =[tex] 7.51\times 10^{-5} \mu m[/tex]
1 pm = 10^{-6} μm
Volume of sphere =[tex]\frac{4}{3}\pi r^3[/tex]
Volume of single iron(II) ion = V
[tex]V=\frac{4}{3}\times 3.14\times (7.51\times 10^{-5} \mu m)^3[/tex]
[tex]V=1.7742\times 10^{-12} \mu m^3[/tex]
Number of iron(II) ions in one hemoglobin structure = 4
Number of hemoglobin structure in blood cell = [tex]2.50\times 10^8[/tex] molecules
Then number of iron (II) ions in [tex]2.50\times 10^8[/tex] molecules of hemoglobin:
[tex]4\times 2.50\times 10^8 ions=10^9 ions[/tex]
Volume of [tex]10^9[/tex] ions of iron = [tex]V\times 10^9[/tex]
Volume of the hemoglobin structure,V' = [tex]95 \mu m^3[/tex]
Percentage volume of iron (II) ions in a single blood cell:
[tex]\frac{10^9\times V}{V'}\times 100[/tex]
[tex]\frac{10^9\times 1.7742\times 10^{-12} \mu m^3}{95 \mu m^3}\times 100[/tex]
=[tex]0.0019\%[/tex]
