Answer: [tex]2(10)^{-9} mg[/tex]
Explanation:
We know the total diameter of the cell (assumed spherical) is:
[tex]d=1.9\mu m=1.9(10)^{-6} m[/tex]
Then its total radius [tex]r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m[/tex]
On the other hand, we know the thickness of the cell wall is [tex]r_{t}=60 nm= 60(10)^{-9} m[/tex] and its density is the same as water ([tex]\rho=997 kg/m^{3}[/tex]).
Since density is the relation between the mass [tex]m[/tex] and the volume [tex]V[/tex]:
[tex]\rho=\frac{m}{V}[/tex]
The mass is: [tex]m=\rho V[/tex] (1)
Now if we are talking about this cell as a thin spherical shell, its volume will be:
[tex]V=\frac{4}{3}\pi R^{3}[/tex] (2)
Where [tex]R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m[/tex]
Then:
[tex]V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}[/tex] (3)
[tex]V=2.952(10)^{-18} m^{3}[/tex] (4)
Substituting (4) in (1):
[tex]m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})[/tex] (5)
[tex]m=2.94(10)^{-15} kg[/tex] (6)
Knowing [tex]1 kg=1000 g[/tex] and [tex]1 mg=0.001 g[/tex]:
[tex]m=2.94(10)^{-15} kg=2(10)^{-9} mg[/tex]
