Although these quantities vary from one type of cell to another, a cell can be 2.2 micrometers in diameter with a cell wall 40 nm thick. If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg

Explanation:

First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:

A = 4πr²

where,

A = Surface Area = ?

r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m

Therefore,

A = 4π (1.1 x 10⁻⁶ m) ²

A = 15.2 x 10⁻¹² m²

Now, we find the volume of the cell wall. For that purpose, we use formula:

V = At

where,

V = Volume of the Cell Wall = ?

t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m

Therefore,

V = (15.2 x 10⁻¹² m²) (4 x 10⁻⁸ m)

V = 60.82 x 10⁻²⁰ m³

Now, to find mass of cell wall, we use formula:

ρ = m/V

m = ρV

where,

ρ = density of water = 1000 kg/m³

m = Mass of Wall = ?

Therefore,

m = (1000 kg/m³) (60.82 x 10⁻²⁰ m³)

m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg