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

The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where

So, V = 4πR²t

Substituting the values of the variables into the equation, we have

V = 4πR²t

V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.

V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.

V = 193.6π × 10⁻²¹ m³

V = 608.21 × 10⁻²¹ m³

V = 6.0821 × 10⁻¹⁹ m³

V ≅ 6.082 × 10⁻¹⁹ m³

We know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.

Making m subject of the formula, we have

m = ρV

Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³

So, m = ρV

m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³

m = 6.082 × 10⁻¹⁶ kg

Converting to mg, we have

m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg

m = 6.082 × 10⁻¹⁰ mg

So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg

Learn more about mass of cell wall here:

the design of a ski jumping suit involves a trade-off between giving the skiers lift to get higher jumps and ensuring the skier does not face a lot of drag before the jump itself. an expansive suit may to lift the skier higher, but at the same time, it will also increase the air resistance. higher air resistance as the skier goes down the slope limits the speed of the jump. a big suit also increases the risk of losing balance or control, especially if there's a strong gust of wind. while technologically enhanced suits a lot, clothing is not the only factor in a ski jumper’s success. gathering speed while skiing down the slope, ensuring good speed on the ramp, and maintaining good body posture and form in the air all contribute to a ski jumper's success.