Question

A certain person had a brain that weighed 1.10 kg and contained 3.76 x 10^10 cells. Assuming that each cell was completely filled with water (density= 1.00g/ml), calculate the length of one side of such a cell if it were a cube.

Answer 1

Weight of brain = 1.1 Kg

Number of cells = 3.76 x 10^10

Density of water = 1g/mL

Let volume in each cell = xmL

So mass of each cell = Density X volume

Mass of total brain = 1KG

So mass of each cell = 1100 grams / total number of cells = 1100 / 3.76 X 10^10

Volume of cell = 1100 / 3.76 x 10^10 X Density = 2.92 X 10^-8 mL

WE know that volume of cube = a^3 (a = length of one side)

Which is equal to = 2.92 X 10^-8mL = 2.92 X 10^-8 X 10^-6 m^3

Length = a = (29.2 X 10^-15 ) ^1/3 = 3.08 X 10^-5 meters