Question

Grains of fine California beach sand are approximately spheres with an average radius of 50.0 μm and are made of silicon dioxide, which has a density of 2.50 × 10³kg/m³. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.20 m on an edge?

Answer 1

Given that,

r = 50 um

$\rho$ = 2.50*10^3 kg/m^3

a = 1.20 m

surface area of one grain = 4*pi*r^2

surface area of cube = 6a^2

number of grains, N = 6a^2 / 4*pi*r^2 = 3a^2 / 2*pi*r^2

Volume of one grain, V = (4/3)*pi*r^3

mass of sand grains ,

m = N $\rho$ V

m = (3a^2 / 2*pi*r^2) * $\rho$ *  (4/3)*pi*r^3

m = 2 $\rho$ ra^2

m = 2 * 2.50*10^3 * 50*10^(-6) * 1.2²

m = 0.36 kg