Grains of fine california beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.6 × 103 kg/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 1.1 m on an edge?

The average radius(r) of each grain is r = 50 nanometers = 50*10^-6 meters

Since it is spherical, so

 Volume=(4/3)*pi*r^3

V= (4/3)*pi*(50*10^-6)^3

V=5.23599*10^-13 m^3

We are given the Density(ρ) =2600kg/m^3

We know that:

Density(p) = mass(m)/volume(V)

m = ρV

So the mass of a single grain is:

m = 5.23599*10^-13 * 2600 = 1.361357*10^-9 kg

The surface area of a grain is:

a = 4*pi*r^2

a = 4*pi*(50*10^-6)^2

a = 3.14*10^-8 m^2

Since we know the surface area and mass of a grain, the conversion factor is:

1.361357*10^-9 kg / 3.14*10^-8 m^2

Find the Surface area of the cube:

cube = 6a^2

cube = 6*1.1^2  = 7.26m^2

multiply this by the converions ratio to get:

total mass of sand grains = (7.26 m^2 * 1.361357*10^-9 kg) / (3.14*10^-8 m^2)

total mass of sand grains = 0.3148 kg = 314.80 g