Question

What is the volume (in km3) of Avogadro's number of sand grains if each grain is a cube and has sides that are 1.6 mm long? km3 (b) How many kilometers of beaches in length would this cover if the beach averages 121 m in width and 11.1 m in depth? Neglect air spaces between grains. km

Answer 1

Part A.

Number of sand grains = 6.022*10^23 grains

Volume of each grain = side^3 = (1.6 mm)^3 = 4.096 mm^3

Since 1 mm = 10^-3 m, and 1 km = 1000 m, So

1 km = 10^6 mm

Now

Total Volume of sand grains = V = 6.022*10^23*4.096 mm^3

V = (6.022*10^23*4.096 mm^3)*(1 km/10^6 mm)^3

V = 6.022*10^23*4.096/10^18 = 2466611.2 km^3

V = 2.47*10^6 km^3

Part B.

When these grains are on the beach of width = 121 m & depth = 11.1 m, then

V = Volume of Beach = L*W*D

L = Length of beach = ?

L = V/(W*D)

Since 1 km = 10^3 m, So

W = 121 m*(1 km/10^3 m) = 0.121 km

D = 11.1 m*(1 km/10^3 m) = 0.0111 km

So,

L = 2466611.2 km^3/(0.121 km*0.0111 km)

L = 2466611.2/(0.121*0.0111) km

L = 1836505993.6 km

L = 1.84*10^9 km = length of beach

Let me know if you've any query.