Question
A cube of aluminum contains 10^21 atoms. how long is its edge(in mm)?
Answers
drwls
Convert 10^21 atoms to g using the atomic weight. Then convert it to volume using the density. Then take the cube root of the volume in cm^3.
Moles = 10^21/6.02*10^23 = 1.661*10^-3 moles
Mass = 23 * 1.661*10^-3 = 0.0382 g
Volume = 0.0382 g/2.70 g/cm^3 = 1.415*10^-2
Edge Length = (1.415*10^-2)^1/3 = 0.24 cm
Moles = 10^21/6.02*10^23 = 1.661*10^-3 moles
Mass = 23 * 1.661*10^-3 = 0.0382 g
Volume = 0.0382 g/2.70 g/cm^3 = 1.415*10^-2
Edge Length = (1.415*10^-2)^1/3 = 0.24 cm