Answer to Question #181760 in General Chemistry for Sheru

Question

Cobalt crystallizes in a HCP unit cell. Its atomic radius is 0.1253nm. Its density is 8,900kg /m³. Using this information, estimate AVOGADRO's number.

Answer 1

Density= 8900Kg/m³= 8.9g/cm

Radius= 0.1253nm= 1.253x10^-8

One mole of Co= 59g

Volume occupied by one mole of Co= mass/density= 59/8.9 = 6.63cm³

Volume of one atom of Co= 4/3πr³

= 4/3 x 3.142x (1.253x10^-8)³

= 8.24x10^-24cm³

number of atoms present in one mole of Cobalt= volume occupied by one mole/volume occupied by one atom

= 6.63/8.24x10²³

= 8.04x10²³ atoms

We can see that this value is close to the Avogadro's number which is 6.02x10²³ atoms.