Question #186386

cobalt crystallizes in a HCP unit cell. it's atomic radius is 0.1253 nm. it's density is 8900 kg/m^3. using this information, estimate avogadro's number.

Expert's answer

Density= 8900Kg/mÂ³= 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= 4/3Ï€r3

= 4/3 x 3.142x (1.253x10-8)Â³

= 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.

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