Question

an element crystalline in body centered cubic lattice with cell edge of 500pm . the density of the element 7.5gm/cmcube . how many atoms are present in 300gm of the element

Answer 1

According to given:
d=7.5g/cm³
a=500pm
mass=300g

The formula to be used is

d=z*m/a³*N

N=z*m/a³*d
Substituting the values
We get

N=2x300/(5*10^-8)³x7.5

By doing further calculation we get
N=6.4 x 10²³
Hence,
The number of atoms present in 300g of the element is 6.4 x 10²³ atoms