Question

A metal forms bcc crystal lattice with density 100 g cm–3. If the edge length of its unit cell is 100 pm then the number of atoms of the metal present in 200 g of the crystal is

Answer 1

We know,

If a metal forms bcc crystal lattice, then the value for z=2.

It means there are 2 atoms per unit cell.

Now,

The formula of density of unit cell is given by

d=(z*M)/(a^3*N) where d= density of unit cell (g/cm^3)

a= edge length of unit cell in cm

M= molar mass in g/mol

N= 6.022*10^23= Avogadro constant

Given that,

a=100 pm= 100*10^(-10) cm= 10^(-8) cm

d= 100 g/cm^3

Now, using the formula of density of unit cell

d=(z*M)/(a^3*N)

100=(2*M)/((10^(-8))^3*6.022*10^23)

100=20*M/6.022

M= 100*6.022/20= 5*6.022

M= 30.111 g/mol

The number of atoms of the metal, n= m/M*N where m= mass of metal

So,

n= 200/30.111*6.022*10^23

n= 3.99987*10^24

Therefore,  the number of atoms of the metal present in 200 g of the crystal is 3.99987*10^24.