Question

6) An atom crystallises in fcc crystal lattice and has
a density of 10 g cm with unit cell edge length
of 100 pm. Calculate number of atoms present in
1 g of crystal. (4 x 1023 atoms)​

Answer 1

Answer:

Heya.............. ❤✌

Here, f= ZM / a^3 no

For FCC z = 4

a= 100 ppm

f= 10gm/ cm^3

NA = Avogadro's no

As we know, M = fa^3 NA/ Z

Hence, no of atoms = given mass ×NA / molar mass

Now, put the value if M

No of atoms = given mass × NA / fa^3 NA

=( 1× 4)/ 10× (10^-8)^3

= 4× 10^23 atoms

Hope it’s helpful....... ☺