Question

A compound from fcc structure .what is the number of octanedral voidc formed in 0.3 mol of it ?​

Answer 1

Answer:

Number of atoms in close packaging = 0.5 mol

1 has 6.022×10

23

particles

So that

Number of close-packed particles

=0.5×6.022×10

23

=3.011×10

23

Number of tetrahedral voids = 2 × number of atoms in close packaging

Plug the values we get

Number of tetrahedral voids =2×3.011×10

23

=6.022×10

23

Number of octahedral voids = number of atoms in close packaging

So that

Number of octahedral voids =3.011×10

23

Total number of voids = Tetrahedral void + octahedral void

=6.022×10

23

+3.011×10

23

=9.03×10

23