Hello,

NaCl crystallizes in such a way that the Cl- ions  occupy all the FCC lattice and Na+ ions occupy all the octahedral voids.

Hence, no. of Cl- ions present per unit cell = (1/8 x no. of corner of unit cell) + ( 1/2 x No. of faces of unit cell )

So, no. of Cl- ions present = 1 + 3 + 4

And No. of Na+ ions present per unit cell = ( 1/4 x no. of edges of unit cell ) + ( 1 at body centre )

So, no. of Na+ ions present = 4

Density = 2.167 g/cc = D

Distance between Na+ and Cl- = 2.814 A = a

No. of atoms per cell = Z = 4

N = Avogadro's No. = ?

M = Mass of Unit cell = ( Mass of Na+ ) + ( mass of Cl- )

So, M = 23 + 35.5 = 58.5

We know that,

M = ( D. N. a^3 ) / Z

So,

58.5 = ( 2.167 x N x ( 2.814 x 10 ^ -8 )^3 ) / 4

So, N = ( 58.5 x 4 ) / 2.167 x 2.184^3

So, N =  4.84 x 10^ 24

Hence, the value of Avogadro's constant is 4.84 x 10 ^ 24

Best Wishes.