In a nacl type crystal distance between na+ and cl- ion is 2.814 amstron and the density of solid is 2.167gcm-3 then find out the value of avogardro constant
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.
