Question

An element A crystallizes in fcc structure. 208g of this element have 4.283 x 1024 atoms. If edge length of the unit cell of this element is 408 pm, calculate the density

Answer 1

Density of Cubic Crystal Lattice =

n * m / N1 * a power(3)


n = particles per unit cell
m = molecular mass / mass
N1 = no. of particles
a= edge length

For FCC, n= 4

So,
n * m / N1 * a power(3)

m = 208 g
n= 4
N1 = 4.283 * 10 power(24)
a = 408pm

Put values in above formula and you'll get answer.

Answer 2

Answer:2.75g/cm^3

Explanation:

We know,

n=w/M =N/Na [n=no. of moles, M=molecular mass, Na=avgadro no., w=given mass]

so,

M/Na =w/N

          = 20/4.283 x 10^24

          = 4.67 x  10^-24

D= (z/a^3)*(M/Na)            [Rewriting d=zM/a^(3)Na]

  =4 *4.67*10^-24 /(408*10^-10)^3

  =2.75g/cm^3

Hope it helps you