Question

An element crystallises as fcc with density as 5.20 g/cc and edge length of the side of unit cell as 300 pm. Calculate mass of that element which contains $3.01\times {10}^{24}$ atoms.

Answer 1

For fcc

z=4

$N=3.1\times {10}^{24}$ atoms

a=300 pm

$=200\times {10}^{-12}$m

$=3\times {10}^{-8}$ cm

$d=\frac{z\times M}{a^3\times N}$

$5.20=\frac{4\times M}{27\times {10}^{-24}\times 3.1\times {10}^{24}}$

$M=108.8g$