Question

An element crystallizes in a f.c.c. lattice with a cell edge of 250 pm. Calculate the density, if 300 g of this element contain $2\times {10}^{24}$ atoms.

Answer 1

Formula of density of the unit cell:

$d=\frac{zM}{a^3{N}_A}$

where,

$z=$ number of particles present per unit cell $=4$

$M=$ Molecular mass $=\frac{300}{2\times {10}^{24}}\times 6.023\times {10}^{23}=90.34g$ ,

$N_A=$ Avogadro’s number $=6.023\times {10}^{23}$

$a=$ Edge length $=250pm=2.5\times {10}^{-8}cm$.

Now, substituting the values we get,

$d=\frac{4\times 90.34}{(2.5{)}^3\times {10}^{-24}\times 6.023\times {10}^{23}}=38.4g/c{m}^3$