Question

Na crystallizes in the cubic lattice (ccp) and the edge length of the unit cell is 430 pm. Calculate the no. of atoms in a unit cell. The density of Na is 0.9623 g(cm${}^3$)

Answer 1

$\rho =\frac{Z\times M_A}{N_A\times a^3}$

$Z=?$

$a=430$ $pm$

$N_A=23$

$\rho =0.9623$ $g/c{m}^3$

$0.9623=\frac{Z\times 23}{6.023\times {10}^{23}\times 79.5\times {10}^{-24}}$

$Z=2$

Therefore, $2$ atoms of sodium are effectively present in a unit cell.