Question

An element has a BCC structure with a cell edge of 288pm.The density of the element is 7.2g/cm3.How many atoms are present in 208g of the element?

Answer 1

solution:

$Formula units per unit cell Z = 2 for BCC

cubic unit cell lattice parameter a = 288 pm = 288 \times10^{-8} cm

Volume V = a3 =2.39\times10^{-23}cm3

Density d = 7.2g/cm3

N¬A = Avogadro constant = 6.022\times10^23

Molecular mass M =?$$Density d = ZM/NA \times a3\\

M = d\timesNA \times a3/Z\\

On Substituting values \\

M= 7.2g/cm3 \times(6.022\times10^23)\times (6.022\times10^23)/2\\

= 51.8gmol-1\\

51.8 g of element contains 6.022\times10^23\\

208g of this element contains=?\\

= 6.022\times10^23\times208/51.8\\

=2.42\times10^24 atoms$