An element ‘A’ of atomic mass 100 having bcc structure has unit cell edge of 400 pm. Calculate the density of ‘A’ and the number of unit cells for 10 g of ‘A’.

Solution

length of the unit cell edge = 400Pm =400x10^-10 cm
volume of the unit cell = ( 400 x10^-10cm)³
= 6.4 x 10^-23 cm^-3

as the element A forms a body centred cubic lattice so no of atom per units cell is 2

z= 2 atoms unit cell

atomic mass of the element =100 g/mol
density of element is given by

$md = \frac{M xZ}{a^{3} xN {}_{A}} d = \frac{100 x 2}{6.4 x 10^{- 23} x 6.023 x 10^{23}} = 51.88 g / {cm}^{3} volume of 10 g of A = \frac{mass}{Density} = \frac{10 g}{5.188 g {cm}^{- 3}} = 1.9275 {cm}^{3} number of unit cell in 1.9275 {cm}^{3} volume = \frac{volume of the substance}{unit cell volume} = 3.0 x 10^{22} unit cell$