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How many atoms are present in 208 gm of element having bcc crystal structure, when the density of metal is 7.2 gm/cm^3 and the length of edge of its unit cell is 288pm?

Varun Prajapati , 7 Years ago
Grade 12
anser 1 Answers
Arun
Formula units per unit cell Z = 2 for BCC
cubic unit cell lattice parameter a = 288 pm = 288 x10-8 cm

Volume V =  a3 =2.39X10-23cm3
Density d = 7.2g/cm3 
 
 N­A = Avogadro constant = 6.022x10²³
Molecular mass M =?
We know that
Density d = ZM/NA X a3
M = dxNA x a3/Z

On Substituting values 
M= 7.2g/cm3 x(6.022x10²³)X (6.022x10²³)/2
= 51.8gmol-1
51.8 g of element contains 6.022X1023
208g of this element contains=?
= 6.022X1023X208/51.8
=2.42X1024 atoms.
Last Activity: 7 Years ago
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