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Asked by Anakha ullas C/o o ullasan

Aug 15, 2015

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?

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?

Answers(2)

Answer

Syeda

Member since Jan 25, 2017

Answer.

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

Swathi Ambati

Member since Apr 28, 2014

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.
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