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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?

Syeda

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 SME Approved

Swathi Ambati

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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