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The density of KBr is 2.73"g cm"^(-3). T...

The density of KBr is `2.73"g cm"^(-3)`. The length of the unit cell is 654 pm. Predict the nature of the unit cell. (Given atomic mass of K = 39, Br = 80)

Text Solution

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We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))or Z=(rhoxxa^(3)xxN_(0)xx10^(-30))/(M)`
Edge length of unit cell (a) = 654 pm
Molecular mass of KBr (M) = `(39+80)=119"g mol"^(-1)`
Density of unit cell `(rho)=2.73"g cm^(-3)`
Avogadro's Number `(N_(0))=6.022xx10^(23)mol^(-1)`
`Z=((2.73"g cm"^(-3))xx(654)^(3)xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))/((119gmol^(-1)))=3.87(~~4)`
Since the unit cell has 4 partiles, it is fcc in nature.
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