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The density of chormium is 7.2"g cm"^(-3...

The density of chormium is `7.2"g cm"^(-3)`. If the unit cell is a cube edge length of 298 pm, determine the type of the unit cell. (Atomic mass of Cr = 52 amu)

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)`
Gram atomic mass of Cr (M) = `52.0"g mol"^(-1)`
Edge length of unit cell (a) = 289 pm = 289
Density of unit cell `(rho)=7.2"g cm"^(-3)`
Avogadro's number `(N_(0))=6.022xx10^(23)mol^(-1)`
`therefore" "Z=((7.2"g cm"^(-3))xx(289)^(3)xx(6.022xx10^(23)mol^(-1))xx(10^(-30)cm^(3)))/((52.0gmol^(-1)))=2`
Since the unit cell has 2 atoms, it is body centre in nature.
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