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An element has a bcc structure with a ce...

An element has a bcc structure with a celledge of `288` pm. The density of the element is `7.2 g cm^(-3)`. How many atoms are present in `208 g` of the element?

Text Solution

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We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orN_(0)=(ZxxM)/(a^(3)xxrhoxx10^(-30))`
Edge length of the unit cell (a) = 288 pm = 288
No. of atoms per unit cell (Z) = 2
Density of the unit cell `(rho)=7.2g//cm^(3)`
Mass of the element (M) = 208g
Number of atoms (N) present may be calculated as :
`N=(2xx(208g))/((288)^(3)xx(7.2"g mol"^(-3))xx(10^(-3)cm^(3)))=24.18xx10^(23)`
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