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An element crystallizes in a fcc lattice with cell edge of 400 pm. The density of the element is 7 g/cm3 . How many atoms one present in 280 g of the element?

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Best answer

Volume of unit cell a3 (a = edge length) = 400 pm

= (400 x 10-12 m)3 

= (400 x 100-10 cm)3 

= 64 x 10-24 cm3

Since each f.c.c. unit cell contains 4 atoms therefore, 

Total number =4 x 0.46 x 1024, 

= 1.84 x 1024 atoms.

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