Question

An element has a bcc structure with
unit cell edge length of 288 pm. How
many unit cells and number of atoms
are present in 200 g of the element?
(1.16×1024, 2.32×1024)
Plz don't give Google answerd

Answer 1

Volume of unit cell = (288 pm)3 = (288 10-10 cm)3

= 2.389 10-23 cm3

Volume of 208 g of the element =  

Number of unit cells =  

= 12.09 1023

For a b. c. c. structure, number of atoms per unit cell = 2

Number of atoms present in 208 g

=No. of atoms per unit cell No. of unit cells

= 2 12.09 1023 = 24.18 1023 = 2.418 1024.