Question

An element has body centered cubic (bcc) structure with edge length 288pm. The density of the element is 7.2g cm-3.How many atoms are present in 208g of the element?​

Answer 1

Explanation:

Volume of Unit cell =$288^{3}$ x $10^{-10} cm$=  2.389×$10^{-23}$  cm

Volume of 208 g of the element= Mass/Density = 208/7.2 = 28.89$cm^{3}$

Number of unit cells= Total Vol/ Volume of unit cell = 28.89/2.389×$10^{-23}$  cm

Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells

For a BCC structure, number of atoms per unit cell =2

Number of atoms present in 208 g =2×12.09×$10^{23}$ = 2.41 x 10^24