An element X crystallises in fcc structure, 208 g of it has 4 . 28 × 10 24 atoms. Calculate the edge of unit cell if density of X is 7 . 2 g / cm 3 .

Given, 208 g correspond to 4 . 28 × 10 24 atoms. Density is = 7 . 2 g / cm 3 Solution: Volume of the unit cell = a 3 ∴ Mass of the unit cell = ( 4 × 208 ) ( 4 . 283 × 10 24 ) = 194 × 10 - 24 g Density = Mass Volume 7 . 2 = ( 194 × 10 - 24 ) a 3 ∴ a 3 = ( 194 × 10 - 24 ) 7 . 2 a 3 = 27 × 10 - 24 a = 27 × 10 - 24 3 a = 3 x 10 - 8 ∴ Edge of the unit cell = 3 × 10 - 8 cm .

An element has a body centered cubic bcc structure with a cell edge 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?

Given, Edge of the unit cell, a = 288 pm Density of the element, ρ = 7 . 2 g / cm 3 Atoms present in element, = 208 g Solution: For the BCC structure, Z = 2 Substituting the values in the expression ρ = Z × M a 3 × N A 7 . 2 g cm - 3 = 2 × M ( 288 × 10 - 10 cm ) 3 × ( 6 . 02 × 10 23 mol - 1 ) or M = 51 . 8 g mol - 1 By mole concept, 51 . 8 g of the element contains = 6 . 02 × 10 23 atoms ∴ 208 g of the element contains = 6 . 02 × 10 23 51 . 8 × 208 atoms = 24 . 7 × 10 23 atoms