Question

An element has a body-centred 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 ?

Answer 1

Volume of the unit cell = `(288 "pm")^(3)`
= `(288 xx 10^(-12) m)^(3) = (288 xx 10^(-10) cm^(3) = 2.39 xx 10^(-23) cm)^(3)` .
Volume of 208 g of the element
`("mass")/("density") = (208 g )/(7.2 g cm^(-3)) = 28.88 cm^(3)`
Number of unit cells in this volume
`= (28.88 cm^(3))/(2.39 xx 10^(-23) cm^(3)// "Unit cell") = 12.08 xx 10^(23)` unit cells .
Since each bcc cubic unit cell contains 2 atoms , therefore , the total number of atoms in 208 g = 2 (atoms / unit cell) `xx 12.08 xx 10^(23)` unit cells
`= 24.16 xx 10^(23)` atoms