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 ?

Explanation

Edge length of the unit cell = 288 pm
`= 288 xx 10^(-10) cm`
Volume of the unit cell = `(288 xx 10^(-10))^(3) cm^(3)`
`= 2.39 xx 10^(-23) cm^(3)`
Mass of element = 208 g
Density of element = `7.2 g cm^(-3)`
Volume of 208 g of the element = `("Mass")/("Density")`
Volume of element = `(208 g)/(7.2 g cm^(-3)) = 28.89 cm^(3)`
Number of unit cells in this volume
`= ("Volume of element")/("Volume of unit cell")`
`= (28.89 cm^3)/(2.39 xx 10^(-23) cm^(-3)//"unit cell")`
`= 12.08 xx 10^(23)` unit cells
Since the structure is bcc, number of atoms present in a unit cell = 2
The number of atoms in 208 g of the element
`= 2 xx 12.08 xx 10^(23) = 24.16 xx 10^(23) ` atoms
or ` = 2.416 xx 10^(24) ` atoms