An element has bcc structures with a cell edge of 288 pm. The density of the element is `7.2 gcm^(-3)`. How many atoms are present in 208 g of the element.

Explanation

Unit cell a = 288pm
Volume of unit cell `= (288 p m)^(3)`
`= ( 288 xx 10^(-8))^(3) cm^(3)`
`= 2.389 xx 10^(-23) cm^(3)`.
Volume of 208g of the element `= ( "Mass")/( "Density")`
`rArr ( 208g)/( 7.2g cm ) = 28.89 cm^(3)`
Number of unit cells `=( "Total volume")/("Volume of a unit cell")`
`= (28.89cm^(3))/( 2.389 xx10^(-23)cm^(3))`
`= 12.09 xx 10^(23)`
No. of atoms = No. of atoms per unit cell present in 208 g `xx` No. of unit cell
` 2 xx 12.09xx 10^(23)`
`= 24.18 xx 10^(23)`
`=2.418 xx 10^(24)`