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

Volume of the unit cell `=(288p m)^(3)`
`=(288xx10^(-23)m)=(288xx10^(-10)cm)^(3)`
`=2.39xx10^(-23)cm^(3)`
Volume of 208 g of the element
`=("mass")/("density")=(208)/("7.2 g cm"^(-3))=28.88cm^(3)`
Number of unit cells in this volume
`=(28.88cm^(3))/(2.39xx10^(-23)cm^(3)//"unit cell")=12.08xx10^(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 × 1023" unit cells "= 24.16xx1023" atoms"`