An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is `"8.9 gcm-3"`. How many atoms are present in 100 g of an element ?

Explanation

`"Mass "=100g`
`"Density"="8.9 g cm"^(-3)`
`"Edge length"="352.4 pm"`
`(a)=352.4xx10^(-10)cm`
`"Volume of the unit cell,"`
`a^(3)=(352.4xx10^(-10)cm)=4.37xx10^(-23)cm^(3)`
`"Volume of 100 g of an element,"`
`=("Mass")/("Density")`
`=(100)/(8.9)cm^(3)=11.23cm^(3)`
Therefore number of unit cells,
`=(11.23)/(4.37xx10^(-23))=2.56xx10^(23)`
Since each Fcc cube contains 4 atoms, therefore total number of atoms in 100 g.
`=4xx(2.56xx10^(23))=10.24xx10^(23)" atoms"`