Face centred cubic crystal lattice of copper has density of `8*966" g. cm"^(-3)`. Calculate the volume of the unit cell.
Given molar mass of copper is `63*5"g. mol"^(-1)` and Avogadro number `N_(A)` is `6*022xx10^(23)"mol^(-1)`.

Explanation

Density of the unit cell
`(d)=("Maxx of the unit cell"(M))/("Volume of the unit cell"(V))` . . . (i)
Let number of atoms in the unit cell be Z and mass of each atom m then `M=Zxxm`
Mass of each atom
`(m)=("Atomic mass")/("Avogadro's no.")`
Thus from equation (i)
`d=(Zxx"Atomic mass")/("Avogadro's no." xx V)` . . . .(ii)
For face centered cubic lattice (FCC), the total no. of atoms in the unit cell
`(Z)=(1)/(8)xx8+(1)/(2)xx6`
`=1+3=4`
So, `8*966"g cm"^(-3)=(4xx63*5"g mol"^(-1))/(6*022xx10^(23)"mol^(-1)xxV)`
or, Volume of the unit cell
`(V)=(254)/(53*99xx10^(23))cm^(3)`
`=4*7xx10^(-23)cm^(3)`
Hence, the volume of the unit cell is `4*7xx10^(-23)cm^(3)`.