An element with molar mass `2.7xx10^(-2)kg*mol^(-1)` forms a cubic unit cell with edge length 405pm. If its density is `2.7xx10^(3)kg*m^(-3)`, what is the nature of the cubic unit cell?

Explanation

The density of a cubic unit cell, `rho=(ZxxM)/(Nxxa^(3))`
where Z = number of atoms per unit cell, M = molar mass of the element, N = Avogadro's number and a = edge length
Given: `M=2.7xx10^(-2)kg*mol^(-1)`
a = 405 pm = `405xx10^(-12)mandrho=2.7xx10^(3)kg*m^(-3)`
`thereforeZ=rhoxx(Nxxa^(3))/M`
`=(2.7xx10^(3)xx6.022xx10^(23)xx(405xx10^(-12))^(3))/(2.7xx10^(-2))=4`
So, the number of particles per unit cell is 4. Hence, the unit cell is face-centred cubic.