An element X (molar mass = 60 g `mol^(-1)`) has a density of 6.23 g `cm^(-3)`. Identify the type of cubic unit cell, if the edge length of the unit cell is `4 xx 10^(-8)` cm.

Explanation

Given, M = 60 g `mol^(-1)` = 0.06 kg `mol^(-1)`
Edge length (a) `= 4 xx 10^(-8) cm = 4 xx 10^(-10) m`
Density (d) = 6.23 g `cm^(-3) = 6.23 xx 10^(3) kg m^(-3)`
Apply the relation,
`d = (Z xx M)/(a^(3) xx N_(A)) rArr Z = (d xx a^(3) xx N_(A))/(M)`
`6.23 xx 10^(3) kg m^(-3) xx (4 xx 10^(-10) m)^(3)`
`= (xx 6.023 xx 10^(23) "atom mol"^(-1))/(0.06 "kg mol"^(-1))`
Z = 4 atom
Since, the number of atoms in the unit cell is four, hence the given cubic unit cell has a face-centred cubic (fcc) or cubic - closed packed (ccp) structure.