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

Explanation

Density , `rho=(ZxxM)/(a^3xxN_0) "or" Z=(rhoxxa^3xxN_A)/M`
Here, M (molar mass of the element )=`2.7xx10^(-2)"kg mol"^(-1)`
a (edge length) =405 pm =`405xx10^(-12) m =4.05xx10^(-10) m`
`rho` (density) =`2.7xx10^3 "kg m"^(-3)`
`N_A` (Avogadro's number )=`6.022xx10^23 "mol"^(-1)`
Substituting these values in expression (i), we get `Z=((2.7xx10^3 "kg m"(-3))(4.05xx10^(-10) m)^3 (6.022xx10^23 "mol"^(-1)))/(2.7xx10^(-2) "kg mol"^(-1))` =3.99=4
Thus, there are 3 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centred or cubic close packed (ccp)