**question** : An element with molar mass 2.7×10^{-2} kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×10^{3} kg m^{-3}, what is the nature of the cubic unit cell ?

**Key points to answer this question** :

First of all we need to calculate the number of atoms present in the unit cell (Z) =

Then find the type of the cubic unit cell on the based of number of Z , as follows –

If Z = 1 then This will simple cell (SC)

If Z = 2 then This will Body-centered cubic (BCC)

If Z = 4 then This will be Face-centred Cubic Unit Cell (FCC)

**solution** : given data –

Molar mass of the given element , M = 2.7×10^{-2} kg mol^{-1}

Edge length of given cubic unit cell , a = 405 pm

Density of given cubic unit cell , d = 2.7×10

^{3}kg m^{-3}also we know that avogadro’s number N

_{A }= 6.02214076 × 10^{23}Mol^{-1}SO we find Z = 4 , it means there are four atoms per unit cell and also we have discussed here above that if Z = 4 means the cubic cell is FCC (face centred cubic unit cell ).