An element with molar mass 27 g mol^{−1} forms a cubic unit cell with edge length 4.05 ✕ 10^{−8} cm. If its density is 2.7 g cm^{−3}, what is the nature of the cubic unit cell?

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#### Solution

Molar mass of the given element, *M *= 27 g mol^{−1} = 0.027 kg mol^{−1}

Edge length, *a* = 4.05 × 10^{−8} cm = 4.05 × 10^{−10} m

Density, *d* = 2.7 g cm^{−3} = 2.7 × 10^{3} kg m^{−3}

Applying the relation,

`d=(ZxxM)/(a^3xxN_A)`

Where, Z is the number of atoms in the unit cell and N_{A} is the Avogadro number.

Thus,

`Z=(`

`=(2.7xx10^3xx(4.05xx10^(-10^3))xx6.022xx10^23)/0.027`

**= 4**

Since the number of atoms in the unit cell is four, the given cubic unit cell has a face-centred cubic (fcc) or cubic-closed packed (ccp) structure.

Concept: Number of Atoms in a Unit Cell

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