#### Question

An element 'X' (At. mass = 40 g mol^{-1}) having f.c.c. the structure has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g of 'X'. (N_{A} = 6.022 × 10^{23} mol^{-1})

#### Solution

Unit cell edge length= 400 pm

= 400 x 100^{-10} cm

The volume of unit cell= a^{3}

= (400 x 10^{-10} cm)^{3}

= 64 x 10^{-24} cm^{3}

Mass of unit cell = No. of atoms in the unit cell × Mass of each atom

Number of atoms in the fcc unit cell = 4

Mass of one atom = `"Atomic Mass"/"Avagadro no" = 40/(6.022 xx 10^23) g mol^(-1)`

Mass of unit cell = `(4xx40)/(6.022 xx 10^23) = 26.57 xx 10^(-23) g mol^(-1)`

Density of unit cell = `"Mass of unit cell"/"Volume of unit cell" = (26.57 xx 10^(-23))/64xx10^(-24) = 4.15 g cm^(-3)`

No of Units cells in `26.57 xx 10^23 g` = 1

No of units cells in 4g = `(1xx4)/(26.57 xx 10^23) = 0.15 xx 10^(-23)`