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?

Answer 1

It is given that density of the element, d = 2.7 ×103kg m-3

Molar mass, M = 2.7 ×10-2 kg mol-1

Edge length, a= 405 pm = 405 ×10-12 m = 4.05 ×10-10 m

It is known that, Avogadro's number, NA= 6.022 ×1023 mol-1

USE THE RELATION IN THE IMAGE



Applying the relation, This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).

Answer 2

$\huge\underline{\bold{\green{Answer : }}}$

The unit cell has 4 atoms. Hence, the nature of the cubic unit cell is ccp or fcc.

$\bigstar\underline{\pink{\bold{Given :- }}}$

Density of the cell = $\bold{ 2.7 \times 10^3 kg\: m^{-3}}$

Molar mass = $\bold{2.7 \times 10^{-2} kg\: mol^{-1}}$

Edge length = $\bold{405 \:pm = 405 \times 10^{-12}m}$

$\bigstar\underline{\orange{\bold{To\: find :-}}}$

Nature of the cubic unit cell.

$\bigstar\underline{\red{\bold{How\: to\: Find :-}}}$

Density, $\bold{\rho =\frac{ Z \times M}{a^3 \times N_0}}$

From this equation, we have to find Z (atomic number) :

Therefore,

$\implies \bold{2.7 \times 10^3 = \frac{Z\times 2.7 \times 10^{-2}}{(405 \times 10^{-12})^3 \times (6.022 \times 10^{23})}}$

$\implies \bold{Z = \frac{(2.7 \times 10^3) \times (405\times10^{-12})^3 \times (6.022\times10^{23})}{2.7\times10^{-2}}}$

$\bold{\therefore\: Z = 4.0}$

Since the unit cell contains 4 atoms, it is cubic close packed structure, ccp or fcc.