Question

An element X (molar mass = 60 g mol⁻¹) has a density of 6.23 g cm⁻³. Identify the type of cubic unit cell, if the edge length of the unit cell is 4 ✕ 10⁻⁸ cm.

Answer 1

Molar mass (M) of element X = 60 g mol⁻¹ = 0.060 kg mol⁻¹
Edge length, a = 4 × 10⁻⁸ cm = 4 × 10⁻¹⁰ m
Density, d = 6.23 g cm⁻³ = 6.23 × 10³ kg m⁻³

^{Applying the relation,}

^{\[d = \frac{Z \times M}{a^3 \times N_A}\]
\[Where, \]
\[Z = \text{Number of atoms in the unit cell}\]
\[ N_A = \text{Avogadro number}\]
\[Z = \frac{d \times a^3 \times N_A}{M} \]
\[ = \frac{6 . 23 \times {10}^3 \times \left( 4 \times {10}^{- 10} \right)^3 \times 6 . 022 \times {10}^{23}}{0 . 060}\]
\[ = 4 . 002 \approx 4\]}

Since the number of atoms in the unit cell is four, the given cubic unit cell is face-centred cubic (fcc).