An element with molar mass \(2.7 × 10^{-2} kg/mol\) forms a cubic unit cell of edge length \(405 pm\). If the density is \(2.7 × 10^3 kg/m^3\). The nature of the cell is?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → FCC
The density of a cubic crystal is the mass of one unit cell divided by its volume. A unit cell with z atoms of molar mass M and edge length a gives
\( \displaystyle d = \frac{z \times M}{a^3 \times N_A} \qquad \Rightarrow \qquad z = \frac{d \times a^3 \times N_A}{M} \)
The number of atoms per unit cell tells the type of cubic cell: z = 1 simple cubic, z = 2 body-centred (bcc), z = 4 face-centred (fcc).
Data in SI units: M = 2.7 × 10−2 kg mol−1, d = 2.7 × 103 kg m−3, a = 405 pm = 405 × 10−12 m = 4.05 × 10−10 m.
\( a^3 = (4.05 \times 10^{-10}\ \text{m})^3 = 66.43 \times 10^{-30}\ \text{m}^3 = 6.643 \times 10^{-29}\ \text{m}^3 \)
\( \displaystyle z = \frac{(2.7 \times 10^{3}) \times (6.643 \times 10^{-29}) \times (6.022 \times 10^{23})}{2.7 \times 10^{-2}} \)
\( \displaystyle z = 10^{5} \times 6.643 \times 6.022 \times 10^{-6} = 0.1 \times 40.0 = 4.0 \)
Four atoms per unit cell means a face-centred cubic (FCC) lattice. (The data are those of aluminium, which does crystallise in an fcc lattice.)
Shortcut: the numbers 2.7 in the density and in the molar mass cancel, so z = 105 × a3 × NA; with a3NA ≈ 6.64 × 10−29 × 6.02 × 1023 ≈ 4.0 × 10−5, z = 4.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Calculations Involving Unit Cell Dimensions – …
