An element with density 5.03 g cm-3 forms a fcc lattice with edge len…

The correct answer is Option (1) → 60.2 u

The density of a cubic crystal equals the density of its unit cell: the mass of the atoms in one unit cell divided by the volume of the cell.

    \( \displaystyle d = \frac{z \times M}{a^3 \times N_A} \qquad\text{so}\qquad M = \frac{d \times a^3 \times N_A}{z} \)

For an fcc lattice z = 4 atoms per unit cell. The data: d = 5.03 g cm⁻³, a = 4.3 × 10⁻⁸ cm, N_A = 6.022 × 10²³ mol⁻¹.

    \( a^3 = (4.3 \times 10^{-8}\,\mathrm{cm})^3 = 79.5 \times 10^{-24}\,\mathrm{cm^3} \)

    \( \displaystyle M = \frac{5.03 \times 79.5 \times 10^{-24} \times 6.022 \times 10^{23}}{4} = \frac{5.03 \times 47.88}{4} = \frac{240.8}{4} = 60.2\,\mathrm{g\,mol^{-1}} \)

So the atomic mass of the element is 60.2 u.

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 – “… Suppose, edge length of a unit cell of a cubic crystal determined by X-ray diffraction is a, d the density of the solid substance and M the molar mass. In case of cubic crystal: Volume of a unit cell = a³ Mass of the unit cell = number of atoms in unit cell × mass of each atom = z × m (Here z is the number of atoms present in one unit cell and m is the mass of a single atom) Mass of an atom present in the unit cell: \(m = \frac{M}{N_A}\) (M is molar mass) Therefore, density of the unit cell \(= \frac{\text{mass of unit cell}}{\text{volume of unit cell}}\) or \(d = \frac{z\,M}{a^3\,N_A}\)”