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Answer: \(\rho = \dfrac{nM}{a^3N_A}\)

Let the edge length of a cubic unit cell be a, so its volume is \(V = a^3\).

If M is the molar mass, the mass of one particle is \(m = \dfrac{M}{N_A}\).

If the unit cell contains n particles, mass of the unit cell \(= n \times \dfrac{M}{N_A}\).

Density \(\rho = \dfrac{\text{mass of unit cell}}{\text{volume of unit cell}} = \dfrac{nM}{a^3N_A}\)

Hence \(M = \dfrac{\rho\,a^3N_A}{n}\). Knowing any three of \(\rho\), a, n and M, the fourth can be found (n = 1, 2, 4 for simple, bcc and fcc cells).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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