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Answer: 2. 63.1 u

Key idea: density \(\rho=\dfrac{ZM}{a^3N_A}\), so \(M=\dfrac{\rho a^3N_A}{Z}\). fcc has Z = 4.

\(a^3=(3.608\times10^{-8})^3=4.697\times10^{-23}\ \mathrm{cm^3}\).

\(M=\dfrac{8.92\times4.697\times10^{-23}\times6.022\times10^{23}}{4}=\dfrac{252.3}{4}\approx63.1\) u.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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