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Answer: 2. 352 pm

Key idea: \(\rho=\dfrac{Z\,M}{N_A a^3}\), with Z = 2 for bcc.

\(\rho=530\ \mathrm{kg\,m^{-3}}=0.530\ \mathrm{g\,cm^{-3}}\).

$$a^3=\frac{2\times6.94}{6.02\times10^{23}\times0.530}=4.35\times10^{-23}\ \mathrm{cm^3}$$$$a=3.52\times10^{-8}\ \mathrm{cm}=352\ \mathrm{pm}$$
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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