alleryStart learning →Gold crystallises in the fcc structure with edge length 408 pm. Its density is (M = 197 g/mol, $N_A = 6.022 \times 10^{23}$):
- 9.6 g/cm³
- 19.3 g/cm³
- 5.0 g/cm³
- 39 g/cm³
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Correct answer: B. 19.3 g/cm³
$\rho = nM/(a^3 N_A)$. fcc: $n=4$.
$a = 408\,\text{pm} = 4.08\times10^{-8}$ cm
$a^3 = 6.79\times10^{-23}$ cm³.
$\rho = 4 \times 197 / (6.79\times10^{-23} \times 6.022\times10^{23}) = 788/40.9 \approx 19.3$ g/cm³ — matches the known density of gold.