Question:
The unit cell of copper corresponds to a face centered cube of edge length $3.596 A$ with one copper atom at each lattice point. The calculated density of copper in $\mathrm{kg} / \mathrm{m}^{3}$ is - [Molar mass of $\mathrm{Cu}: 63.54 \mathrm{~g}$; Avogadro Number $=6.022$ $\left.\times 10^{23}\right]$
Solution:
$\mathrm{FCC}$,
$\mathrm{d}=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N}_{\mathrm{A}} \times \mathrm{a}^{3}}=\frac{4 \times 63.54}{1000 \times 6.022 \times 10^{23} \times\left(3.596 \times 10^{-10}\right)^{3}}$
$=9076 \mathrm{~kg} / \mathrm{m}^{3}$
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