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JEE MAIN 2021
25-02-2021 S2
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 $\left.=6.022 \times 10^{23}\right]$
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Solution
$$ \begin{aligned} &\text { FCC, }\\ &\begin{aligned} & d=\frac{Z \times M}{N_A \times a^3}=\frac{4 \times 63.54}{1000 \times 6.022 \times 10^{23} \times\left(3.596 \times 10^{-10}\right)^3} \\ & =9076 \mathrm{~kg} / \mathrm{m}^3 \end{aligned} \end{aligned} $$
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