Q24 / 90 · JEE Main 2021 (25 Feb Shift 2) · Chemistry

The unit cell of copper corresponds to a face centered cube of edge length $3.596 Å$ with one copper atom at each lattice point. The calculated density of copper in $kg / m^{3}$ is ___ . [Molar mass of $Cu : 63.54 g;$ Avogadro Number $= 6.022 \times 10^{23} J$ ]

Correct answer: 9076

Solution

FCC,

$d = \frac{Z \times M}{N_{A} \times a^{3}} = \frac{4 \times 63.54}{1000 \times 6.022 \times 10^{23} \times {(3.596 \times 10^{- 10})}^{3}}$

$= 9076 kg / m^{3}$

Hence, the correct answer is 9076.