Copper crystallises in an fcc lattice with edge length $361 pm$ . Taking the atomic mass of Cu as $63.5$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , calculate the density in $g c m^{- 3}$ (two decimals).

Numerical answer type, answer in g cm^-3, tolerance 0.02.

Answer: 8.97 g cm^-3

ρ = ZM / (N_{A} a^{3})

with

Z = 4

ρ = \frac{4 \times 63.5}{6.022 \times 1 0 ^{23} \times ( 3.61 \times 1 0 ^{- 8} ) ^{3}} = 8.9654 \approx 8.97 g c m^{- 3}