A metal crystallizes in a face-centred cubic (fcc) lattice. A 160 g...

Total volume of the sample $V = \dfrac{\text{Mass}}{\text{Density}} = \dfrac{160}{5.0} = 32 \text{ cm}^3$

Since the metal crystallizes in an fcc lattice, the number of atoms per unit cell is $Z = 4$.

Total number of unit cells in the sample $= \dfrac{\text{Total number of atoms}}{Z} = \dfrac{2 \times 10^{24}}{4} = 5 \times 10^{23}$

Volume of one unit cell $a^3 = \dfrac{\text{Total volume}}{\text{Total number of unit cells}} = \dfrac{32}{5 \times 10^{23}} = 6.4 \times 10^{-23} \text{ cm}^3 = 64 \times 10^{-24} \text{ cm}^3$

Edge length $a = (64 \times 10^{-24})^{1/3} \text{ cm} = 4 \times 10^{-8} \text{ cm}$

Converting to picometers:
$a = 4 \times 10^{-8} \times 10^{10} \text{ pm} = 400 \text{ pm}$

Answer: $400 \text{ pm}$