X-ray data show that an fcc metal ( $M = 27 g mo l^{- 1}$ ) has an edge of $405 pm$ , but its measured density is $2.66 g c m^{- 3}$ because some lattice sites are vacant. Use $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ . Calculate the percentage of lattice sites that are vacant (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

Answer: 1.47 %

Ideal density (all sites filled,

Z = 4

ρ = \frac{4 \times 27}{6.022 \times 1 0 ^{23} \times ( 4.05 \times 1 0 ^{- 8} ) ^{3}} = 2.6997 g c m^{- 3}

. Measured density is proportional to the actual number of atoms per cell, so the vacant fraction is

1 - \frac{2.66}{2.6997} = 0.0147

, i.e. 1.47%.