Answer
$\frac{N_{v}}{N}=4.41\times 10^{-4}$
Work Step by Step
given that the equilibrium fraction of lattice sites that are vacant in metal at $ 900^\circ$ is
${N_{v}}=2.3 \times 10^{25}m^{-3}$ and $\rho_{m}=7.40$ also$A_{m}=85.5$
Value of N for metal is
$N=\frac{\rho_{m}\times N_{m}}{A_{m}}$
$N=\frac{7.40\times 6.023\times 10^{23}}{85.5}$
$N=5.21\times 10^{28}atom.m^{-3}$
now
$\frac{N_{v}}{N}=\frac{2.3 \times 10^{25}}{5.21\times 10^{28}}$
$\frac{N_{v}}{N}=4.41\times 10^{-4}$
