Answer
$N_{v}=1.16\times 10^{23}vacancies.m^{-3}$
Work Step by Step
given that the equilibrium fraction of lattice sites that are vacant in silver (Ag) at $ 700^\circ$ is
$\frac{N_{v}}{N}=2 \times 10^{6}$
So
$N_{v}=2 \times 10^{6}\times N$
Value of N for Ag is
$N=\frac{\rho_{Ag}\times N_{A}}{A_{Ag}}$
$N=\frac{10.35\times 6.023\times 10^{23}}{107.87}$
$N=5.78\times 10^{28}atom.m^{-3}$
$N_{v}=2 \times 10^{6}\times5.78\times 10^{28}atom.m^{-3}$
$N_{v}=1.16\times 10^{23}vacancies.m^{-3}$
