Calcium crystallizes in a face-centred cubic $(FCC)$ lattice with an edge length of $556 \ pm$. Calculate the density if:
$(i)$ There is $0.1 \%$ Schottky defect.
$(ii)$ There is $0.1 \%$ Frenkel defect.

(N/A) For an $FCC$ lattice, the number of atoms per unit cell is $Z = 4$. The molar mass of $Ca$ is $M = 40.08 \ g \ mol^{-1}$. The edge length is $a = 556 \ pm = 5.56 \times 10^{-8} \ cm$. The Avogadro constant is $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$(i)$ In a Schottky defect, some lattice sites are vacant, reducing the number of effective atoms per unit cell. The new number of atoms is $Z' = 4 \times (1 - 0.001) = 3.996$. The density is $\rho = \frac{Z' \times M}{N_A \times a^3} = \frac{3.996 \times 40.08}{6.022 \times 10^{23} \times (5.56 \times 10^{-8})^3} \approx 1.543 \ g \ cm^{-3}$.
$(ii)$ In a Frenkel defect, ions shift to interstitial sites, but the total number of ions in the crystal remains the same. Thus, the number of atoms per unit cell remains $Z = 4$. The density is $\rho = \frac{Z \times M}{N_A \times a^3} = \frac{4 \times 40.08}{6.022 \times 10^{23} \times (5.56 \times 10^{-8})^3} \approx 1.545 \ g \ cm^{-3}$.