Calcium metal crystallizes in a face-centred cubic lattice with edge length of 0.556 nm. Calculate the density of the metal if it contains (i)0.5% Frenkel defects (ii)0.2% Schottky defects.

Explanation

(i)Frenkel defects have no effect on the density because number of atoms per unit cell remains the same .
For FCC, Z=4. For Ca, M=`40 g mol^(-1)`.
Also, a=0.556 nm =`0.556xx10^(-7)` cm
`therefore ` Density, `rho=(ZxxM)/(a^3xxN_0)=(4xx40 "g mol"^(-1))/((0.556xx10^(-7)cm)^3xx(6.022xx10^23 mol^(-1)))=1.5458 g cm^(-3)`
(ii)If there were not defect (or there were only Frenkel defects) , number of atoms per unit cell=4 (being FCC). There would have been no change in density .
Due to 0.2 % Schottky defect, number of atoms per unit cell decreases. Thus, now number of atoms per unit cell =`4-0.2/100xx4=3.992`
`therefore` Density , `rho=(ZxxM)/(a^3xxN_0)=((3.992)(40 g mol^(-1)))/((0.556 xx 10^(-7) cm)^3 (6.022xx10^23 mol^(-1)))=1.5427 g cm^(-3)`