chem-density

If the side of the cubic unit cell is 0.545nm, calculate the density of the calcium fluoride.
( there are 4 Ca^2+ ions and 8 F^- ions in the unit cell of the calcium fluoride.)
(Ca=40.08, F=19.00, Avogadro's no. 6.02 x 10^23 mol^-1)

Molar mass of CaF2 = 40.08 + 19.00x2 = 78.08 g mol^-1
Each formula unit contains 1 Ca^2+ ion and 2 F^- ions.
No. of CaF2 formula units in a unit cell = 4/1 = 8/2 = 4
No. of moles of CaF2 in a unit cell = 4/(6.02 x 10^23) mol
Mass of a unit cell = 78.08 x [4/(6.02 x 10^23)] = 5.188 x 10^-22 g

Volume of a unit cell = (0.545 x 10^-7 cm)^3 = 1.619 x 10^-22 cm^3

Density of CaF2 = (5.188 x 10^-22)/( 1.619 x 10^-22) = 3.204 g cm^-3