JEE Advanced Model Paper 8 with solutions for online practice

Length of edge of the unit cell = 412.1 pm, = 412.1 × 10−10 cm
Volume of the unit cell = (412.1×10−10)3 = 7.00 × 10−23cm^3$$
Formula mass of caesium chloride = 168.5 a.m.u
The number of formula units per unit cell of caesium chloride is 1 as it has bodycentered cubic lattice.
∴ Z = 1 formula units / unit cell
Density = Formula mass × Z / unit cell volume (cm)3 × Navo =

168.5×1

7.00×10−23×6.023×1023

= 4 g / cm3