Determine the density of caesium chloride which crystallises in a bcc type structure with the edge length $412.1$ p.m. The atomic masses of Cs and Cl are $133$ and $35.5$ respectively.

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Solution

The molecular weight of CsCl,
$M = 133 + 35.5 = 168.5 g / m o l$ .
A bcc unit cell. It has 1 atoms of Cs and one atom of Cl per unit cell, $Z = 1$
Avogadro's number $N_{0} = 6.023 \times 10^{23} m o l e c u l e s / m o l$
The edge length $a = 412.1 p m = 412.1 \times 10^{- 10} c m$
The density
$ρ = \frac{Z \times M}{N_{0} \times a^{3}}$
$ρ = \frac{1 \times 168.5 g / m o l}{6.023 \times 10^{23} m o l e c u l e s / m o l \times (412.1 \times 10^{- 10} c m^{3})^{3}}$
$ρ = 4.0 g / c m^{3}$

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