Determine the density of caesium chloride which crystallises in a $b c c$ type structure with the edge length $412.1 p m$ . The atomic masses of $C s$ and $C l$ are $133$ and $35.5$ respectively. $(4.0 g c m^{- 3})$

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Solution

The molecular weight of CsCl,

$M = 133 + 35.5 = 168.5 g / m o l$ .

A bcc unit cell has 2 atoms per unit cell( $Z = 2$ )

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

$ρ = \frac{Z \times M}{N_{0} \times a^{3}}$

$ρ = \frac{2 \times 168.5 g / m o l}{6.023 \times 10^{23} m o l e c u l e s / m o l \times (412.1 \times 10^{- 10} c m^{3})^{3}}$

$ρ = 8.0 g / c m^{3}$

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