# Caesium chloride may be considered to form interpenetrating simple primitive cubic crystals. The edge length of unit cell is $412\;pm$. Determine the density of $CsCl$

$\begin{array}{1 1} 4.0\;g\;cm^{-3} \\ 7 \;g\;cm^{-3} \\ 8\;g\;cm^{-3} \\ 9\;g\;cm^{-3} \end{array}$

Answer :$4.0\;g\;cm^{-3}$
The interpenetrating simple primitive cubic crystals means the unit cell is body- centred where $Cl^{-}$ ions occupy corners and $Cs^{+}$ ion occupy body center of the cube.
There is one $Cs^{+}$ ion and one $Cl^{-}$ ion (or one molecule of CsCl) per unit cell.
From the expression $\rho = \large\frac{N}{a^3} \bigg( \large\frac{M}{N_A}\bigg)$
We get, $\rho =\large\frac{1}{(412 \times 10^{-12} m)^3}\bigg( \large\frac {168.5 \;g\;mol^{-1} }{6.023 \times 10^{23} \;mol^{-1}} \bigg)$
$\qquad= 0.4 \times 10^{7} \;g m^{-3} =4.0 \;g\;cm^{-3}$
answered Aug 8, 2014 by
edited May 9, 2017 by meena.p