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Caesium chloride may be considered to form interpenetrating simple primitive cubic crystals.The edge length of unit cell is 412pm.Determine the density of CsCl.Given :M(Cs)=$133gmol^{-1}$

$\begin{array}{1 1}4.0gcm^{-3}\\3.0gcm^{-3}\\4.2gcm^{-3}\\2.0gcm^{-3}\end{array} $

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A)
Answer : $4.0gcm^{-3}$
The interpenetrating simple primitive cubic crystals means the unit cell is body-centred where $Cl^-$ ions occupy corners and $Cs^+$ ion occupy body centre of the cube.There is one $Cs^+$ ion and one $Cl^-$ ( or one molecule of CsCl) per unit cell.
From the expression
$\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
$\rho=\large\frac{1}{(412\times 10^{-12}m)^3}\big(\frac{168.5gmol^{-1}}{6.023\times 10^{23}mol^{-1}}\big)$$=0.4\times 10^7gm^{-3}=4.0gcm^{-3}$
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