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$CsBr$ crystallises in a body centered cubic lattice. The unit cell length is $436.6\; pm$ . Given that the atomic mass of $Cs = 133$ and of $Br = 80\; amu$. Avogadro's number being $6.02\times10^{23}mol^{-1}$ The density of $CsBr$ is

$(a)\;8.50g/cm^3\qquad(b)\;4.25g/cm^3\qquad(c)\;42.5g/cm^3\qquad(d)\;0.425g/cm^3$

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Density = $\large\frac{Z\times mol.wt}{a^3\times N_A}$
$\therefore$ Z = 2 for b.c.c and mol.wt of CsBr = 213
$\rho = \large\frac{2\times 213}{(4.366\times10^{-8})^3\times6.032\times10^{23}}$
$= 8.50 g/cm^3$
Hence answer is (a)
answered Feb 27, 2014 by sharmaaparna1
edited Jun 22 by sharmaaparna1
 

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