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CsBr crystallises in a body centred cubic lattice. The net cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being $6 .02 \times 10^{23} {mol}^{- 1}$ , the density of CsBr is

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8.25 $g / {cm}^{3}$

4.25 $g / {cm}^{3}$

42.5 $g / {cm}^{3}$

0.425 $g / {cm}^{3}$

answer is B.

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Detailed Solution

Z for CsBr = 1

(Since, one effective molecule is present) one is Cs atom and one is Br atom.

$d = \frac{ZM}{N_{0} a^{3}}$

$= \frac{1 \times 213}{{(4 .366 \times 10^{- 8})}^{3} \times 6 .02 \times 10^{23}} = 4.25 g / {cm}^{3}$

for unit cell, $\frac{8 .50}{2} g / {cm}^{3}$

$= 4 .25 g / {cm}^{3}$

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