The Solid State - Result Question 25
####25. $CsBr$ crystallises in a body centered cubic lattice. The unit cell length is $436.6 pm$. Given that the atomic mass of $Cs=133$ and that of $Br=80 amu$ and Avogadro number being $6.02 \times 10^{23} mol^{-1}$, the density of $CsBr$ is
(a) $0.425 g / cm^{3}$
(b) $8.25 g / cm^{3}$
(c) $4.25 g / cm^{3}$
(d) $42.5 g / cm^{3}$
[2006]
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Solution:
- (c) In body centred cubic lattice one molecule of $CsBr$ is within one unit cell.
Atomic mass of unit cell $=133+80=213$ a.m.u
Volume of cell $=(436.6 \times 10^{-10})^{3} cm^{3}$
$ \begin{aligned} \text{ Density } & =\frac{Z \times \text{ at.wt. }}{\text{ Av.no. } \times \text{ vol.of unit cell }} \\ \text{ Density } & =\frac{1 \times 213}{6.02 \times 10^{23} \times(436.6)^{3} \times 10^{-30}} \\ & =\frac{213 \times 10^{7}}{6.02 \times(436.6)^{3}}=4.25 g / cm^{3} \end{aligned} $
In the density formula, $Z$ is number of particles present in a unit cell and $M$ is molar mass of that particle. For any diatomic molecule like $CsBr, Z=$ 2 but for any mono atomic molecule like $Li, Z=1$ for bec lattice.
