Question

$CsBr$ crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 u and that of Br = 80 u and Avogadro number being $\mathrm{6.0.2}\times {10}^{23}mo{l}^{-1},$ the density of $CsBr$ is

• A. $42.5g/c{m}^3$
• B. $0.425g/c{m}^3$
• C. $8.25g/c{m}^3$
• D. $4.25g/c{m}^3$

Answer 1

Answer: (D)

Density of $CsBr=\frac{Z\times M}{a^3\times N_0}$
where
$Z=$ no. of atoms in the bcc unit cell $=2$
$M=$ molar mass of $CsBr=133+80=213$
$a=$ edge length of unit cell = 436.6 pm $<br/>=436.6\times {10}^{-10}cm$
$\therefore$ Density $=\frac{2\times 213}{{(436.6\times {10}^{-10})}^3\times 6.02\times {10}^{23}}$
$=8.50g/c{m}^3$
For a unit cell density $=\frac{8.50}{2}=4.25g/c{m}^3$