Question

$CsBr$ crystallizes in a body centered cubic lattice. The unit cell length is $436.6pm$ . Given that the atomic mass of $Cs=133amu$ and that of $Br=80amu$ and Avogadro number being $6.02\times {10}^{23}mo{l}^{-1}$ , the density of $CsBr$ is

• A. $42.5g/c{m}^3$
• B. $4.25g/c{m}^3$
• C. $8.5g/c{m}^3$
• D. $0.425g/c{m}^3$

Answer 1

The correct option is C $8.5g/c{m}^3$

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A\times a^3}$
where,
Given Z= two formula unit of $CsBr$ in 1 cubic unit
$Z=2$
$M=\text{Molar mass}=213N_A=\text{Avagadro number}$
$a=\text{length of the unit cell}=436.6pm=436.6\times {10}^{-10}cm$
$\rho =\frac{1\times 213}{6.022\times {10}^{23}\times (436.6\times {10}^{-10}{)}^3}=8.5g/c{m}^3$