Question

$CsBr$ crystals are formed with bcc crystal lattice where $B{r}^{-}$ occupies the corners and $C{s}^{+}$ occupies the centre of the cube with a unit cell length of nearly $400pm$. What will be density of $CsBr$?
Given molecular mass of CsBr is 213 g/mol and
$N_A=6\times {10}^{23}$

• A. $42.5g/c{m}^3$
• B. $5.54g/c{m}^3$
• C. $10.46g/c{m}^3$
• D. $1.25g/c{m}^3$

Answer 1

The correct option is B $5.54g/c{m}^3$

In bcc, the ions are present in the corners and body centre of the cube.
$\text{Number of}B{r}^{-}\text{per unit cell}=\frac{1}{8}\times 8=1$
$\text{Number of}C{s}^{+}\text{per unit cell}=1$
Hence, an unit cell of $CsBr$ comprise of one ion pair of $CsBr$
Thus, Z = 1 for $CsBr$ in an unit cell
Density is the ratio of mass to volume of the unit cell.
In a unit cell one molecule of CsBr is present.

$a=400pm=400\times {10}^{-12}m=400\times {10}^{-10}cm$
Volume of the unit cell = $a^3$ = V = $(400\times {10}^{-10}{)}^3$
$\text{Density of unit cell,}\rho =\frac{ZM}{V{N}_A}$
where,
Z is the number of atoms in unit cell
M is the molar mass
$N_A$ is the avagadro number.
V is the volume of unit cell.
$\rho =\frac{Z\times M}{a^3\times N_A}$

$\rho =\frac{1\times 213}{(400\times {10}^{-10}{)}^3\times 6\times {10}^{23}}$
$\rho =5.54g/c{m}^3$