Question

$CsBr$ crystallises in a bcc 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 no. being 6.02310 per mol the density of $CsBr$ is? 42.5 g/cc 0.425 g/cc 8.25 g/cc 4.25 g/cc

Answer 1

Number of atoms of

$C{s}^{+}$

= 1 (

$\because$

Cs atom occupies body centre position)

Number of atoms of ${Br}^{-}$ = $\frac{1}{8}\times 8$ ($\because$ Br atom occupies corners of the lattice)

$\therefore$ Number of formula unit of CsBr in lattice = 1

$=\frac{1\times 213}{6.023\times {10}^{23}\times {(436.6\times {10}^{-3})}^3}$

$=\frac{213}{6.023\times {10}^{23}\times {(4.366\times {10}^{-8})}^3}$

= 4.25 g/cc