Question

$CsBr$ crystallises in a body centred cubic lattice. The unit cell length is $436.6pm$. Given that the atomic mass of $Cs=133u$ and that of $Br=80u$ and Avogadro number being $6.023\times 10^{23}mo{l}^{-1}$, the density of $CsBr$ is :

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

Answer 1

Answer: (C)

Denisty $=\frac{Z\times M}{a^3\times N_A}$
Given $Z=$ two formula unit of $CsBr$ in 1 cubic unit $=2$ and edge length, $a=$ $436.6pm=436.6\times 10^{-10}cm$
Molecular weight of $CsBr=133+80=213g/mol$.
Upon substituting the values in the above density equation :
Density $=\frac{2\times 213}{{\left(436.6\times 10^{-10}\right)}^3\times 6.022\times 10^{23}}=8.5g/c{m}^3$