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 $6.023 \times 10^{23} mol^{-1}$, the density of $CsBr$ is :

• A. $42.5 \,g /cm^3$
• B. $0.425\, g /cm^3$
• C. $8.5\, g /cm^3$
• D. $4.25\, g /cm^3$

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, $a =$ $436.6 pm =436.6 \times 10^{-10} cm$
Molecular weight of $CsBr =133+80=213 g / mol$.
Upon substituting the values in the above density equation :
Density $=\frac{2 \times 213}{\left(436.6 \times 10^{-10}\right)^3 \times 6.022 \times 10^{23}}=8.5 g / cm ^3$