$CsBr$ 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 :
- $42.5 \,g /cm^3$
- $0.425\, g /cm^3$
- $8.5\, g /cm^3$
- $4.25\, g /cm^3$
The Correct Option is C
Solution and Explanation
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 $=\frac{2 \times 213}{\left(436.6 \times 10^{-10}\right)^{3} \times 6.022 \times 10^{23}}=8.5 g / cm ^{3}$
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