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 $ 6.0.2\times {{10}^{23}}mo{{l}^{-1}}, $ the density of $ CsBr $ is

• A. $ 42.5g/c{{m}^{3}} $
• B. $ 0.425g/c{{m}^{3}} $
• C. $ 8.25\text{ }g/c{{m}^{3}} $
• D. $ 4.25\text{ }g/c{{m}^{3}} $

Answer 1

Answer: (D)

Density of $ CsBr=\frac{Z\times M}{{{a}^{3}}\times {{N}_{0}}} $
where
$ Z= $ no. of atoms in the bcc unit cell $ =2 $
$ M= $ molar mass of $ CsBr=133+80=213 $
$ a= $ edge length of unit cell = 436.6 pm $
=436.6\times {{10}^{-10}}cm $
$ \therefore $ Density $ =\frac{2\times 213}{{{(436.6\times {{10}^{-10}})}^{3}}\times 6.02\times {{10}^{23}}} $
$ =8.50g/c{{m}^{3}} $
For a unit cell density $ =\frac{8.50}{2}=4.25\text{ }g/c{{m}^{3}} $