`CsBr` crystallises in a body`-` centred cubic lattice. The unit cell length is `436.6 p m`. Given that `:` the atomic mass of `Cs=133` and that of `Br=80 am u` and Avogadro's number being `6.02xx10^(23)mol^(-1)`, the density of `CsBr` is `:`

Options

A. `42.5g//cm^(3)`
B. `0.425g//cm^(3)`
C. `8.25g//cm^(3)`
D. `4.25g//cm^(3)`

Answer

D

Explanation

`"Density of CsBr"=(ZxxM)/(a^(3)xxN_(0))`
`Zrarr"number of atoms in the bcc unit cell"=2`
`M rarr"molar mass of CsBr"=133+80=213`
`ararr "edge length of unit cell"=436.6"pm"`
`=436.6xx10^(-10)cm`
`therefore" Density"=(2xx213)/((436.6xx10^(-10))^(3)xx6.023xx10^(23))`
`=8.49xx10^(-7)xx10^(7)g//cm^(3)`
`=8.50g//cm^(3)`
For a unit cell `=(8.50)/(2)=4.25g//cm^(3)`