CsBr crystallises in a body centred cubic lattice. The unit cell length is 436 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avagadro number being `6.02xx10^(23)mol^(-1)` the density of CsBr is

Options

A. `4.25g cm^(-3)`
B. `42.5gcm^(-3)`
C. `0.425 g cm^(-3)`
D. `8.31g cm^(-3)`

Answer

1

Explanation

Density of `CsBr="density of unit cell"`
`=("mass of unit cell")/("volume of unit cell")`
`= (ZM)/(a^(3)N_(A))`
`Z`(number of formula unit per bcc unit cell)=`1`
Note that bcc unit cell contains one `Cs^(+)` and one `Br^(-)`.
M(molar mas of `CsBr`)`=(133+80)gmol^(-1)`
`=213g mol^(-1)`
`a` (edge length of unit cell)=`436.6p m`
`=436.6xx10^(-12)m`
`=436.6xx10^(-10)cm`
Therefore,
density of `CsBr`
`((1"formula unit")(213 gmol^(-1)))/((436.6xx10^(-10)cm)^(3)(6.023xx10^(23)"formula units/mol"))`
`= 4.25 g cm^(-3)`