CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being `6.02 xx 10^23 "mol"^(-1)` the density of CsBr is

Options

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

Answer

B

Explanation

Density `=("Z. mol wt")/(a^(3).N_(A))`
Trick : Z for bcc usually 2 but for this case
Z = 1 because 1 bcc has 2 atoms which here means 1 atom of Cs and 1 atom of Br
giving a total of 1 molecule and Z is the no. of molecule per bcc.
`rho = (1 xx 213)/((4.366 xx 10^(-8))^(3) xx 6.022 xx 10^(23))`
`= 4.25 g//cm^(3)`