The density of KBr is `2.75"gm cm"^(-3)`. The edge length of unit cell is 654 pm. Predict the type of cubic lattice to which unit cell of KBr belongs. `(N_(0)=6xx023xx10^(23)"mol"^(-1))`.
Atomic mass of K = 39, Br = 80.

Explanation

`d=2.75"g cm"^(-3)`
`a=654"pm"=6.54xx10^(-8)"cm"Z=?`
`N_(A)=6.023xx10^(23)"mol"^(-1)`
`M=(39+80)"g mol"^(-1)`
`=119"g mol"^(-1)`
`d=(ZxxM)/(N_(A)xxa^(3))`
`Z=(d xxN_(A)xxa^(3))/(M)`
`=((2.75"g cm"^(-3))xx(6.023xx10^(23)"mol"^(-1))xx(6.54xx10^(-8))^(3)"cm"^(3))/(119"g mol"^(-1))`
`=(2.75xx6.023xx279.73)/(1190)`
`~=4` (f.c.c. lattice)