KBr crystallises in a face-centred cubic crystal. The density of KBr crystal is `2.65g*cm^(-3)`. If formula mass of KBr is `119g*mol^(-1)`, then find the distance between `K^(+)andBr^(-)` in KBr crystal.

Explanation

We know, `rho=(ZxxM)/(Nxxa^(3))`
Given: `rho=2.65g*cm^(-3),M=119g*mol^(-1)`
Since the unit cell belongs to fcc system, Z = 4
`therefore" "a^(3)=(ZxxM)/(rhoxxN)=(4xx119)/(2.65xx6.022xx10^(23))=2.982xx10^(-22)cm^(3)`
`therefore" "a=6.68xx10^(-8)cm=6.68Å`
`therefore" "`The distance between `K^(+)andBr^(-)` in KBr crystal
`=1/2xx6.68Å=3.34Å.`