KF has NaCl structure. What is the distance between `K^(+)` and `F^(-)` in KF, if the density is `1.5g cm^(-3)`?

Explanation

Since KF has NaCl structure, i.e., `Cl^(-), F^(-)` ions are present at all six faces and all corners of the cube. Similarly, like `Na^(+)`, all `K^(+)` ions occupy the middle position of 12 edges of the cube and centre of the cube.
`a=2(r_(K^(+))+r_(F^(-)))`
or `r_(K^(+))+r_(F^(-))=(a)/(2)`
As we know
Density `(rho) = (n xx M_(m))/(N_(o)xx a^(3))`
Here, n = 4
`M_(m)=` Molar mass of `KF = (39+19)gm = 58 gm`
`rho = (4xx58)/(6.023xx10^(23)xx a^(3))`
or `a^(3)=(4xx58)/(6.023xx10^(23)xx2.48)`
or `a = 5.37xx10^(-8)cm`
`= 5.37 Å`
`therefore r_(K^(+))+r_(F^(-))=2.685 Å`