`KCl` has `NaCl` type face centred cubic crystal structure and `CsF` has `CsCl` type cubic crystal structure. Calculate the ratio of densities of `CsF` and `KCl` it is given that the molar mass of `CsF` is two that of `KCl` and edge length of `KCl` unit cell of `1.5` times that for `CsF`.

Options

A. `1.68`
B. `2.72`
C. `3.12`
D. `4.62`

Answer

1

Explanation

`a_(KCl)=1.5a_(cal)`
`2M_(KCl)=M_(CaF)`
`rho_(KCl)=(4M_(KCl))/(N_(A)a_(KCl)^(3))`
`rho_(CaF)=(1M_(CaF))/(N_(A)a_(CaF)^(3))`
`(rho_(CaF))/(rho_(KCl))=(M_(CaF))/(a_(CaF)^(3))=(a_(KCl)^(3))/(4M_(KCl))=(2)/(4)xx((3)/(2))^(3)=(27)/(16)=1.6875`