`KCI` crystallizes in the same type of lacttice as does `NaCl`. Given that
`(r_(Na^(o+)))/(r_(Cl^(Θ))) = 0.5` and `(r_(Na^(o+)))/(r_(K^(o+))) = 0.7`
Calculate (a) the ratio of side of the unit cell for `KCl` to that for `NaCl`, and (b) the ratio of density of `NaCl` to that `KCl`.

Explanation

`(a) (""^(r)N_(a^(+)))/(""^(r)C_(?^(-)))=0.5 " "(""^(r)N_(a^(+)))/(""^(r)K^(+))=0.7`
`(a_(KC?))/(a_(Nacl))=(2[r_(K^(+))+r_(c?)])/(2[r_(Na^(+))+r_(c.^(-))])` -…..(i)
`r_(Na^(+))/(r_(C?^(-)))+1=1+0.5, (r_(Na^(+))+r_(c?^(-)))/(r_(C?^(-)))=1.5...(ii)`
`(r_(Na^(+)))/(r_(c?^(-))) div r_(Na^(+))/(r_(k^(+)))=(5)/(7),r_(k^(+))/(r_(C?^(-)))=(5)/(7),r_(K^(+))/r_(C?^(-)+1` )
`(5)/(7)+1,(r_(k)+r_(c?^(-)))/(r_(c?^(-)))=(12)/(7)`....(iii)
From eq. (i)
`(a_(k?))/(a_(Nacl))=(12//7)/(1.5)=(1.714)/(1.5)=1.143`
(b)`d_(Nacl)=(M_(Nacl))/(M_(Kcl))((a_(aKCl))/(a_(NaCl)))^(3)=(58.5)/(74.5) .(1.143)^(3)`
1.172