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KCI crystallizes in the same type of lac...

`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`.

Text Solution

Verified by Experts

`NaCl` crystallizes in the face-centred cubic unit cell, such that
`r_(Na^(o+)) + r_(Cl^(Θ)) = a//2`
where `a` is the edge length of unit cell. Now since
`r_(Na^(o+)//r_(Cl^(Θ)) = 0.5` and `r_(Na^(o+))//r_(K^(o+)) = 0.7`, we will have
`r_(Na^(o+) + r_(Cl^(Θ)))/(r_(Cl^(Θ))) = 1.5, [r_(Na^(o+))/r_(Cl^(Θ)) + 1= 0.5 + 1 ]`
and `r_(Na^(o+))/r_(Cl^(Θ)) = r_(K^(o+))/(r_(Cl^(Θ))//0.5)= (0.5)/(r_(Na^(o+))//r_(K^(o+))) = (0.5)/(0.7)`
[Add `1` to both sides]
or ` (r_(K^(o+)) + r_(Cl^(Θ)))/(r_(Na^(o+)) + r_(Cl^(Θ))) = (1.2)/(0.7) xx (1)/(1.5)`
or `or (a_(KCl)//2)/(a_(NaCl)//2) = (1.2)/(0.7 xx 1.5)`
or `(a_(KCl))/(a_(NaCl)) = (1.2)/(1.05 )= 1.143`
Now since `rho = (Z_(eff))/(a^(3)) ((Mw)/(N_(A)))`
We will have `rho_(NaCl)/(rho_(KCl)) = ((a_(KCl))/(a_(NaCl)))^(3) (Mw_(NaCl))/(Mw_(KCl))`
`= (1.143)^(3) ((58.5)/(74.5)) = 1.172`
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