# $KCl$ crystallizes in the same type of lattice as does $NaCl$.Givn that $\large\frac{r_{Na^+}}{r_{Cl^-}}$$=0.5 and \large\frac{r_{Na^+}}{r_{K^+}}$$=0.7$.Calculate the ratio of the side unit cell for $KCl$ to that for $NaCl$,and the ratio of density of $NaCl$ to that of $KCl$

$\begin{array}{1 1}1.143,1.172\\2.143,2.172\\3.143,3.172\\4.143,4.172\end{array}$

why do we need consider 1.5 and 1.2

NaCl crystallizes in the face-centred cubic unit cell,such that
$r_{Na^+}+r_{Cl^-}=\large\frac{a}{2}$
Where $a$ is the edge length of unit cell.Now since $r_{Na^+}/r_{Cl^-}$$=0.5 and r_{Na^+}/r_{K^+}$$=0.7$,we will have
$\large\frac{r_{Na^+}+r_{Cl^-}}{r_{Cl^-}}$$=1.5 And \large\frac{r_{K^+}}{r_{Cl^-}}=\frac{r_{K^+}}{r_{Na^+}/0.5}=\frac{0.5}{r_{Na^+}r_K^+}=\frac{0.5}{0.7} \large\frac{r_k^++r_{Cl^-}}{r_{Na^+}+r_{Cl^-}}=\frac{1.2}{0.7}\times \frac{1}{1.5} Or \large\frac{a_{KCl}/2}{a_{NaCl}/2}=\frac{1.2}{0.7\times 1.5} Or \large\frac{a_{KCl}}{a_{NaCl}}=\frac{1.2}{1.05}$$=1.143$
Now since $\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$,we will have
$\large\frac{\rho_{NaCl}}{\rho_{KCl}}=\big(\frac{a_{KCl}}{a_{NaCl}}\big)^3\big(\frac{M_{NaCl}}{M_{KCl}}\big)=$$(1.143)^2\big(\large\frac{58.5}{74.5}\big)$$=1.172$