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KCl crystallizes in the same type of lattice as does $NaCl$ . Given that $\large\frac{r_{Na^{+}}}{r_{cl^{-}}}$$=0.5$ and $\large\frac{r_{Na^{+}}}{r_{K^{+}}}$$=0.5$ calculate the ratio of density of NaCl to that KCl

$\begin{array}{1 1} 103.5 \\ 1.143 \\ 8.4 \\ 1.172 \end{array} $

1 Answer

Answer :$ 1.172$
$\rho = \large\frac{N}{a^3} \bigg( \frac{M}{N_A}\bigg)$, we will have
$\large\frac{\rho_{NaCl}}{KCl}$$=\bigg( \large\frac{a_{KCl}}{a_{NaCl}}\bigg)^3\bigg(\frac{M_{NaCl}}{M_{KCl}} \bigg) $$=(1.143 )^2 \bigg( \large\frac{58.5}{74.5}\bigg) $$=1.172$
answered Aug 11, 2014 by meena.p
 

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