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

$\begin{array}{1 1}1.143,1.172\\2.143,3.172\\3.143,0.172\\0.143,0.172\end{array} $

1 Answer

NaCl crystallizes in the face centered cubic unit cell such that
$r_{Na^+}+r_{Cl^-}=\large\frac{a}{2}$,where a is the edge length of unit cell .Now since
$\large\frac{r_{Na^+}}{r_{Cl^-}}=$$0.5,\large\frac{r_{Na^+}}{r_{K^+}}=$$0.7$
We will have,$\large\frac{r_{Na^+}+r_{Cl^-}}{r_{Cl^-}}$$=1.5$
$\large\frac{r_{K^+}}{r_{Cl^-}}=\frac{r_{K^+}}{r_{Na^+}/0.5}=\frac{0.5}{r_{Na^+}/r_{K^+}}=\frac{0.5}{0.7}$
Hence $\large\frac{r_{K^+}+r_{Cl^-}}{r_{Na^+}+r_{Cl^-}}=\frac{1.2}{0.7}\times \frac{1}{1.5}$
$\large\frac{a_{KCl/2}}{a_{NaCl/2}}=\frac{1.2}{0.7\times 1.5}=\frac{a_{Kcl}}{a_{NaCl}}$$=1.143$
Now since ,$\rho=\large\frac{n}{a^3}\big[\large\frac{M}{N_A}\big]$
$\therefore$ We will have
$\large\frac{\rho_{NaCl}}{\rho_{KCl}}=\big(\large\frac{a_{KCl}}{a_{NaCl}}\big)^3\big(\large\frac{M_{NaCl}}{M_{KCl}}\big)$
$\Rightarrow (1.143)^3\big(\large\frac{58.5}{74.5}\big)$$=1.172$
answered Aug 1, 2014 by sreemathi.v
 
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