Question

The distance between $Na ^{+}$and $Cl ^{-}$ions in solid $NaCl$ of density $43.1 \,g \,cm ^{-3}$ is ________$\times 10^{-10} m$. (Nearest Integer)
(Given : $N _{ A }=6.02 \times 10^{23} \,mol ^{-1}$ )

Answer 1

Unit cell formula $- Na _{4} Cl _{4}$
Mass per unit cell $=\frac{ Z \times M \cdot M .}{ N _{ A }} g$
$=\frac{4 \times 58.5}{N_{ A }} g$
$d _{\text {unit cell }}=\frac{ m }{ V }=\frac{ m }{ a ^{3}}$
$\Rightarrow \frac{4 \times 58.5}{ N _{ A } \cdot a ^{3}}=43.1$
$\Rightarrow a ^{3}=9.02 \times 10^{-24} cm ^{3}$
$\Rightarrow a =2.08 \times 10^{-8} cm$
$\Rightarrow a =2.08 \times 10^{-10} m$
Also $a =2\left( r _{ Na ^{+}}+ r _{ Cl ^{-}}\right)$
$\Rightarrow r _{ Na ^{+}}+ r _{ Cl ^{-}}=1.04 \times 10^{-10} m$
$\therefore$ The answer is $1$