Question

The distance between $N{a}^{+}$and $C{l}^{-}$ions in solid $NaCl$ of density $43.1gc{m}^{-3}$ is ________$\times 10^{-10}m$. (Nearest Integer)
(Given : $N_A=6.02\times 10^{23}mo{l}^{-1}$ )

Answer 1

Answer: 1

Unit cell formula $-N{a}_{4}C{l}_4$
Mass per unit cell $=\frac{Z\times M\cdot M.}{N_A}g$
$=\frac{4\times 58.5}{N_A}g$
$d_{\text{unit cell }}=\frac{m}{V}=\frac{m}{a^3}$
$\Rightarrow \frac{4\times 58.5}{N_A\cdot a^3}=43.1$
$\Rightarrow a^3=9.02\times 10^{-24}c{m}^3$
$\Rightarrow a=2.08\times 10^{-8}cm$
$\Rightarrow a=2.08\times 10^{-10}m$
Also $a=2\left(r_{N{a}^{+}}+r_{C{l}^{-}}\right)$
$\Rightarrow r_{N{a}^{+}}+r_{C{l}^{-}}=1.04\times 10^{-10}m$
$\therefore$ The answer is $1$