Question

The distance between $N{a}^{+}$ and $C{l}^{-}$ ions in $NaCl$ with a density $3.165gc{m}^{-3}$ is

• A. $497pm$
• B. $248.5pm$
• C. $234pm$
• D. $538.5pm$

Answer 1

Answer: (B)

$d=\frac{Z\times M}{a^3\times N_A}$
$a^3=\frac{Z\times M}{d\times N_A}$
$=\frac{4\times 58.5}{3.165\times 6.023\times 10^{23}}$
$a^3=122.77\times 10^{-24}c{m}^3$
$a=4.97\times 10^{-8}cm$ or $497pm$
Distance between $N{a}^{+}$ and $C{l}^{-}=\frac{a}{2}=\frac{497}{2}$
$=248.5pm$