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Q. The distance between $Na^+$ and $Cl^-$ ions in $NaCl$ with a density $3.165\, g\, cm^{-3}$ is

The Solid State

Solution:

$d=\frac{Z\times M}{a^{3}\times N_{A}}$
$a^{3}=\frac{Z\times M}{d\times N_{A}}$
$=\frac{4\times58.5}{3.165\times6.023\times10^{23}}$
$a^{3}=122.77\times10^{-24}\,cm^{3}$
$a=4.97\times10^{-8}\,cm$ or $497\,pm$
Distance between $Na^{+}$ and $Cl^{-} =\frac{a}{2}=\frac{497}{2}$
$=248.5\,pm$