Question

The distance between $Na^+$ and $Cl^–$ ions in solid $NaCl$ of density $43.1\ g\ cm^{–3}$ is ____ $× 10^{–10}\ m$. (Nearest Integer)
(Given : $N_A = 6.02 × 10^{23} mol^{–1}$)

Answer 1

$ρ = \frac {Z\times M}{a^3\times N_A}$

$43.1 = \frac {4×58.5}{a^3×6.02×10^{23}}$

$a^3 = \frac {4×58.5}{43.1×6.02×10^{23}}$
$a^3 = 0.9 × 10^{–23}$
$a^3= 9 × 10^{–24}$
$a = 2.08 × 10^{–8} cm$
$a= 2.08 × 10^{–10} m$
for $NaCl$, distance between $Na^+$ and $Cl^–$ = $\frac a2$
$= \frac {2.04 × 10^{–10}}{2} m$
$= 1.02 \times 10^{–10} m$

So, the answer is $1$.