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

Correct 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 \(1\).