The distance between Na+ and Cl$-$ ions in solid NaCl of density 43.1 g cm$-$3 is _______________ $\times$ 10$-$10 m. (Nearest Integer)
(Given : NA = 6.02 $\times$ 1023 mol$-$1)
Answer (integer)
1
Solution
$\rho=\frac{Z \times M}{a^{3} \times N_{A}}$
<br/><br/>
$43.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}}$
<br/><br/>
$$
\begin{aligned}
& a^{3}=0.9 \times 10^{-23} \\\\
& =9 \times 10^{-24}
\end{aligned}
$$
<br/><br/>
$a=2.08 \times 10^{-8} \mathrm{~cm}$
<br/><br/>
$=2.08 \times 10^{-10} \mathrm{~m}$
<br/><br/>
for $\mathrm{NaCl}$, distance between $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}=\frac{a}{2}$
<br/><br/>
$=1.04 \times 10^{-10} \mathrm{~m}$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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