Verified Solution • 2025–26 Session moderate

QJEEM22Ch0579: JEE Main 2022 Chemistry - Class 12 JEEM22Ch0579 Solution: The distance between Na + and Cl ions in solid NaCl of density…

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)

Step-by-Step Verified Solution

ρ=Z×Ma3×NA\rho=\frac{Z \times M}{a^{3} \times N_{A}}

43.1=4×58.5a3×6.02×102343.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}}

amp;a3=0.9×10−23amp;=9×10−24\begin{aligned}& a^{3}=0.9 \times 10^{-23} \\\\& =9 \times 10^{-24}\end{aligned}

a=2.08×10−8 cma=2.08 \times 10^{-8} \mathrm{~cm}

=2.08×10−10 m=2.08 \times 10^{-10} \mathrm{~m}

for NaCl\mathrm{NaCl} , distance between Na+\mathrm{Na}^{+} and Cl−=a2\mathrm{Cl}^{-}=\frac{a}{2}

=1.04×10−10 m=1.04 \times 10^{-10} \mathrm{~m}

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