JEE Main 2022 · 25 Jun · Shift 1 · Chemistry Q82

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ChemistryInteger+4 / -1Question 82 of 90

The distance between $\mathrm{Na^+}$ and $\mathrm{Cl^-}$ ions in solid NaCl of density 43.1 g cm $^{-3}$ is ______ $\times10^{-10}$ m. (Nearest Integer)

(Given : $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$ )

Your answer

Answer and worked solution

Official answer: 1

Unit cell (fcc, $Z=4$ , $M=58.5$ ): $a^3=\dfrac{4\times58.5}{43.1\times6.02\times10^{23}}\approx9.02\times10^{-24}$ cm $^3$ , so $a\approx2.08\times10^{-8}$ cm.

Na-Cl distance. $\frac a2\approx1.04\times10^{-8}$ cm $\approx1\times10^{-10}$ m. (The density 43.1 g cm $^{-3}$ is as printed; real NaCl is about 2.16.)

Answer: 1

Topics:Density of crystals

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JEE Main 2022 · 25 Jun · Shift 1, Chemistry question 82