The density of NaCl is 2.165 , g , cm^{-3}. T | vedclass

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(C) For $NaCl$ crystal,the density $d = 2.165 \, g \, cm^{-3}$.$NaCl$ has an $fcc$ structure,so the number of formula units per unit cell $Z = 4$.The

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$6.09 	imes 10^{23}$

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(C) For $NaCl$ crystal,the density $d = 2.165 \, g \, cm^{-3}$.$NaCl$ has an $fcc$ structure,so the number of formula units per unit cell $Z = 4$.The distance between $Na^+$ and $Cl^-$ ions is $d_{ion} = a/2$,where $a$ is the edge length.$281 \, pm = a/2 \implies a = 562 \, pm = 562 	imes 10^{-10} \, cm$.The molar mass of $NaCl$ is $M = 23 + 35.5 = 58.5 \, g \, mol^{-1}$.The density formula is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Substituting the values: $2.165 = \frac{4 	imes 58.5}{(562 	imes 10^{-10})^3 	imes N_A}$.$N_A = \frac{4 	imes 58.5}{(562 	imes 10^{-10})^3 	imes 2.165}$.$N_A = \frac{234}{1.775 	imes 10^{-22} 	imes 2.165} \approx 6.09 	imes 10^{23} \, mol^{-1}$.

The density of $NaCl$ is $2.165 \, g \, cm^{-3}$. The distance between $Na^+$ and $Cl^-$ ions in the $NaCl$ crystal is $281 \, pm$. Based on this information,the value of Avogadro's number is ...

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