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(A) For an $NaCl$ crystal (fcc structure),the distance between $Na^+$ and $Cl^-$ ions $(d)$ is given by $d = a/2$,where $a$ is the edge length of the

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$6.05 	imes 10^{23}$

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(A) For an $NaCl$ crystal (fcc structure),the distance between $Na^+$ and $Cl^-$ ions $(d)$ is given by $d = a/2$,where $a$ is the edge length of the unit cell.Given $d = 2.814 \ \overset{\circ}{A}$,so $a = 2 	imes 2.814 = 5.628 \ \overset{\circ}{A} = 5.628 	imes 10^{-8} \ cm$.The density formula is $ho = (Z 	imes M) / (N_A 	imes a^3)$.For $NaCl$,$Z = 4$ (number of formula units per unit cell) and $M = 58.5 \ g/mol$.Substituting the values: $2.167 = (4 	imes 58.5) / (N_A 	imes (5.628 	imes 10^{-8})^3)$.$N_A = (4 	imes 58.5) / (2.167 	imes 178.28 	imes 10^{-24})$.$N_A = 234 / (386.33 	imes 10^{-24}) \approx 6.057 	imes 10^{23} \ mol^{-1}$.

In a $NaCl$ crystal,the distance between $Na^+$ and $Cl^-$ ions is $2.814 \ \overset{\circ}{A}$ and the density of the crystal is $2.167 \ g/cm^3$. Calculate the value of Avogadro's number $(N_A)$.

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