KF has a NaCl type structure. What is the dis | vedclass

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(A) $1$. The density formula for a cubic unit cell is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.$2$. For $NaCl$ type structure,$Z = 4$. The molar mas

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$268.8 \ pm$

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(A) $1$. The density formula for a cubic unit cell is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.$2$. For $NaCl$ type structure,$Z = 4$. The molar mass of $KF$ is $M = 39 + 19 = 58 \ g \ mol^{-1}$.$3$. Given $ho = 2.48 \ g \ cm^{-3}$ and $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.$4$. $a^3 = \frac{Z 	imes M}{ho 	imes N_A} = \frac{4 	imes 58}{2.48 	imes 6.022 	imes 10^{23}} = 155.3 	imes 10^{-24} \ cm^3$.$5$. $a = \sqrt[3]{155.3 	imes 10^{-24}} = 5.375 	imes 10^{-8} \ cm = 537.5 \ pm$.$6$. In $NaCl$ type structure,the distance between cation and anion is $d = \frac{a}{2}$.$7$. $d = \frac{537.5 \ pm}{2} = 268.75 \ pm \approx 268.8 \ pm$.

$KF$ has a $NaCl$ type structure. What is the distance between $K^+$ and $F^-$ in $KF$ if the density is $2.48 \ g \ cm^{-3}$?

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