KF has a structure similar to NaCl. If the de | vedclass

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(B) For $NaCl$ type structure, $Z = 4$. The molar mass of $KF$ is $M = 39 + 19 = 58 \, g/mol$.Using the density formula: $d = \frac{Z 	imes M}{N_A

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$270$

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(B) For $NaCl$ type structure, $Z = 4$. The molar mass of $KF$ is $M = 39 + 19 = 58 \, g/mol$.Using the density formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$$2.48 = \frac{4 	imes 58}{6.022 	imes 10^{23} 	imes a^3}$$a^3 = \frac{232}{2.48 	imes 6.022 	imes 10^{23}} = \frac{232}{14.934 	imes 10^{23}} = 15.535 	imes 10^{-23} \, cm^3 = 155.35 	imes 10^{-24} \, cm^3$$a = \sqrt[3]{155.35} 	imes 10^{-8} \, cm \approx 5.375 	imes 10^{-8} \, cm = 537.5 \, pm$.In $NaCl$ type structure, the distance between cation and anion is $r_+ + r_- = \frac{a}{2}$.$r_+ + r_- = \frac{537.5}{2} \approx 268.75 \, pm \approx 270 \, pm$.

$KF$ has a structure similar to $NaCl$. If the density of the crystal is $2.48 \, g/cm^3$, the distance between $K^+$ and $F^-$ ions will be ............. $pm$. (Atomic masses: $K = 39 \, amu, F = 19 \, amu$)

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