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(C) For $CsCl$ in $bcc$ structure,the number of formula units per unit cell $Z = 1$.The density formula is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Giv

Vedclass · Accepted Answer

$4.1 	imes 10^{-8} \, cm$

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(C) For $CsCl$ in $bcc$ structure,the number of formula units per unit cell $Z = 1$.The density formula is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $d = 3.988 \, g/cm^3$,$M = 168.4 \, g/mol$,$N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Rearranging for $a^3$: $a^3 = \frac{Z 	imes M}{N_A 	imes d} = \frac{1 	imes 168.4}{6.022 	imes 10^{23} 	imes 3.988}$.$a^3 \approx \frac{168.4}{24.015 	imes 10^{23}} \approx 7.012 	imes 10^{-23} \, cm^3$.$a = \sqrt[3]{70.12 	imes 10^{-24}} \approx 4.12 	imes 10^{-8} \, cm$.

If the density of a $CsCl$ crystal,which crystallizes in a $bcc$ structure,is $3.988 \, g/cm^3$,then the edge length of the unit cell will be ........ $(CsCl = 168.4 \, g/mol)$.

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