The density of KBr crystal is 2.75 , g/cm^3 a | vedclass

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(C) Given: Density $(d) = 2.75 \, g/cm^3$,Edge length $(a) = 654 \, pm = 654 	imes 10^{-10} \, cm$,Molar mass of $KBr (M) = 39 + 80 = 119 \, g/mol$,A

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$Z = 4$,Face Centered Cubic

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(C) Given: Density $(d) = 2.75 \, g/cm^3$,Edge length $(a) = 654 \, pm = 654 	imes 10^{-10} \, cm$,Molar mass of $KBr (M) = 39 + 80 = 119 \, g/mol$,Avogadro number $(N_A) = 6.022 	imes 10^{23} \, mol^{-1}$.Using the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$.$Z = \frac{d 	imes a^3 	imes N_A}{M} = \frac{2.75 	imes (654 	imes 10^{-10})^3 	imes 6.022 	imes 10^{23}}{119}$.$Z = \frac{2.75 	imes 279.77 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{119} \approx 3.89 \approx 4$.Since $Z = 4$,the unit cell is $fcc$ (Face Centered Cubic).

The density of $KBr$ crystal is $2.75 \, g/cm^3$ and the edge length of the unit cell is $654 \, pm$. Calculate the number of formula units per unit cell and identify the type of unit cell. $(K = 39 \, u, Br = 80 \, u)$.

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