In an fcc lattice,the edge length of a silver | vedclass

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(A) Given: Edge length $a = 4.077 	imes 10^{-8} \ cm$,Density $d = 10.5 \ g \ cm^{-3}$,for $fcc$ lattice,number of atoms per unit cell $Z = 4$,and Av

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$107.14$

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(A) Given: Edge length $a = 4.077 	imes 10^{-8} \ cm$,Density $d = 10.5 \ g \ cm^{-3}$,for $fcc$ lattice,number of atoms per unit cell $Z = 4$,and Avogadro's number $N_A = 6.023 	imes 10^{23} \ mol^{-1}$.Using the density formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$Rearranging for molar mass $M$: $M = \frac{d 	imes a^3 	imes N_A}{Z}$$M = \frac{10.5 	imes (4.077 	imes 10^{-8})^3 	imes 6.023 	imes 10^{23}}{4}$$M = \frac{10.5 	imes 67.76 	imes 10^{-24} 	imes 6.023 	imes 10^{23}}{4}$$M = \frac{428.56}{4} = 107.14 \ g \ mol^{-1}$.

In an $fcc$ lattice,the edge length of a silver unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$. Calculate the atomic mass of silver.

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