Metal M crystallizes in fcc lattice. If the e | vedclass

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

Vedclass · Accepted Answer

$107.15$

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(A) For an $fcc$ lattice,the number of atoms per unit cell,$Z = 4$. Given: Edge length $a = 4.077 	imes 10^{-8} \ cm$,Density $d = 10.5 \ g \ cm^{-3}$,Avogadro's number $N_A = 6.022 	imes 10^{23} \ mol^{-1}$. The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$. Rearranging for atomic mass $M$: $M = \frac{d 	imes a^3 	imes N_A}{Z}$. $M = \frac{10.5 	imes (4.077 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{4}$. $M = \frac{10.5 	imes 67.77 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{4}$. $M = \frac{428.55}{4} \approx 107.14 \ g \ mol^{-1}$.

Metal $M$ crystallizes in $fcc$ lattice. If the edge length of the unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$,the atomic mass of the metal is:

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