Metal M crystallises in fcc lattice. If the u | vedclass

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(B) The formula for density is $d = \frac{Z 	imes M}{N_{A} 	imes a^{3}}$.Rearranging for atomic mass $(M)$: $M = \frac{d 	imes N_{A} 	imes a^{3}}{

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$107.1$

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(B) The formula for density is $d = \frac{Z 	imes M}{N_{A} 	imes a^{3}}$.Rearranging for atomic mass $(M)$: $M = \frac{d 	imes N_{A} 	imes a^{3}}{Z}$.For an $fcc$ lattice,the number of atoms per unit cell $Z = 4$.Given: $d = 10.5 \ g \ cm^{-3}$,$a = 4.077 \ \mathring{A} = 4.077 	imes 10^{-8} \ cm$,and $N_{A} = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $M = \frac{10.5 	imes 6.022 	imes 10^{23} 	imes (4.077 	imes 10^{-8})^{3}}{4}$.$M = \frac{10.5 	imes 6.022 	imes 10^{23} 	imes 67.77 	imes 10^{-24}}{4}$.$M = \frac{428.4}{4} \approx 107.1 \ g \ mol^{-1}$.

Metal $M$ crystallises in $fcc$ lattice. If the unit cell has an edge length of $4.077 \ \mathring{A}$ and the density is $10.5 \ g \ cm^{-3}$,then the atomic weight of the metal is:

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