A metal has a fcc lattice. The edge length of | vedclass

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(D) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ lattice, the number of atoms per unit cell $Z = 4$.Given edge l

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$27$

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(D) The formula for density is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ lattice, the number of atoms per unit cell $Z = 4$.Given edge length $a = 404 \, pm = 404 	imes 10^{-10} \, cm$.Given density $ho = 2.72 \, g \, cm^{-3}$ and $N_A = 6.02 	imes 10^{23} \, mol^{-1}$.Substituting the values: $2.72 = \frac{4 	imes M}{6.02 	imes 10^{23} 	imes (404 	imes 10^{-10})^3}$.Solving for $M$: $M = \frac{2.72 	imes 6.02 	imes 10^{23} 	imes 66.0 	imes 10^{-24}}{4} \approx 27 \, g \, mol^{-1}$.

$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )

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