An element crystallising in fcc lattice has a | vedclass

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(B) Given: Density $(d) = 8.92 \ g \ cm^{-3}$,Edge length $(a) = 3.61 	imes 10^{-8} \ cm$,Avogadro's number $(N_A) = 6.022 	imes 10^{23} \ mol^{-1}$

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$63.178$

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(B) Given: Density $(d) = 8.92 \ g \ cm^{-3}$,Edge length $(a) = 3.61 	imes 10^{-8} \ cm$,Avogadro's number $(N_A) = 6.022 	imes 10^{23} \ mol^{-1}$.For an $fcc$ lattice,the number of atoms per unit cell $(Z) = 4$.The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$,where $M$ is the atomic weight.Rearranging for $M$: $M = \frac{d 	imes a^3 	imes N_A}{Z}$.Substituting the values: $M = \frac{8.92 	imes (3.61 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{4}$.$M = \frac{8.92 	imes 47.0458 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{4}$.$M = \frac{252.712}{4} = 63.178 \ u$.

An element crystallising in $fcc$ lattice has a density of $8.92 \ g \ cm^{-3}$ and edge length of $3.61 \times 10^{-8} \ cm$. What is the atomic weight of the element (in $u$)? $(N_A = 6.022 \times 10^{23} \ mol^{-1})$

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