A face-centered cubic (FCC) solid of an eleme | vedclass

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Question

(A) For a face-centered cubic $(FCC)$ unit cell,the number of atoms per unit cell $(z)$ is $4$.The atomic mass $(M)$ is $60 \, g/mol$.The edge length

Vedclass · Accepted Answer

$6.25 \, g/cm^3$

Vedclass · Answer

(A) For a face-centered cubic $(FCC)$ unit cell,the number of atoms per unit cell $(z)$ is $4$.The atomic mass $(M)$ is $60 \, g/mol$.The edge length $(a)$ is $4 	imes 10^{-8} \, cm$.Avogadro's number $(N_A)$ is $6 	imes 10^{23} \, mol^{-1}$.The formula for density $(d)$ is given by $d = \frac{z 	imes M}{a^3 	imes N_A}$.Substituting the values: $d = \frac{4 	imes 60}{(4 	imes 10^{-8})^3 	imes (6 	imes 10^{23})}$.$d = \frac{240}{64 	imes 10^{-24} 	imes 6 	imes 10^{23}} = \frac{240}{384 	imes 10^{-1}} = \frac{240}{38.4} = 6.25 \, g/cm^3$.

$A$ face-centered cubic $(FCC)$ solid of an element (atomic mass $60$) has a cubic edge length of $4 \times 10^{-8} \, cm$. If Avogadro's number is $6 \times 10^{23} \, mol^{-1}$,then the density of the solid is:

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