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(A) For a simple cubic structure, the number of atoms per unit cell, $Z = 1$. The edge length $a = 336 \ pm = 3.36 	imes 10^{-8} \ cm$. The molar mas

Vedclass · Accepted Answer

$8.98$

Vedclass · Answer

(A) For a simple cubic structure, the number of atoms per unit cell, $Z = 1$. The edge length $a = 336 \ pm = 3.36 	imes 10^{-8} \ cm$. The molar mass $M$ can be calculated from the given data: $2.64 	imes 10^{23}$ atoms have a mass of $90 \ g$. Therefore, $6.022 	imes 10^{23}$ atoms (Avogadro's number, $N_A$) have a mass $M = \frac{90 	imes 6.022 	imes 10^{23}}{2.64 	imes 10^{23}} \approx 205.3 \ g \ mol^{-1}$. Using the density formula $d = \frac{Z 	imes M}{a^3 	imes N_A}$: $d = \frac{1 	imes 205.3}{(3.36 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}} \approx 8.98 \ g \ cm^{-3}$.

Calculate the density of a metal which forms a simple cubic structure with an edge length of unit cell $336 \ pm$. ($90 \ g$ of metal contains $2.64 \times 10^{23}$ atoms) (in $g \ cm^{-3}$)

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