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(C) For an $fcc$ unit cell, the number of atoms per unit cell $Z = 4$.Given: Density $d = 8.92 	ext{ g cm}^{-3}$, edge length $a = 3.608 	imes 10^{-

Vedclass · Accepted Answer

$63$

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(C) For an $fcc$ unit cell, the number of atoms per unit cell $Z = 4$.Given: Density $d = 8.92 	ext{ g cm}^{-3}$, edge length $a = 3.608 	imes 10^{-8} 	ext{ cm}$, $N_A = 6.022 	imes 10^{23} 	ext{ mol}^{-1}$.The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Rearranging for molar mass $M$: $M = \frac{d 	imes a^3 	imes N_A}{Z}$.$M = \frac{8.92 	imes (3.608 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{4}$.$M = \frac{8.92 	imes 46.97 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{4}$.$M = \frac{252.36}{4} \approx 63.09 	ext{ g/mol}$.Thus, the atomic mass is approximately $63 	ext{ g/mol}$.

An element crystallises in an $fcc$ unit cell with a cell edge length of $3.608 \times 10^{-8} \text{ cm}$. The density of the element is $8.92 \text{ g cm}^{-3}$. Calculate the atomic mass of the element $(N_A = 6.022 \times 10^{23} \text{ mol}^{-1})$. (in $\text{ g/mol}$)

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