Calculate the number of atoms in 5.4 g of a | vedclass

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(A) For an $fcc$ structure,the number of atoms per unit cell $(Z)$ is $4$. The density formula is $\varrho = \frac{Z 	imes M}{N_A 	imes a^3}$,which

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$3.0 	imes 10^{22}$

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(A) For an $fcc$ structure,the number of atoms per unit cell $(Z)$ is $4$. The density formula is $\varrho = \frac{Z 	imes M}{N_A 	imes a^3}$,which can be rearranged as $M = \frac{\varrho 	imes a^3 	imes N_A}{Z}$. Given $\varrho 	imes a^3 = 7.2 	imes 10^{-22} \ g$,$Z = 4$,and $N_A = 6.022 	imes 10^{23} \ mol^{-1}$. The molar mass $M = \frac{7.2 	imes 10^{-22} 	imes 6.022 	imes 10^{23}}{4} \approx 108.4 \ g/mol$. The number of moles $n = \frac{	ext{mass}}{M} = \frac{5.4}{108.4} \approx 0.0498 \ mol$. The number of atoms $= n 	imes N_A = 0.0498 	imes 6.022 	imes 10^{23} \approx 3.0 	imes 10^{22}$ atoms.

Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

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