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(B) For a $bcc$ unit cell,the number of atoms per unit cell $(Z)$ is $2$. The mass of the unit cell is calculated as $m = \frac{Z 	imes M}{N_A}$,wher

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$27.8$

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(B) For a $bcc$ unit cell,the number of atoms per unit cell $(Z)$ is $2$. The mass of the unit cell is calculated as $m = \frac{Z 	imes M}{N_A}$,where $M = 50 \, g \, mol^{-1}$ and $N_A = 6.022 	imes 10^{23} \, mol^{-1}$. $m = \frac{2 	imes 50}{6.022 	imes 10^{23}} \approx 16.606 	imes 10^{-23} \, g = 166.06 	imes 10^{-24} \, g$. The volume of the unit cell $(V)$ is given by $V = \frac{m}{d}$,where $d = 5.96 \, g \, cm^{-3}$. $V = \frac{166.06 	imes 10^{-24} \, g}{5.96 \, g \, cm^{-3}} \approx 27.86 	imes 10^{-24} \, cm^3$. Thus,the volume is approximately $27.8 	imes 10^{-24} \, cm^3$.

$A$ metal (atomic mass $= 50$) has a body-centered cubic $(bcc)$ crystal structure. If the density of the metal is $5.96 \, g \, cm^{-3}$,then the volume of the unit cell is ............ $\times 10^{-24} \, cm^3$.

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