Which is the volume of unit cell of a metal ( | vedclass

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(C) The density formula for a unit cell is given by: $ho = \frac{Z 	imes M}{N_A 	imes a^3}$,where $a^3$ is the volume of the unit cell $(V)$.For a

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$2.76 	imes 10^{-23} \ cm^3$

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(C) The density formula for a unit cell is given by: $ho = \frac{Z 	imes M}{N_A 	imes a^3}$,where $a^3$ is the volume of the unit cell $(V)$.For a $BCC$ structure,the number of atoms per unit cell $(Z)$ is $2$.Given: Atomic mass $(M)$ = $25 \ g \ mol^{-1}$,Density $(ho)$ = $3 \ g \ cm^{-3}$,$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Rearranging the formula for volume $(V)$: $V = \frac{Z 	imes M}{ho 	imes N_A}$.Substituting the values: $V = \frac{2 	imes 25}{3 	imes 6.022 	imes 10^{23}}$.$V = \frac{50}{18.066 	imes 10^{23}} \approx 2.767 	imes 10^{-23} \ cm^3$.

Which is the volume of unit cell of a metal (atomic mass $25 \ g \ mol^{-1}$) having $BCC$ structure and density $3 \ g \ cm^{-3}$?

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