The number of atoms in 2.4 g of body-centred | vedclass

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(C) For a $BCC$ crystal,the number of atoms per unit cell,$Z = 2$.The density formula is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 10 \ g \

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$6 	imes 10^{22}$

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(C) For a $BCC$ crystal,the number of atoms per unit cell,$Z = 2$.The density formula is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 10 \ g \ cm^{-3}$,$a = 200 \ pm = 200 	imes 10^{-10} \ cm = 2 	imes 10^{-8} \ cm$,$N_A = 6 	imes 10^{23} \ mol^{-1}$,$Z = 2$.Calculating molar mass $(M)$:$M = \frac{d 	imes a^3 	imes N_A}{Z} = \frac{10 	imes (2 	imes 10^{-8})^3 	imes 6 	imes 10^{23}}{2} = \frac{10 	imes 8 	imes 10^{-24} 	imes 6 	imes 10^{23}}{2} = 24 \ g \ mol^{-1}$.Number of moles in $2.4 \ g$ = $\frac{2.4 \ g}{24 \ g \ mol^{-1}} = 0.1 \ mol$.Number of atoms = $	ext{moles} 	imes N_A = 0.1 	imes 6 	imes 10^{23} = 6 	imes 10^{22} \ atoms$.

The number of atoms in $2.4 \ g$ of body-centred cubic $(BCC)$ crystal with edge length $200 \ pm$ is (density = $10 \ g \ cm^{-3}$,$N_A = 6 \times 10^{23} \ atoms \ mol^{-1}$)

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