How many atoms of niobium are present in 2.43 | vedclass

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(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$

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$2.0 	imes 10^{22}$

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(D) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N_A$ is Avogadro's number,and $a^3$ is the volume of the unit cell $(V)$.Alternatively,$d = \frac{Z 	imes 	ext{mass of one atom}}{V}$.For a $bcc$ structure,the number of atoms per unit cell is $Z = 2$.Given: $d = 9 \ g \ cm^{-3}$,$V = 2.7 	imes 10^{-23} \ cm^3$,and total mass $= 2.43 \ g$.Let $N$ be the total number of atoms in $2.43 \ g$.The mass of one atom is $\frac{2.43}{N}$.Substituting the values into the density formula: $9 = \frac{2 	imes (2.43 / N)}{2.7 	imes 10^{-23}}$.Rearranging for $N$: $N = \frac{2 	imes 2.43}{9 	imes 2.7 	imes 10^{-23}}$.$N = \frac{4.86}{24.3 	imes 10^{-23}} = 0.2 	imes 10^{23} = 2.0 	imes 10^{22}$ atoms.

How many atoms of niobium are present in $2.43 \ g$ if it forms $bcc$ structure with density $9 \ g \ cm^{-3}$ and volume of unit cell $2.7 \times 10^{-23} \ cm^3$?

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