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$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$. Given: density $d = 6.8 \ g \ cm^{-3}$, edge length $a = 290 \ pm = 290 	imes

Vedclass · Accepted Answer

$2.4 	imes 10^{24}$

Vedclass · Answer

$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $n = 2$. Given: density $d = 6.8 \ g \ cm^{-3}$, edge length $a = 290 \ pm = 290 	imes 10^{-10} \ cm$. Using the formula $d = \frac{n 	imes M}{a^3 	imes N_A}$, where $M$ is the molar mass: $6.8 = \frac{2 	imes M}{(290 	imes 10^{-10})^3 	imes 6.022 	imes 10^{23}}$. $M = \frac{6.8 	imes (2.4389 	imes 10^{-23}) 	imes 6.022 	imes 10^{23}}{2} \approx 50 \ g \ mol^{-1}$. Number of atoms in $200 \ g = \frac{	ext{mass}}{M} 	imes N_A = \frac{200}{50} 	imes 6.022 	imes 10^{23} = 4 	imes 6.022 	imes 10^{23} = 2.4088 	imes 10^{24} \approx 2.4 	imes 10^{24}$ atoms.

An element has a density of $6.8 \ g \ cm^{-3}$ and crystallizes in a $bcc$ structure with a unit cell edge length of $290 \ pm$. The number of atoms in $200 \ g$ of the element is:

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