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Question

$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$. The edge length $(a)$ is $288 \ pm = 288 	imes 10^{-10} \ cm = 2.88 	im

Vedclass · Accepted Answer

$51.78 \ g \ mol^{-1}$

Vedclass · Answer

$(A)$ For a $bcc$ unit cell, the number of atoms per unit cell $(n)$ is $2$. The edge length $(a)$ is $288 \ pm = 288 	imes 10^{-10} \ cm = 2.88 	imes 10^{-8} \ cm$. The density formula is $ho = \frac{n 	imes M}{a^{3} 	imes N_{A}}$. Rearranging for molar mass $(M)$: $M = \frac{ho 	imes a^{3} 	imes N_{A}}{n}$. Substituting the values: $M = \frac{7.2 	imes (2.88 	imes 10^{-8})^{3} 	imes 6.022 	imes 10^{23}}{2}$. $M = \frac{7.2 	imes 23.8878 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{2} \approx \frac{103.56}{2} \approx 51.78 \ g \ mol^{-1}$. Thus, the correct option is $A$.

An element has a $bcc$ structure with a cell edge of $288 \ pm$. The density of the element is $7.2 \ g \ cm^{-3}$. What is the atomic mass of the element?

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