An element occurs in the body-centred cubic ( | vedclass

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(A) $1$. For a $BCC$ unit cell, the number of atoms per unit cell $(Z)$ is $2$. $2$. The edge length $(a)$ is $288 \ pm = 288 	imes 10^{-10} \ cm$. $

Vedclass · Accepted Answer

$24.2 	imes 10^{23}$

Vedclass · Answer

(A) $1$. For a $BCC$ unit cell, the number of atoms per unit cell $(Z)$ is $2$. $2$. The edge length $(a)$ is $288 \ pm = 288 	imes 10^{-10} \ cm$. $3$. The volume of the unit cell $(V)$ is $a^3 = (288 	imes 10^{-10} \ cm)^3 \approx 2.39 	imes 10^{-23} \ cm^3$. $4$. The density $(ho)$ is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. $5$. Rearranging to find the molar mass $(M)$: $M = \frac{ho 	imes N_A 	imes a^3}{Z} = \frac{7.2 	imes 6.022 	imes 10^{23} 	imes 2.39 	imes 10^{-23}}{2} \approx 51.8 \ g \ mol^{-1}$. $6$. The number of moles in $208 \ g$ is $n = \frac{	ext{mass}}{M} = \frac{208}{51.8} \approx 4.015 \ mol$. $7$. The number of atoms is $n 	imes N_A = 4.015 	imes 6.022 	imes 10^{23} \approx 24.18 	imes 10^{23} \approx 24.2 	imes 10^{23}$.

An element occurs in the body-centred cubic $(BCC)$ structure with an edge length of $288 \ pm$. The density of the element is $7.2 \ g \ cm^{-3}$. The number of atoms present in $208 \ g$ of the element is nearly:

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