An element has a body-centered cubic (BCC) st | vedclass

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(C) For a $BCC$ structure, the number of atoms per unit cell $(Z)$ is $2$.Given: Density $(d)$ = $7.2 \ g/cm^3$, Edge length $(a)$ = $288 \ pm = 288 \

Vedclass · Accepted Answer

$24 	imes 10^{23}$

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(C) For a $BCC$ structure, the number of atoms per unit cell $(Z)$ is $2$.Given: Density $(d)$ = $7.2 \ g/cm^3$, Edge length $(a)$ = $288 \ pm = 288 	imes 10^{-10} \ cm$, Mass $(m)$ = $208 \ g$.The volume of the unit cell is $a^3 = (288 	imes 10^{-10} \ cm)^3 = 2.3887872 	imes 10^{-23} \ cm^3$.The volume of $208 \ g$ of the element is $V = \frac{	ext{mass}}{	ext{density}} = \frac{208 \ g}{7.2 \ g/cm^3} = 28.888 \ cm^3$.The number of unit cells is $\frac{V}{a^3} = \frac{28.888}{2.3887872 	imes 10^{-23}} \approx 1.209 	imes 10^{24}$.Since each $BCC$ unit cell contains $2$ atoms, the total number of atoms = $2 	imes 1.209 	imes 10^{24} \approx 2.418 	imes 10^{24} = 24.18 	imes 10^{23} \approx 24 	imes 10^{23}$ atoms.

An element has a body-centered cubic $(BCC)$ structure with an edge length of $288 \ pm$. The density of the element is $7.2 \ g/cm^3$. How many atoms are present in $208 \ g$ of the element?

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