An element has a body-centred cubic (bcc) str | vedclass

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$(A)$ Volume of the unit cell $= (288 \, pm)^{3} = (288 	imes 10^{-10} \, cm)^{3} = 2.3887872 	imes 10^{-23} \, cm^{3}$.Volume of $208 \, g$ of the

Vedclass · Accepted Answer

$2.416 	imes 10^{24} \, 	ext{atoms}$

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$(A)$ Volume of the unit cell $= (288 \, pm)^{3} = (288 	imes 10^{-10} \, cm)^{3} = 2.3887872 	imes 10^{-23} \, cm^{3}$.Volume of $208 \, g$ of the element $= \frac{	ext{mass}}{	ext{density}} = \frac{208 \, g}{7.2 \, g \, cm^{-3}} = 28.888 \, cm^{3}$.Number of unit cells in this volume $= \frac{28.888 \, cm^{3}}{2.3887872 	imes 10^{-23} \, cm^{3}/	ext{unit cell}} \approx 12.093 	imes 10^{23} \, 	ext{unit cells}$.Since each $bcc$ unit cell contains $2$ atoms, the total number of atoms $= 2 	imes 12.093 	imes 10^{23} \approx 2.418 	imes 10^{24} \, 	ext{atoms}$.

An element has a body-centred cubic $(bcc)$ structure with a cell edge 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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