Calculate the number of atoms present in 1.62 | vedclass

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(D) Step $1$: Identify the number of atoms per unit cell for $bcc$. For $bcc$, $Z = 2$.Step $2$: Use the density formula: $ho = \frac{Z 	imes M}{N_

Vedclass · Accepted Answer

$1.0 	imes 10^{22}$

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(D) Step $1$: Identify the number of atoms per unit cell for $bcc$. For $bcc$, $Z = 2$.Step $2$: Use the density formula: $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, which can be rearranged as $M = \frac{ho 	imes a^3 	imes N_A}{Z}$.Step $3$: Calculate the molar mass $M$: $M = \frac{3.25 	imes 10^{-22} 	imes 6.022 	imes 10^{23}}{2} = \frac{195.715}{2} \approx 97.86 	ext{ g/mol}$.Step $4$: Calculate the number of moles $n$: $n = \frac{	ext{mass}}{M} = \frac{1.625}{97.86} \approx 0.0166 	ext{ mol}$.Step $5$: Calculate the number of atoms: $	ext{Atoms} = n 	imes N_A = 0.0166 	imes 6.022 	imes 10^{23} \approx 1.0 	imes 10^{22}$ atoms.

Calculate the number of atoms present in $1.625 \text{ g}$ of a metal that forms a $bcc$ unit cell structure. Given: $[\rho \times a^3 = 3.25 \times 10^{-22} \text{ g}]$.

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