Calculate the number of atoms present in 90 t | vedclass

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(B) $1$. For a $bcc$ structure, the number of atoms per unit cell, $Z = 2$.$2$. The mass of one unit cell is given by $m = ho 	imes V = ho 	imes

Vedclass · Accepted Answer

$1.0 	imes 10^{24}$

Vedclass · Answer

(B) $1$. For a $bcc$ structure, the number of atoms per unit cell, $Z = 2$.$2$. The mass of one unit cell is given by $m = ho 	imes V = ho 	imes a^3$.$3$. Given $ho 	imes a^3 = 1.8 	imes 10^{-22} 	ext{ g}$, so the mass of one unit cell is $1.8 	imes 10^{-22} 	ext{ g}$.$4$. The number of unit cells in $90 	ext{ g}$ of the metal is $N_{uc} = \frac{	ext{Total mass}}{	ext{Mass of one unit cell}} = \frac{90 	ext{ g}}{1.8 	imes 10^{-22} 	ext{ g}} = 50 	imes 10^{22} = 5.0 	imes 10^{23}$.$5$. Total number of atoms = $Z 	imes N_{uc} = 2 	imes 5.0 	imes 10^{23} = 1.0 	imes 10^{24}$ atoms.

Calculate the number of atoms present in $90 \text{ g}$ of a metal that forms a $bcc$ structure. Given: $[\rho \times a^3 = 1.8 \times 10^{-22} \text{ g}]$, where $\rho$ is density and $a$ is the edge length of the unit cell.

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