Calculate the number of atoms present in 1 te | vedclass

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(A) Step $1$: Identify the number of atoms per unit cell $(Z)$ for a simple cubic structure. For a simple cubic unit cell, $Z = 1$.Step $2$: Use the d

Vedclass · Accepted Answer

$1.504 	imes 10^{22}$

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(A) Step $1$: Identify the number of atoms per unit cell $(Z)$ for a simple cubic structure. For a simple cubic unit cell, $Z = 1$.Step $2$: Use the density formula: $ho = \frac{Z 	imes M}{N_A 	imes V}$, where $ho$ is density, $M$ is molar mass, $N_A$ is Avogadro's number, and $V$ is the volume of the unit cell.Step $3$: Rearrange to find the mass of one unit cell: $ho 	imes V = \frac{Z 	imes M}{N_A}$. Given $ho 	imes V = 66.5 	imes 10^{-24} 	ext{ g}$, this represents the mass of one unit cell.Step $4$: Calculate the number of unit cells in $1 	ext{ g}$ of metal: $	ext{Number of unit cells} = \frac{	ext{Total mass}}{	ext{Mass of one unit cell}} = \frac{1 	ext{ g}}{66.5 	imes 10^{-24} 	ext{ g}} \approx 1.5037 	imes 10^{22}$.Step $5$: Since each unit cell contains $1$ atom $(Z=1)$, the total number of atoms is $1.5037 	imes 10^{22} 	imes 1 = 1.504 	imes 10^{22}$.

Calculate the number of atoms present in $1 \text{ g}$ of a metal that forms a simple cubic unit cell structure, given that the product of density and volume of the unit cell is $66.5 \times 10^{-24} \text{ g}$.

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