Calculate the number of atoms in 1 text{ g} o | vedclass

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(C) For a $bcc$ crystal structure, the number of atoms per unit cell, $Z = 2$.Density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$,

Vedclass · Accepted Answer

$3.03 	imes 10^{21}$

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(C) For a $bcc$ crystal structure, the number of atoms per unit cell, $Z = 2$.Density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $M$ is the molar mass and $N_A$ is Avogadro's number.Rearranging for the mass of one unit cell: $m_{cell} = ho 	imes a^3 = \frac{Z 	imes M}{N_A}$.Given $ho 	imes a^3 = 6.6 	imes 10^{-22} 	ext{ g}$, this represents the mass of one unit cell.The number of unit cells in $1 	ext{ g}$ of metal is $n = \frac{1 	ext{ g}}{6.6 	imes 10^{-22} 	ext{ g/unit cell}} \approx 1.515 	imes 10^{21} 	ext{ unit cells}$.Since each $bcc$ unit cell contains $2$ atoms, the total number of atoms $= 2 	imes 1.515 	imes 10^{21} = 3.03 	imes 10^{21} 	ext{ atoms}$.

Calculate the number of atoms in $1 \text{ g}$ of a metal that forms a $bcc$ crystal structure. Given that the product of density and unit cell volume is $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$.

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