Calculate the number of atoms in 1 g of a me | vedclass

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Question

(A) For an $fcc$ crystal structure,the number of atoms per unit cell $(Z)$ is $4$. The density formula is $\varrho = \frac{Z 	imes M}{N_A 	imes a^3}

Vedclass · Accepted Answer

$2.315 	imes 10^{22}$

Vedclass · Answer

(A) For an $fcc$ crystal structure,the number of atoms per unit cell $(Z)$ is $4$. The density formula is $\varrho = \frac{Z 	imes M}{N_A 	imes a^3}$,where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 	imes 10^{23} \ mol^{-1})$. Rearranging the formula,we get $M = \frac{\varrho 	imes a^3 	imes N_A}{Z}$. Substituting the given values: $M = \frac{1.728 	imes 10^{-22} 	imes 6.022 	imes 10^{23}}{4} = \frac{104.06}{4} = 26.015 \ g/mol$. The number of atoms in $1 \ g$ of the metal is calculated as: $	ext{Number of atoms} = \frac{	ext{mass}}{M} 	imes N_A = \frac{1}{26.015} 	imes 6.022 	imes 10^{23} \approx 2.315 	imes 10^{22}$ atoms.

Calculate the number of atoms in $1 \ g$ of a metal if it forms an $fcc$ crystal structure,given that $\varrho \times a^3 = 1.728 \times 10^{-22} \ g$.

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