Calculate the number of atoms in 5 g of a me | vedclass

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(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$. Edge length $(a)$ $= 336 \ pm = 3.36 	imes 10^{-8} \ cm$. Density $(

Vedclass · Accepted Answer

$1.4 	imes 10^{22}$

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(A) For a simple cubic unit cell, the number of atoms per unit cell $(z)$ $= 1$. Edge length $(a)$ $= 336 \ pm = 3.36 	imes 10^{-8} \ cm$. Density $(d)$ $= \frac{z 	imes M}{a^3 	imes N_A}$, where $M$ is the molar mass and $N_A$ is Avogadro's number $(6.022 	imes 10^{23} \ mol^{-1})$. $9.4 = \frac{1 	imes M}{(3.36 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}$. $M = 9.4 	imes (3.793 	imes 10^{-23}) 	imes 6.022 	imes 10^{23} \approx 214.65 \ g \ mol^{-1}$. Number of atoms in $5 \ g = \frac{	ext{mass}}{	ext{molar mass}} 	imes N_A = \frac{5}{214.65} 	imes 6.022 	imes 10^{23} \approx 1.4 	imes 10^{22}$ atoms.

Calculate the number of atoms in $5 \ g$ of a metal that crystallises in a simple cubic unit cell structure with an edge length of $336 \ pm$. (Density of the metal $= 9.4 \ g \ cm^{-3}$)

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