An element crystallizes in a body-centred cub | vedclass

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(D) For a body-centred cubic $(BCC)$ lattice, the number of atoms per unit cell is $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{

Vedclass · Accepted Answer

$5.0 	imes 10^{24}$

Vedclass · Answer

(D) For a body-centred cubic $(BCC)$ lattice, the number of atoms per unit cell is $Z = 2$. The density formula is given by $ho = \frac{Z 	imes M}{N_A 	imes a^3}$. Given: $ho = 5.0 \ g \ cm^{-3}$, $a = 200 \ pm = 200 	imes 10^{-10} \ cm = 2 	imes 10^{-8} \ cm$. $5.0 = \frac{2 	imes M}{6.022 	imes 10^{23} 	imes (2 	imes 10^{-8})^3}$. $M = \frac{5.0 	imes 6.022 	imes 10^{23} 	imes 8 	imes 10^{-24}}{2} = 12.044 \ g \ mol^{-1}$. Number of moles in $100 \ g = \frac{100}{M} = \frac{100}{12.044} \approx 8.303 \ mol$. Number of atoms $= 	ext{moles} 	imes N_A = \frac{100}{M} 	imes N_A$. Substituting $M = \frac{5.0 	imes N_A 	imes 8 	imes 10^{-24}}{2}$: Number of atoms $= \frac{100 	imes 2}{5.0 	imes 8 	imes 10^{-24}} = \frac{200}{40 	imes 10^{-24}} = 5 	imes 10^{24}$.

An element crystallizes in a body-centred cubic lattice. The edge length of the unit cell is $200 \ pm$ and the density of the element is $5.0 \ g \ cm^{-3}$. Calculate the number of atoms in $100 \ g$ of this element.

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