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(B) For a $bcc$ lattice, the number of atoms per unit cell, $n = 2$. Given: edge length $a = 500 \ pm = 5 	imes 10^{-8} \ cm$, density $ho = 7.5 \

Vedclass · Accepted Answer

$6.4 	imes 10^{23} \ 	ext{atoms}$

Vedclass · Answer

(B) For a $bcc$ lattice, the number of atoms per unit cell, $n = 2$. Given: edge length $a = 500 \ pm = 5 	imes 10^{-8} \ cm$, density $ho = 7.5 \ g \ cm^{-3}$, mass $m = 300 \ g$. Volume of unit cell $a^{3} = (5 	imes 10^{-8} \ cm)^{3} = 125 	imes 10^{-24} \ cm^{3}$. Using the density formula $ho = \frac{nM}{a^{3} N_{A}}$, we find the molar mass $M$: $M = \frac{ho 	imes a^{3} 	imes N_{A}}{n} = \frac{7.5 	imes 125 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{2} \approx 282.3 \ g \ mol^{-1}$. Number of atoms in $300 \ g$ of metal = $\frac{	ext{mass}}{	ext{molar mass}} 	imes N_{A} = \frac{300}{282.3} 	imes 6.022 	imes 10^{23} \approx 6.4 	imes 10^{23} \ 	ext{atoms}$.

An element crystallises in a $bcc$ lattice with a cell edge of $500 \ pm$. The density of the element is $7.5 \ g \ cm^{-3}$. How many atoms are present in $300 \ g$ of the metal?

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