What is the atomic mass of an element with BC | vedclass

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(C) For a $BCC$ unit cell, the number of atoms per unit cell is $n = 2$.Density formula: $ho = \frac{M 	imes n}{a^3 	imes N_A}$Rearranging for mol

Vedclass · Accepted Answer

$81.3 \ g \ mol^{-1}$

Vedclass · Answer

(C) For a $BCC$ unit cell, the number of atoms per unit cell is $n = 2$.Density formula: $ho = \frac{M 	imes n}{a^3 	imes N_A}$Rearranging for molar mass $M$: $M = \frac{ho 	imes a^3 	imes N_A}{n}$Given: $ho = 10 \ g \ cm^{-3}$, $a = 300 \ pm = 3 	imes 10^{-8} \ cm$, $N_A = 6.022 	imes 10^{23} \ mol^{-1}$, $n = 2$.$M = \frac{10 	imes (3 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}{2}$$M = \frac{10 	imes 27 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{2}$$M = \frac{162.594}{2} \approx 81.3 \ g \ mol^{-1}$.

What is the atomic mass of an element with $BCC$ structure and density $10 \ g \ cm^{-3}$ having an edge length of $300 \ pm$?

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