An element (molar mass 180 g mol^{-1}) has | vedclass

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(C) Given: Molar mass $M = 180 \ g \ mol^{-1}$,Density $ho = 18 \ g \ cm^{-3}$,Avogadro constant $N_A \approx 6.022 	imes 10^{23} \ mol^{-1}$.For a

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$\sqrt[3]{33.2} 	imes 10^{-8} \ cm$

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(C) Given: Molar mass $M = 180 \ g \ mol^{-1}$,Density $ho = 18 \ g \ cm^{-3}$,Avogadro constant $N_A \approx 6.022 	imes 10^{23} \ mol^{-1}$.For a $BCC$ crystal,the number of atoms per unit cell $z = 2$.The formula for density is $ho = \frac{M 	imes z}{a^3 	imes N_A}$.Rearranging for the edge length $a$: $a^3 = \frac{M 	imes z}{ho 	imes N_A}$.Substituting the values: $a^3 = \frac{180 	imes 2}{18 	imes 6.022 	imes 10^{23}} \approx \frac{360}{108.396 	imes 10^{23}} \approx 3.32 	imes 10^{-23} \ cm^3$.To match the format,we write $a^3 = 33.2 	imes 10^{-24} \ cm^3$.Therefore,$a = \sqrt[3]{33.2} 	imes 10^{-8} \ cm$.

An element (molar mass $180 \ g \ mol^{-1}$) has a $BCC$ crystal structure with a density of $18 \ g \ cm^{-3}$. What is the edge length of the unit cell?

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