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(D) Given: Edge length $a = 250 \ pm = 2.5 	imes 10^{-8} \ cm$. Atomic mass $M = 90.3 \ g \ mol^{-1}$. For $fcc$ lattice,the number of atoms per unit

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$38.40$

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(D) Given: Edge length $a = 250 \ pm = 2.5 	imes 10^{-8} \ cm$. Atomic mass $M = 90.3 \ g \ mol^{-1}$. For $fcc$ lattice,the number of atoms per unit cell $n = 4$. Avogadro's number $N_{A} = 6.022 	imes 10^{23} \ mol^{-1}$. The density formula is $ho = \frac{n 	imes M}{a^{3} 	imes N_{A}}$. Substituting the values: $ho = \frac{4 	imes 90.3}{(2.5 	imes 10^{-8})^{3} 	imes 6.022 	imes 10^{23}}$. $ho = \frac{361.2}{15.625 	imes 10^{-24} 	imes 6.022 	imes 10^{23}} = \frac{361.2}{9.409} \approx 38.40 \ g \ cm^{-3}$.

An element crystallises in a $fcc$ lattice with cell edge $250 \ pm$. Calculate the density of the element (atomic mass $= 90.3 \ g \ mol^{-1}$) (in $g \ cm^{-3}$)

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