An element (atomic mass 100 g/mol) having bc | vedclass

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Question

(B) For a $bcc$ structure, the number of atoms per unit cell $(n)$ is $2$.Given: Atomic mass $(M)$ = $100 \ g/mol$, edge length $(a)$ = $400 \ pm = 40

Vedclass · Accepted Answer

$5.188$

Vedclass · Answer

(B) For a $bcc$ structure, the number of atoms per unit cell $(n)$ is $2$.Given: Atomic mass $(M)$ = $100 \ g/mol$, edge length $(a)$ = $400 \ pm = 400 	imes 10^{-10} \ cm$, Avogadro's number $(N_A)$ = $6.022 	imes 10^{23} \ mol^{-1}$.The formula for density $(ho)$ is $ho = \frac{n 	imes M}{a^3 	imes N_A}$.Substituting the values: $ho = \frac{2 	imes 100}{(400 	imes 10^{-10})^3 	imes 6.022 	imes 10^{23}}$.$ho = \frac{200}{64 	imes 10^{-24} 	imes 6.022 	imes 10^{23}} = \frac{200}{38.54} \approx 5.189 \ g/cm^3$.Thus, the correct option is $(B)$.

An element (atomic mass $100 \ g/mol$) having $bcc$ structure has unit cell edge $400 \ pm$. Then density of the element is (in $g/cm^3$)

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