A diatomic molecule X_2 has a body-centred cu | vedclass

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(C) The density formula is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z = 2$ for a $bcc$ structure.Given $ho = 6.17 \ g \ cm^{-3}$, $a = 300

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$4$

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(C) The density formula is $ho = \frac{Z 	imes M}{N_A 	imes a^3}$, where $Z = 2$ for a $bcc$ structure.Given $ho = 6.17 \ g \ cm^{-3}$, $a = 300 \ pm = 300 	imes 10^{-10} \ cm$, and $N_A = 6 	imes 10^{23} \ mol^{-1}$.Substituting the values: $6.17 = \frac{2 	imes M}{6 	imes 10^{23} 	imes (300 	imes 10^{-10})^3}$.$6.17 = \frac{2 	imes M}{6 	imes 10^{23} 	imes 27 	imes 10^{-24}} = \frac{2 	imes M}{16.2}$.$M = \frac{6.17 	imes 16.2}{2} \approx 50 \ g \ mol^{-1}$.Number of molecules $= \frac{	ext{mass}}{M} 	imes N_A = \frac{200 \ g}{50 \ g \ mol^{-1}} 	imes N_A = 4 \ N_A$.

$A$ diatomic molecule $X_2$ has a body-centred cubic (bcc) structure with a cell edge of $300 \ pm$. The density of the molecule is $6.17 \ g \ cm^{-3}$. The number of molecules present in $200 \ g$ of $X_2$ is (Avogadro constant $N_A = 6 \times 10^{23} \ mol^{-1}$) (in $N_A$)

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