An element M crystallises in a body-centred c | vedclass

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(D) For a body-centred cubic $(BCC)$ unit cell,the number of atoms per unit cell is $Z = 2$.The density formula is given by $ho = \frac{Z 	imes M_{

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$22$

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(D) For a body-centred cubic $(BCC)$ unit cell,the number of atoms per unit cell is $Z = 2$.The density formula is given by $ho = \frac{Z 	imes M_{atomic}}{N_A 	imes a^3}$.Given: $ho = 6.0 \, g \, cm^{-3}$,$a = 300 \, pm = 300 	imes 10^{-10} \, cm = 3 	imes 10^{-8} \, cm$.$6.0 = \frac{2 	imes M_{atomic}}{6.022 	imes 10^{23} 	imes (3 	imes 10^{-8})^3}$.$6.0 = \frac{2 	imes M_{atomic}}{6.022 	imes 10^{23} 	imes 27 	imes 10^{-24}} = \frac{2 	imes M_{atomic}}{16.2594}$.$M_{atomic} = \frac{6.0 	imes 16.2594}{2} = 48.7782 \, g \, mol^{-1}$.Number of moles in $180 \, g = \frac{180}{48.7782} \approx 3.6902 \, mol$.Number of atoms $= 	ext{moles} 	imes N_A = 3.6902 	imes 6.022 	imes 10^{23} \approx 22.22 	imes 10^{23}$.The nearest integer is $22$.

An element $M$ crystallises in a body-centred cubic unit cell with a cell edge of $300 \, pm$. The density of the element is $6.0 \, g \, cm^{-3}$. The number of atoms present in $180 \, g$ of the element is $............ \times 10^{23}$ (Nearest integer).

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Fill in the blanks:
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$2.$ Mass of atoms present in unit cell $(m) = ........$

$NaCl$ has $4$ formula units per unit cell. If the edge length of the unit cell is $0.564 \, nm$,then the density of the crystal is ............ $\text{g/cm}^3$. $(NaCl = 58.5 \, \text{g/mol})$

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