A $b c c$ element (atomic mass $65$ ) has a cell edge of $420 p m$ . Calculate its density in $g m s / c m^{3}$ .

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Solution

Density, $ρ = \frac{Z \times M}{a^{3} \times N_{A}}$
given $Z = 2$ (for bcc arrangement)
$M = 65$
$a = 420 p m = 420 \times 10^{- 10} c m$
$N_{A} = 6.023 \times 10^{23}$
$∴ ρ = \frac{2 \times 65}{(420 \times 10^{- 10})^{3} \times 6.023 \times 10^{23}}$

$= \frac{2 \times 65 \times 10^{4}}{(42)^{3} \times 6.023}$

$= 2.91 g c m^{- 3}$
Density of given element is $2.91 g c m^{- 3}$ .

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A bcc element (atomic mass 65) has a cell edge of 420 pm. Calculate its density in gms/cm3.

Density, ρ=Z×Ma3×NAgiven Z=2 (for bcc arrangement)M=65a=420 pm=420×10−10cmNA=6.023×1023∴ρ=2×65(420×10−10)3×6.023×1023 =2×65×104(42)3×6.023 =2.91 g cm−3Density of given element is 2.91 g cm−3.

A $b c c$ element (atomic mass $65$ ) has a cell edge of $420 p m$ . Calculate its density in $g m s / c m^{3}$ .