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If each edge of a cubic unit cell of an element having atomic mass 120 and density $6.25 \mathrm{~g} \mathrm{cc}^{-1}$ measures $400 \mathrm{pm}$, then the crystal lattice is
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The correct answer is:
body centered
Atomic weight $(m)=120$
$N_A=6.022 \times 10^{23}$
Unit cell $(a)=400 \mathrm{pm}$
$=400 \times 10^{-10} \mathrm{~m}$
As we know
$d=\frac{Z M}{N_A \times a^3}$
$6.25=\frac{Z \times 120}{6.022 \times 10^{23} \times\left(400 \times 10^{-10}\right)^3}$
$Z=2$
So, the crystal lattice is body centered.
$N_A=6.022 \times 10^{23}$
Unit cell $(a)=400 \mathrm{pm}$
$=400 \times 10^{-10} \mathrm{~m}$
As we know
$d=\frac{Z M}{N_A \times a^3}$
$6.25=\frac{Z \times 120}{6.022 \times 10^{23} \times\left(400 \times 10^{-10}\right)^3}$
$Z=2$
So, the crystal lattice is body centered.
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