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An element has a body centered cubic structure with a unit cell edge length of $400 \mathrm{pm}$. Atomic mass of an element is $24 \mathrm{~g} \mathrm{~mol}^{-1}$. What is the density of the element?
$$
\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23} \mathrm{~mol}^{-1}\right)
$$
Options:
$$
\left(\mathrm{N}_{\mathrm{A}}=6 \times 10^{23} \mathrm{~mol}^{-1}\right)
$$
Solution:
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Verified Answer
The correct answer is:
$1.25 \mathrm{~g} \mathrm{~cm}^{-3}$
Given,
Body centered cubic (bcc) $Z=2$
Edge length $(a)=400 \mathrm{pm}=400 \times 10^{-10}$
$M($ Atomic mass $)=24 \mathrm{~g} \mathrm{~mol}^{-1}$
$\because \quad$ Density $(d)=\frac{Z \times M}{a^3 \times N_A}$
$d=\frac{2 \times 24}{\left(400 \times 10^{-10}\right)^3 \times 6.022 \times 10^{23}}$
$d=125 \mathrm{~g} \mathrm{~cm}^{-3}$
Body centered cubic (bcc) $Z=2$
Edge length $(a)=400 \mathrm{pm}=400 \times 10^{-10}$
$M($ Atomic mass $)=24 \mathrm{~g} \mathrm{~mol}^{-1}$
$\because \quad$ Density $(d)=\frac{Z \times M}{a^3 \times N_A}$
$d=\frac{2 \times 24}{\left(400 \times 10^{-10}\right)^3 \times 6.022 \times 10^{23}}$
$d=125 \mathrm{~g} \mathrm{~cm}^{-3}$
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