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Calculate the density of an element having molar mass $27 \mathrm{~g} \mathrm{~mol}^{-1}$ that forms fcc unit cell. $\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=38.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
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Verified Answer
The correct answer is:
$2.8 \mathrm{~g} \mathrm{~cm}^{-3}$
Density, $\rho=\frac{M \text { n }}{a^3 N_A}$
$$
\begin{aligned}
\rho & =\frac{27 \mathrm{~g} \mathrm{~mol}^{-1} \times 4 \text { atom }}{38.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
& =2.8 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}
$$
$$
\begin{aligned}
\rho & =\frac{27 \mathrm{~g} \mathrm{~mol}^{-1} \times 4 \text { atom }}{38.5 \mathrm{~cm}^3 \mathrm{~mol}^{-1}} \\
& =2.8 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}
$$
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