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Calculate the molar mass of an element having density $7.8 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms bec unit cell. $\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=16.2 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
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The correct answer is:
$63.18 \mathrm{~g} \mathrm{~mol}^{-1}$
For bec unit cell, $\mathrm{n}=2$.
$$
\begin{aligned}
& \text { Density }(\rho)=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\
& 7.8=\frac{\mathrm{M} \times 2}{16.2} \\
& \mathrm{M}=\frac{7.8 \times 16.2}{2}=63.18 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Density }(\rho)=\frac{\mathrm{Mn}}{\mathrm{a}^3 \mathrm{~N}_{\mathrm{A}}} \\
& 7.8=\frac{\mathrm{M} \times 2}{16.2} \\
& \mathrm{M}=\frac{7.8 \times 16.2}{2}=63.18 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
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