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Calculate the molar mass of metal having density $9.3 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms simple cubic unit cell. $\left[\mathrm{a}^3 \cdot \mathrm{N}_{\mathrm{A}}=22.6 \mathrm{~cm}^3 \mathrm{~mol}^{-1}\right]$
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The correct answer is:
$210.2 \mathrm{~g} \mathrm{~mol}^{-1}$
For simple cubic unit cell, $\mathrm{n}=1$.
$\text { Density }(\rho)=\frac{M n}{a^3 N_A}$
$\begin{aligned}
& 9.3=\frac{\mathrm{M} \times 1}{22.6} \\
& \mathrm{M}=\frac{9.3 \times 22.6}{1}=210.2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}$
$\text { Density }(\rho)=\frac{M n}{a^3 N_A}$
$\begin{aligned}
& 9.3=\frac{\mathrm{M} \times 1}{22.6} \\
& \mathrm{M}=\frac{9.3 \times 22.6}{1}=210.2 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}$
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