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An element has a bcc structure with cell edge of $288 \mathrm{pm}$. The density of element is $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$. What is the atomic mass of an element?
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The correct answer is:
$51.78$
(D)
$\begin{array}{l}
\mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm} \\
\therefore \mathrm{a}^{3}=\left(2.88 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=2.39 \times 10^{-23} \mathrm{~cm}^{3} \\
\rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{n}=2(\text { for bcc cell }) \\
\therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{7.2 \times 2.39 \times 10^{-23} \times 6.022 \times 10^{23}}{2}=51.81 \mathrm{~g}
\end{array}$
$\begin{array}{l}
\mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm} \\
\therefore \mathrm{a}^{3}=\left(2.88 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=2.39 \times 10^{-23} \mathrm{~cm}^{3} \\
\rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{n}=2(\text { for bcc cell }) \\
\therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}=\frac{7.2 \times 2.39 \times 10^{-23} \times 6.022 \times 10^{23}}{2}=51.81 \mathrm{~g}
\end{array}$
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