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A compound has fcc structure. If density of unit cell is $3 \cdot 4 \mathrm{~g} \mathrm{~cm}^{-3}$, what is the edge length of unit cell? (Molar mass $=98.99$ )
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The correct answer is:
$5 .783 A^{\circ}$
(D)
$\begin{array}{l}
\mathrm{n}=4, \rho=3.4 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{M}=98.99, \quad \mathrm{a}=? \\
\mathrm{a}^{3}=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{A}}=\frac{4 \times 98.99}{3.4 \times 6.022 \times 10^{23}} \\
\mathrm{a}^{3}=193.4 \times 10^{-24} \mathrm{~cm}^{3} \\
\therefore \mathrm{a}=5.783 \times 10^{-8} \mathrm{~cm} \\
\therefore \mathrm{a}=5.783 Å
\end{array}$
$\begin{array}{l}
\mathrm{n}=4, \rho=3.4 \mathrm{~g} \mathrm{~cm}^{-3}, \mathrm{M}=98.99, \quad \mathrm{a}=? \\
\mathrm{a}^{3}=\frac{\mathrm{n} \times \mathrm{M}}{\rho \times \mathrm{N}_{A}}=\frac{4 \times 98.99}{3.4 \times 6.022 \times 10^{23}} \\
\mathrm{a}^{3}=193.4 \times 10^{-24} \mathrm{~cm}^{3} \\
\therefore \mathrm{a}=5.783 \times 10^{-8} \mathrm{~cm} \\
\therefore \mathrm{a}=5.783 Å
\end{array}$
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