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An element with density $2 \cdot 8 \mathrm{~g} \mathrm{~cm}^{-3}$ forms fec unit cell having edge length $4 \times 10^{-8} \mathrm{~cm}$. Calculate molar mass of the element.
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The correct answer is:
$27 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}$
$\rho=2.8 \mathrm{gcm}^{-3}, \quad \mathrm{a}=4 \times 10^{-8} \mathrm{~cm}$,
$\mathrm{n}=4($ for fcc),$\quad \mathrm{M}=?$
$\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}} \quad \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}$
$\therefore M=\frac{2.8 \times\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{4}=26.97 \approx 27 \mathrm{~g} \mathrm{~mol}^{-1}$
$\mathrm{n}=4($ for fcc),$\quad \mathrm{M}=?$
$\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}} \quad \therefore \mathrm{M}=\frac{\rho \times \mathrm{a}^{3} \times \mathrm{N}_{\mathrm{A}}}{\mathrm{n}}$
$\therefore M=\frac{2.8 \times\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{4}=26.97 \approx 27 \mathrm{~g} \mathrm{~mol}^{-1}$
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