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Calculate the volume of unit cell of element with density $7.2 \mathrm{~g} \mathrm{~cm}^{-3}$ that forms bcc structure. (288 gram of this element $3.35 \times 10^{24}$ atoms)
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The correct answer is:
$4.18 \times 10^{23} \mathrm{~cm}^3$
$\mathrm{d}=\frac{\mathrm{Z}}{\mathrm{V}} \times \frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}$
$7.2=\frac{2}{V} \times \frac{288}{3.35 \times 10^{24}}$
$\mathrm{V}=4.18 \times 10^{22} \mathrm{~cm}^3$
$7.2=\frac{2}{V} \times \frac{288}{3.35 \times 10^{24}}$
$\mathrm{V}=4.18 \times 10^{22} \mathrm{~cm}^3$
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