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Question

Iron has a body-centered cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm3. Use this information to calculate Avogadro's number. (At. Mass of Fe=55.845 amu)

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Solution

Givenbccunitcellno.ofatomsinunitcell=2a=286.65pmd=7.87gcm3atomicmassofFe=56.0μd=Z×Ma3×N0[whered=density,z=no.ofatomsinunitcell,a3=Fevolume,M=atomicmass,N0=avogadrono.]No=Z×Ma3×d=2×56(286.65×1010)×(7.87)=6.022×1023mol1

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