An element crystallizes in a FCC lattice with cell edge of 400 pm. The density of the element is 7gcm−3. How many atoms are present in 280 g of the element?
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Solution
Volume of the unit cell = (400pm)3 =6.4×107pm3 =6.4×10−23cm3 Volume of 280 g of the element =2807=40cm3 No. of unit cell in this volume =40cm36.2×10−23cm3/unit cell =6.4×1023unit cell Since, each FCC cubic unit cell contains 4 atoms.