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CBSE Class 12-science Answered

An element has a body centred cubic (bbc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are there in 208 g of the element?
Asked by Topperlearning User | 17 Jun, 2016, 10:06: AM
answered-by-expert Expert Answer
Volume = 288 pm = (288 x 10-10 cms)3 = 2.39 x 10 -23 cm 3
 Volume of 208 g of the element = mass/density = 208 g / 7.2 g cm-3 = 28.88 cm3
 Number of unit cells in this volume = 28.88 cm3/2.39 x 10 -23 cm3/unit cell = 12.08 x1023 unit cells.
Since each bcc cubic unit cell contains 2 atoms, therefore, the total number of atoms in 208g               = 2 atoms /unit cell x 12.08 1023 unit cells = 24.16 x 1023atoms
Answered by | 17 Jun, 2016, 12:06: PM

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