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An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is $7.2 g/cm^3$. How many atoms are present in 208 g of the element?

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1 Answer

Volume of unit cell = $(288×10^{-10})^3 cm^3 = 2.39×10^{−23}cm^3$
Volume of 208 g of the element = $\large\frac{\text{mass}}{\text{volume}}=\frac{208}{7.2}$$=28.88cm^3$
No of unit cells in this volume = $\large\frac{28.88}{2.39\times 10^{-23}}$$=12.08\times 10^{23}$
Since each bcc unit cell contains 2 atoms
$\therefore$ No of atom in 208g=$2\times 12.08\times 10^{23}=24.16\times 10^{23}$ atom
Hence (A) is the correct answer.
answered Jun 2, 2014 by sreemathi.v
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