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An element crystallises in fcc type of unit cell. The volume of one unit cell is $24.99 \times 10^{-24} \mathrm{~cm}^{3}$ and density of the element $7 \cdot 2 \mathrm{~g} \mathrm{~cm}^{-3} .$ Calculate the number of unit cells in $36 \mathrm{~g}$ of pure sample of element?
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The correct answer is:
$2 \cdot 0 \times 10^{23}$
(C)
Vol. of element $=\frac{\text { Mass }}{\text { Density }}=\frac{36 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=5 \mathrm{~cm}^{3}$
No. of unit cells in $36 \mathrm{~g}$ of pure sample of element $=\frac{\text { Total Vol. of element }}{\text { Vol. of one unit cell }}$
$=\frac{5 \mathrm{~cm}^{3}}{24.99 \times 10^{-24} \mathrm{~cm}^{3}}=2.0 \times 10^{23}$
Vol. of element $=\frac{\text { Mass }}{\text { Density }}=\frac{36 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=5 \mathrm{~cm}^{3}$
No. of unit cells in $36 \mathrm{~g}$ of pure sample of element $=\frac{\text { Total Vol. of element }}{\text { Vol. of one unit cell }}$
$=\frac{5 \mathrm{~cm}^{3}}{24.99 \times 10^{-24} \mathrm{~cm}^{3}}=2.0 \times 10^{23}$
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