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Calculate the volume of unit cell having four particles in it with density $19.0 \mathrm{~g} \mathrm{~cm}^{-3}$ [molar mass of element $=190 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
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The correct answer is:
$6.64 \times 10^{-23} \mathrm{~cm}^3$
$\begin{aligned} d & =\frac{Z \times M}{V \times N_A} \\ 19 & =\frac{4 \times 190}{V \times 6.02 \times 10^{23}} \\ V & =\frac{40}{6.02} \times 10^{-23} \\ & =6.6 \times 10^{-23} \mathrm{~cm}^3\end{aligned}$
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