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If $V_0$ is the volume of a standard unit cell of germanium crystal containing $N_0$ atoms, then the expression for the mass $m$ of a volume $V$ in terms of $V_0, N_0, M_{\text {mol }}$ and $N_A$ is [here, $M$ is the molar mass of germanium and $N_A$ is the Avogadro's constant]
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The correct answer is:
$M \frac{V}{V_0} \frac{N_0}{N_A}$
Number of unit cells in volume $V$
$=\frac{\text { Total volume }}{\text { Volume of a unit cell }}=\frac{V}{V_0}$
Number of atoms in volume $V$
$=$ Number of unit cells $\times$ Number of atoms in 1 unit cell $=\frac{V}{V_0} \times N_0$
Number of moles in volume $V$
$=\frac{\text { Number of a toms }}{\text { Avagadro number }}=\frac{V}{V_0} \times \frac{N_0}{N_A}$
Mass $n$ of given sample volume
$=$ Number of moles $\times$ Molar mass
$\Rightarrow m=\frac{V}{V_0} \times \frac{N_0}{N_A} \times M$
$=\frac{\text { Total volume }}{\text { Volume of a unit cell }}=\frac{V}{V_0}$
Number of atoms in volume $V$
$=$ Number of unit cells $\times$ Number of atoms in 1 unit cell $=\frac{V}{V_0} \times N_0$
Number of moles in volume $V$
$=\frac{\text { Number of a toms }}{\text { Avagadro number }}=\frac{V}{V_0} \times \frac{N_0}{N_A}$
Mass $n$ of given sample volume
$=$ Number of moles $\times$ Molar mass
$\Rightarrow m=\frac{V}{V_0} \times \frac{N_0}{N_A} \times M$
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