Discussion Forum



1)

 If $V_{0}$ is the volume of a standard unit cell of germanium for crystal containing $N_{0}$ atoms, then the expression for the mass m of volume V in terms of $V_{0}$, $N_{0}$, $M_{mol}$ and $N_{A}$  is [here, M is the molar mass of germanium and $N_{A}$ is the Avogadro's constant ]


A) M $\frac{V}{V_{0}}\frac{N_{A}}{N_{0}}$

B) $\frac{N_{A}}{N_{0}} \frac{V_{0}}{V}$ M

C) M $\frac{V}{V_{0}}\frac{N_{0}}{N_{A}}$

D) M $\frac{V_{0}}{V_{}}\frac{N_{0}}{N_{A}}$

Answer:

Option C

Explanation:

Number of unit cells in volume V

= Total volume/ Volume of a unit cell = $\frac{V}{V_{0}}$

 Nu,ber of atoms in volume V

 = Number of unit cells x number of atoms in 1 unit cell =$\frac{V}{v_{0}} \times N_{0}$

Number of moles in volume V

     = Number  of atoms/ Avagradro number= $\frac{V}{V_{0}} \times \frac{N_{0}}{N_{A}}$

 Mass n of  given sample volume= Number of moles x Molar mass

 $\Rightarrow$    m=$\frac{V}{V_{0}} \times \frac{N_{0}}{N_{A}} \times M$