A metallic element crystallises in simple cubic lattice. If edge legth of the unit cell is $3 A^{\circ}$,…
A metallic element crystallises in simple cubic lattice. If edge legth of the unit cell is $3 A^{\circ}$, with density $8 \mathrm{~g} / \mathrm{cc}$, what is the number of unit cells in $100 \mathrm{~g}$ of the metal? $($ Molar mass of metal $=108 \mathrm{~g} / \mathrm{mol}$ )
$1.33 \times 10^{20}$
$2 \times 10^{24}$
$2.7 \times 10^{22}$
$5 \times 10^{23}$
Solution
$(\mathrm{B})$
$\mathrm{a}=3 Å=3 \times 10^{-8} \mathrm{~cm}$
Volume of the unit cell $\left(\mathrm{a}^{3}\right)=\left(3 \times 10^{-8}\right)^{3} \mathrm{~cm}^{3}=27 \times 10^{-24} \mathrm{~cm}^{3}$
Mass of unit cell $=27 \times 10^{-24} \mathrm{~cm}^{3} \times 8 \mathrm{~g} \mathrm{~cm}^{-3}$
$=216 \times 10^{-24} \mathrm{~g}$
$\begin{aligned} 216 \times 10^{-24} \mathrm{~g}=1 \text { unit cell } & \\ \therefore 108 \mathrm{~g}=& \frac{108}{216 \times 10^{-24}}=5 \times 10^{23} \text { unit cells. } \end{aligned}$