Question

Regarding graphite the following information is available: The density of graphite $=2.25g/c{m}^3$. What is $C-C$ bond distance in graphite?

• A. $1.68\mathring{A}$
• B. $1.545\mathring{A}$
• C. $2.852\mathring{A}$
• D. $1.426\mathring{A}$

Answer 1

Answer: (D)

Let $C-C$ bond distance in graphite is $d\mathring{A}$
$\therefore$ Surface area $=6\times \frac{\sqrt{3}}{4}\times d^2=6\times \frac{\sqrt{3}}{4}\times d^2\times 10^{-16}c{m}^2$
$\therefore$ Vol. of the unit cell $=6\times \frac{\sqrt{3}}{4}\times d^2\times 10^{-16}\times 3.35\times 10^{-8}c{m}^3$
$\therefore$ Mass of the unit cell $=6\times \frac{\sqrt{3}}{4}\times 3.35\times d^2\times 2.25\times 10^{-24}g.$
$\therefore 6\times \frac{\sqrt{3}}{4}\times 3.35\times 2.25\times 10^{-24}\times d^2=12\times \frac{10}{6}\times 10^{-24}\times 2$
$\therefore d^2=\frac{10}{3}\times 4\frac{1\times 2}{\sqrt{3}\times 3.35\times 2.25}=2.04265$ or $d=1.429\mathring{A}$