Question

Regarding graphite the following information is available: The density of graphite $=2.25 \,g / cm ^{3}$. 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 $=6 \times \frac{\sqrt{3}}{4} \times d^{2}=6 \times \frac{\sqrt{3}}{4} \times d^{2} \times 10^{-16} \,cm ^{2}$
$\therefore$ Vol. of the unit cell $=6 \times \frac{\sqrt{3}}{4} \times d^{2} \times 10^{-16} \times 3.35 \times 10^{-8} \,cm ^{3}$
$\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 $d=1.429 \,\mathring{A}$