Question

Graphite has hcp arrangement of carbon atoms and the parallel planes are $3.35$ $\mathring{A}$ apart. The density (in g cm${}^{-3}$) of graphite is :

• A. $2.46$
• B. $2.12$
• C. $1.20$
• D. $1.41$

Answer 1

The correct option is B $2.12$

One unit cell of graphite contains $4$ C atoms. Thus, $Z=4$
The formula mass of graphite is $12$ g/mol.
Avogadro's number is $N_o=6.023\times {10}^{23}$
The edge length is $a=3.35A^o$
The density of graphite $=\frac{Z\times \text{formula mass}}{N_o\times a^3}=\frac{4\times 12}{6.023\times {10}^{23}\times (3.35\times {10}^{-8}{)}^3}=2.12$ g/cc