Graphite has h.c.p arrangements of carbon atoms and the parallel planes are 3.55 $\dot{A}$ apart. Determine density of graphite :

In hcp alrangement total height

$h = 4\times \sqrt{\frac{2}{3}}r$

as per given 

$\frac{h}{2}=2\sqrt{\frac{2}{3}}r=3.35 \dot{A} = 3.35\times {10}^{-8}cm.$

Volume of unit cell = 24 $\sqrt{2}{r}^3$

= $24\sqrt{2}{\left(\sqrt{\frac{3}{2}\times \frac{1}{2}\times 3.35\times {10}^{-8}}\right)}^3$

$= 2.93\times {10}^{-22}c{m}^3$

Effective no. of atoms in hexagonal unit cell = 6

$\because density = \frac{6\times 12}{6.023\times {10}^{23}\times 2.93\times {10}^{-22}}= 0.41 g/cc$