Question

Ice crystallises in a hexagonal lattice. At low temperature at which the structure was determined, the lattice constants were $a=4.53\, \mathring{A}$ and $c=7.41\, \mathring{A}$. How many $H _{2} O$ molecules are contained in a unit cell? $\left(d_{\text {ice }}=0.92\, g\, /\, cm ^{3}\right) ?$

Answer 1

Density of ice $=0.92\, g / cm ^{3}$

Volume of unit cell $=a \times a \sin 60^{\circ} \times c=a^{2} \sin 60^{\circ} \times c$

$=\left(4.53 \times 10^{-8}\right)^{2} \frac{\sqrt{3}}{2} \times 7.41 \times 10^{-8}\, cm ^{3}$

$=131.688 \times 10^{-24}\, cm ^{3}$

$d=\frac{M}{a^{3} N_{0}}$

$\therefore M$ (unit cell) $=d a^{3}\, N_{0}$

Molar mass of $H _{2} O =18\, g\, mol ^{-1}$

Thus, number of $H _{2} O$ molecules in one unit cell

$=\frac{72.97}{18} \approx 4$