Ice crystallizes in a hexagonal lattice. At the low temperatures at which the structure was determin — Solid State Chemistry Question
Question
Ice crystallizes in a hexagonal lattice. At the low temperatures at which the structure was determined, the lattice constants were $a = 4.53 \text{ \AA}$ and $b = 7.41 \text{ \AA}$ as shown in the figure. How many $\text{H}_2\text{O}$ molecules are contained in a unit cell? (Given that density of ice = $0.92 \text{ g/cm}^3$)
💡 Solution & Explanation
For a hexagonal unit cell, volume $V = \frac{\sqrt{3}}{2} a^2 c$. Using $a = 4.53 \times 10^{-8} \text{ cm}$ and $c = 7.41 \times 10^{-8} \text{ cm}$, $V = \frac{1.732}{2} \times (4.53 \times 10^{-8})^2 \times 7.41 \times 10^{-8} \approx 1.316 \times 10^{-22} \text{ cm}^3$. Mass of unit cell $= d \times V = 0.92 \times 1.316 \times 10^{-22} = 1.21 \times 10^{-22} \text{ g}$. Molecules $Z = \frac{\text{Mass} \times N_A}{M} = \frac{1.21 \times 10^{-22} \times 6.022 \times 10^{23}}{18} \approx 4.05 \approx 4$. Therefore, correct answer is 4.