CsBr has CsCl type structure. Its density is 4.26 g/ml. The side of unit cell is (Cs = 133, Br = 80) — Solid State Chemistry Question

💡 Solution & Explanation

For CsCl type, $Z=1$. Molar mass $M = 133 + 80 = 213$ g/mol. Density $d = \frac{Z \times M}{N_A \times a^3} \implies 4.26 = \frac{1 \times 213}{6.022 \times 10^{23} \times a^3}$. Solving for $a^3$ gives approximately $83 \times 10^{-24}$ cm$^3$, so $a \approx 4.368$ Å. Therefore, correct answer is B.