A dilute solution contains mole of solute 'A' in 1 kg of a solvent with molal elevation constant . T — Solutions and Colligative Properties Chemistry Question
Question
A dilute solution contains $n$ mole of solute 'A' in 1 kg of a solvent with molal elevation constant $K_b$. The solute A undergoes dimerization as $2\text{A} \rightleftharpoons \text{A}_2$. If $\Delta T_b$ is elevation in boiling point of given solution and molality = molarity, the equilibrium constant, $K_c$, for dimer formation is
💡 Solution & Explanation
Initial moles of A = $n$. At equilibrium, $[\text{A}] = n(1-\alpha)$ and $[\text{A}_2] = n\alpha/2$. Total moles $= n(1-\alpha/2)$. Observed $\Delta T_b = K_b \times n(1-\alpha/2)$. Thus, $1-\alpha/2 = \Delta T_b / (nK_b)$, which yields $\alpha = 2(nK_b - \Delta T_b)/(nK_b)$. The equilibrium constant $K_c = [\text{A}_2]/[\text{A}]^2 = (n\alpha/2) / [n(1-\alpha)]^2$. Substituting $\alpha$ into the expression simplifies to $K_c = \frac{K_b(n \cdot K_b - \Delta T_b)}{(2\Delta T_b - n \cdot K_b)^2}$. Therefore, correct answer is B.