At , the vapour pressure of 0.1 molal aqueous solution of urea is 0.03 mm less than that of water an — Solutions and Colligative Properties Chemistry Question
Question
At $20^\circ\text{C}$, the vapour pressure of 0.1 molal aqueous solution of urea is 0.03 mm less than that of water and the vapour pressure of 0.1 molal solution of KCl is 0.0594 mm less than that of water. The apparent percentage dissociation of KCl in water at the given temperature is (Neglect the moles of solute particles in comparison to the moles of water in both solutions.)
💡 Solution & Explanation
The lowering of vapour pressure ($\Delta P$) is directly proportional to the effective molality ($i \times m$) for very dilute solutions. Urea is a non-electrolyte ($i=1$), so $\Delta P_1 \propto 1 \times 0.1$. KCl dissociates, so $\Delta P_2 \propto i \times 0.1$. Taking the ratio: $\Delta P_2 / \Delta P_1 = i / 1 \implies 0.0594 / 0.03 = i \implies i = 1.98$. For KCl, $i = 1 + \alpha \implies 1.98 = 1 + \alpha \implies \alpha = 0.98$. The percentage dissociation is 98%. Therefore, correct answer is 0098.