To one litre of 0.1 M-HCl solution, 0.025 mole of solid is added. Assuming complete dissociation of — Solutions and Colligative Properties Chemistry Question
Question
To one litre of 0.1 M-HCl solution, 0.025 mole of solid $\text{NH}_4\text{Cl}$ is added. Assuming complete dissociation of solutes, the freezing point of solution is ($K_f$ of water = $1.86 \text{ K-kg mol}^{-1}$)
💡 Solution & Explanation
1 L of 0.1 M HCl contains 0.1 mol of HCl, which dissociates into 0.2 mol of ions ($0.1 \text{ H}^+ + 0.1 \text{ Cl}^-$). Adding 0.025 mol of solid $\text{NH}_4\text{Cl}$ yields 0.05 mol of ions ($0.025 \text{ NH}_4^+ + 0.025 \text{ Cl}^-$). Total moles of particles = $0.2 + 0.05 = 0.25 \text{ mol}$. Assuming 1 L of solution has 1 kg of water, molality $m \approx 0.25 \text{ m}$. Depression $\Delta T_f = K_f \times m = 1.86 \times 0.25 = 0.465^\circ\text{C}$. Freezing point = $-0.465^\circ\text{C}$. Therefore, correct answer is A.