A non-volatile solute 'X' completely dimerizes in water, if the temperature is below and the solute — Solutions and Colligative Properties Chemistry Question
Question
A non-volatile solute 'X' completely dimerizes in water, if the temperature is below $-3.72^\circ\text{C}$ and the solute completely dissociates as $\text{X} \rightarrow \text{Y} + \text{Z}$, if the temperature is above $100.26^\circ\text{C}$. In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of 'X' is dissolved in 1.0 kg water ($K_b = 0.52 \text{ K-kg/mol}, K_f = 1.86 \text{ K-kg/mol}$). Identify the incorrect information related with the solution.
💡 Solution & Explanation
The molality is $m=1$. (a) For freezing, $T$ drops. The solute only dimerizes below $-3.72^\circ\text{C}$. At freezing, if $i=1$, $\Delta T_f = 1.86^\circ\text{C}$. The FP is $-1.86^\circ\text{C}$ (which is above $-3.72$), so it doesn't dimerize. (b) To boil, $T$ rises. If $i=1$, BP is $100.52^\circ\text{C}$, which is $>100.26^\circ\text{C}$. Since $100.52 > 100.26$, the solute completely dissociates before boiling, so $i=2$. $\Delta T_b = 2(0.52)(1) = 1.04^\circ\text{C}$, BP is $101.04^\circ\text{C}$. (c) At $-7.44^\circ\text{C}$, it is below $-3.72^\circ\text{C}$ so it dimerizes ($i=0.5$). $\Delta T_f = 7.44 = 0.5(1.86)/W_{kg} \Rightarrow W_{kg} = 0.125 \text{ kg} = 125 \text{ g}$ water left. Ice separated $= 1000 - 125 = 875 \text{ g}$, which is 87.5%, not 75%. Thus (c) is incorrect. (d) At $102.08^\circ\text{C}$ (above $100.26$), $i=2$. $\Delta T_b = 2.08 = 2(0.52)/W_{kg} \Rightarrow W_{kg} = 0.5 \text{ kg} = 500 \text{ g}$. Vapor escaped = 50%. Therefore, correct answer is C.