An amount of 1 mole of a non-volatile solid is dissolved in 200 moles of water. The solution is cool — Solutions and Colligative Properties Chemistry Question
Question
An amount of 1 mole of a non-volatile solid is dissolved in 200 moles of water. The solution is cooled to a temperature 'T' K (lower than the freezing point of the solution) to cause ice formation. After removal of ice, the remaining solution is heated to 373 K, where the vapour pressure of solution is observed to be 740 mm Hg. Identify the correct information(s) ($K_f \text{ of water} = 2.0 \text{ K-kg/mol}$)
💡 Solution & Explanation
At 373 K, pure water $P^\circ = 760 \text{ mm Hg}$. For the final solution, $(760 - 740) / 740 = n_{\text{solute}} / n_{\text{water}} \Rightarrow 20 / 740 = 1 / n_{\text{water}} \Rightarrow n_{\text{water}} = 37$ moles remaining. Ice formed $= 200 - 37 = 163$ moles (option a). Molality at $T$ K is $m = 1 / (37 \times 0.018)$. $\Delta T_f = K_f \times m = 2.0 \times (1 / 0.666) = 2000 / (37 \times 18)$. Thus, $T = 273 - 2000/(37 \times 18)$ (option b). For the original solution, $m = 1 / (200 \times 0.018) = 1/3.6$. $\Delta T_f = 2.0 / 3.6 = 20/36 = 10/18^\circ\text{C}$, so freezing point is $-10/18^\circ\text{C}$ (option c). Relative lowering of final solution is $(760 - 740) / 760 = 20/760 = 1/38$ (option d). All statements are correct. Therefore, correct answer is A,B,C,D.