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Chapter 10: Problem 88

The freezing-point depression of a 0.091-m solution of CsCl is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CaCl}_{2}\) is \(0.440^{\circ} \mathrm{C}\). In which solution does ion association appear to be greater? Explain.

Short Answer

Expert verified
Ion association appears to be greater in the 0.091-m CsCl solution compared to the 0.091-m CaCl2 solution, as the freezing-point depression value is lower for the CsCl solution (0.320°C) than for the CaCl2 solution (0.440°C).

Step by step solution

01

First, compare the freezing-point depression values for each solution: - 0.320°C for 0.091-m CsCl - 0.440°C for 0.091-m CaCl2 Since the molality is the same for both solutions, we can directly compare their freezing-point depression values. #Step 2: Determine which solution has the greater ion association#

The greater ion association corresponds to a lower freezing-point depression value. Given the comparison in Step 1, it's clear that the 0.091-m CsCl solution has a lower freezing-point depression value (0.320°C) than the 0.091-m CaCl2 solution (0.440°C). Thus, ion association appears to be greater in the CsCl solution. In conclusion, ion association appears to be greater in the 0.091-m CsCl solution compared to the 0.091-m CaCl2 solution.

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