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Chapter 11: Problem 94

The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CsCl}\) is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-\mathrm{m}\) 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
The greater ion association appears to be in the CsCl solution. This is because the experimental Van't Hoff factor (i) for CsCl is lower than that for CaCl₂, indicating a higher degree of ion association in the CsCl solution. The experimental Van't Hoff factors were calculated using the formula: i = (observed freezing-point depression) / (theoretical freezing-point depression), and the values found were 0.320°C / 2 for CsCl and 0.440°C / 3 for CaCl₂. Since the experimental Van't Hoff factor for the CsCl solution is lower, it has greater ion association.

Step by step solution

01

Calculate the observed freezing-point depression for each solution

We are given the values for the observed freezing-point depression for both solutions: ΔTf_CsCl = 0.320°C ΔTf_CaCl₂ = 0.440°C
02

Calculate the theoretical freezing-point depression for each solution

For CsCl, there is one cation (Cs⁺) and one anion (Cl⁻) per formula unit, so the total number of ions is 2. Therefore, the theoretical van't Hoff factor for CsCl is i_CsCl = 2. For CaCl₂, there is one cation (Ca²⁺) and two anions (2 Cl⁻) per formula unit, so the total number of ions is 3. Therefore, the theoretical van't Hoff factor for CaCl₂ is i_CaCl₂ = 3. Now, we can use the formula ΔTf = Kf * molality * i to calculate the theoretical freezing-point depression for each solution: ΔTf_CsCl = Kf * (0.091 m) * i_CsCl ΔTf_CaCl₂ = Kf * (0.091 m) * i_CaCl₂
03

Calculate the experimental Van't Hoff factors for each solution

We can now calculate the experimental Van't Hoff factors for each solution using the formula mentioned in the analysis: i_CsCl = (observed freezing-point depression) / (theoretical freezing-point depression) i_CsCl = 0.320°C / (Kf * (0.091 m) * 2) i_CaCl₂ = 0.440°C / (Kf * (0.091 m) * 3)
04

Compare the experimental van't Hoff factors to determine the ion association

The solution with the lower experimental Van't Hoff factor will have a greater degree of ion association since the expected freezing-point depression will not be achieved. We can see that i_CsCl and i_CaCl₂ are both divided by the same Kf and molality (0.091 m), so by comparing their observed freezing-point depression values (0.320°C for CsCl and 0.440°C for CaCl₂), we can conclude that the solution with the greater ion association is the one with the lower observed freezing-point depression: 0.320°C / 2 < 0.440°C / 3 Since 0.320°C / 2 (CsCl) is less than 0.440°C / 3 (CaCl₂), the greater ion association is observed in the CsCl solution.

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Most popular questions from this chapter

a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{lll}{\mathrm{CaCl}_{2}(s)} & {-2247 \mathrm{k} / \mathrm{mol}} & {-46 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Cal}_{2}(s)} & {-2059 \mathrm{k} / \mathrm{mol}} & {-104 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-},\) is more strongly attracted to water?

Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

A solution at \(50^{\circ} \mathrm{C}\) containing 2.0 \(\mathrm{mol}\) of liquid \(\mathrm{A}\) and 3.0 \(\mathrm{mol}\) of liquid \(\mathrm{B}\) has a total vapor pressure of \(240 .\) torr. If pure A has a vapor pressure of 150 . torr at \(50^{\circ} \mathrm{C},\) what is the vapor pres- sure of pure \(\mathrm{B}\) at \(50^{\circ} \mathrm{C} ?\)

The term proof is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is 95\(\%\) (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92 proof ethanol-water solution? Assume the density of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) , is 0.79 \(\mathrm{g} / \mathrm{cm}^{3}\) and the density of water is 1.0 $\mathrm{g} / \mathrm{cm}^{3}$ .

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C} )\) b. a solution of glucose in water with $\chi_{\mathrm{CH}_{2} \mathrm{O}_{6}}=0.01$ c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\) (Consider the vapor pressure of both methanol \([143 \text { torr at }\) \(25^{\circ} \mathrm{C} ]\) and water.
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