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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 CaCl鈧, 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掳C / 2 for CsCl and 0.440掳C / 3 for CaCl鈧. 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: 螖Tf_CsCl = 0.320掳C 螖Tf_CaCl鈧 = 0.440掳C
02

Calculate the theoretical freezing-point depression for each solution

For CsCl, there is one cation (Cs鈦) and one anion (Cl鈦) 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 CaCl鈧, there is one cation (Ca虏鈦) and two anions (2 Cl鈦) per formula unit, so the total number of ions is 3. Therefore, the theoretical van't Hoff factor for CaCl鈧 is i_CaCl鈧 = 3. Now, we can use the formula 螖Tf = Kf * molality * i to calculate the theoretical freezing-point depression for each solution: 螖Tf_CsCl = Kf * (0.091 m) * i_CsCl 螖Tf_CaCl鈧 = Kf * (0.091 m) * i_CaCl鈧
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掳C / (Kf * (0.091 m) * 2) i_CaCl鈧 = 0.440掳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_CaCl鈧 are both divided by the same Kf and molality (0.091 m), so by comparing their observed freezing-point depression values (0.320掳C for CsCl and 0.440掳C for CaCl鈧), we can conclude that the solution with the greater ion association is the one with the lower observed freezing-point depression: 0.320掳C / 2 < 0.440掳C / 3 Since 0.320掳C / 2 (CsCl) is less than 0.440掳C / 3 (CaCl鈧), the greater ion association is observed in the CsCl solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

van't Hoff factor
The van't Hoff factor, often denoted as "i," is a critical concept in understanding colligative properties such as freezing-point depression. It's a measure of the degree of ionization or dissociation a solute undergoes in solution. When a solute like an ionic compound dissolves in a solvent, it typically separates into its respective ions.

For instance, the compound CsCl splits into Cs鈦 and Cl鈦 ions. Therefore, in the case of CsCl, the expected van't Hoff factor would be 2, indicating each formula unit breaks into two particles. Similarly, CaCl鈧 dissociates into one Ca虏鈦 and two Cl鈦 ions, resulting in a theoretical van't Hoff factor of 3.

It's important to note that the van't Hoff factor alters the freezing-point depression calculation by multiplying the original molality with the factor. Thus, a higher van't Hoff factor indicates more particles in the solution, which leads to a greater change in the freezing point. However, ion association can complicate this prediction, as we'll explore further.
ion association
Ion association is when ions in a solution are not completely dissociated but instead, some of them remain paired. This affects the effective number of particles contributing to colligative properties like freezing-point depression.

When you calculate the theoretical van't Hoff factor based on complete dissociation, you assume all ionic bonds break. However, in reality, some ions may still interact, reducing the number of distinct species in the solution. This creates a scenario where the experimental van't Hoff factor is less than expected.
  • Higher ion association implies fewer free particles in solution, leading to a lower experimental van't Hoff factor.
  • This results in a smaller observed effect on freezing-point depression compared to theoretical predictions.
Thus, a significant ion association indicates strong interactions between ions, resulting in a less effective contribution to changing the solution's freezing point.
molality
Molality is a concentration term used in chemistry particularly involving colligative properties. It is defined as the number of moles of solute per kilogram of solvent, and is represented as "m."

Molality plays a crucial role because it is independent of temperature and volume changes, making it a preferred measurement in colligative property calculations. In the context of the freezing-point depression, the change in temperature, \( \Delta T_f \), is directly proportional to the molality of the solution. The formula used is:\[\Delta T_f = K_f \times \text{molality} \times i\]

Here, \( K_f \) is the cryoscopic constant (a property of the solvent), and \( i \) is the van't Hoff factor. Thus, by knowing a solution's molality, you can accurately predict its expected change in freezing point, provided the ideal dissociation scenario (with the van't Hoff factor) holds true.

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Most popular questions from this chapter

A forensic chemist is given a white solid that is suspected of being pure cocaine \(\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}, \text { molar mass }=303.35 \mathrm{g} / \mathrm{mol}\right)\) She dissolves \(1.22 \pm 0.01 \mathrm{g}\) of the solid in \(15.60 \pm 0.01 \mathrm{g}\) benzene. The freezing point is lowered by \(1.32 \pm 0.04^{\circ} \mathrm{C}\) a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calculate the uncertainty in the molar mass. b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}, \text { molar }\right.\) mass \(=299.36 \mathrm{g} / \mathrm{mol}\) )? c. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?

Which of the following statements is(are) true? Correct the false statements. a. The vapor pressure of a solution is directly related to the mole fraction of solute. b. When a solute is added to water, the water in solution has a lower vapor pressure than that of pure ice at \(0^{\circ} \mathrm{C} .\) c. Colligative properties depend only on the identity of the solute and not on the number of solute particles present. d. When sugar is added to water, the boiling point of the solution increases above \(100^{\circ} \mathrm{C}\) because sugar has a higher boiling point than water.

Consider the following solutions: 0.010\(m \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water 0.020 \(m \mathrm{CaBr}_{2}\) in water 0.020 \(m \mathrm{KCl}\) in water 0.020 \(m \mathrm{HF}\) in water \((\mathrm{HF} \text { is a weak acid.) }\) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as 0.040 \(\mathrm{m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is a nonelectrolyte. b. Which solution would have the highest vapor pressure at \(28^{\circ} \mathrm{C} ?\) c. Which solution would have the largest freezing-point depression?

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 \(\mathrm{g} / \mathrm{cm}^{3}\) .

Liquid A has vapor pressure \(x\) , and liquid \(\mathrm{B}\) has vapor pressure \(y .\) What is the mole fraction of the liquid mixture if the vapor above the solution is \(30 . \% \mathrm{A}\) by moles? \(50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\) ? (Calculate in terms of \(x\) and \(y . )\) Liquid A has vapor pressure \(x,\) liquid \(\mathrm{B}\) has vapor pressure y. What is the mole fraction of the vapor above the solution if the liquid mixture is \(30 . \% \mathrm{A}\) by moles? \(50 . \% \mathrm{A} ? 80 . \% \mathrm{A}\) ? (Calculate in terms of \(x\) and \(y . )\)

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