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Chapter 8: Problem 19

The "Tip of the Week" in a local newspaper suggested using a fertilizer such as ammonium nitrate or ammonium sulfate instead of salt to melt snow and ice on sidewalks, because salt can damage lawns. Which of the following compounds would give the largest freezing point depression when \(100 \mathrm{~g}\) is dissolved in \(1 \mathrm{~kg}\) of water? (a) \(\mathrm{NaCl}\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(NH鈧�)鈧係O鈧� will give the largest freezing point depression.

Step by step solution

01

Understand Freezing Point Depression

The freezing point depression occurs when a solute is added to a solvent, reducing the temperature at which the solvent freezes. It is given by \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of ions in solution), \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.
02

Calculate the Molality for Each Compound

The molality \(m\) is defined as the moles of solute per kilogram of solvent. For each compound, we will assume that its entire mass is dissolved in \(1\) kg of water. Thus, \[m = \frac{\text{mass of solute in grams}}{\text{molar mass of solute in g/mol}}.\]For all compounds, the mass of solute is \(100\) g.
03

Determine the van't Hoff Factor

The van't Hoff factor \(i\) represents the number of particles the compound dissociates into in solution. Assess this for each compound:- NaCl dissociates into 2 ions: Na鈦� and Cl鈦�, so \(i = 2\).- NH鈧凬O鈧� dissociates into 2 ions: NH鈧勨伜 and NO鈧冣伝, so \(i = 2\).- (NH鈧�)鈧係O鈧� dissociates into 3 ions: 2 NH鈧勨伜 and SO鈧劼测伝, so \(i = 3\).
04

Calculate Freezing Point Depression for Each

Assume the same \(K_f\) for simplicity as it does not change the relative comparison.- For NaCl:\[\Delta T_f = i \cdot m = 2 \cdot m\]- For NH鈧凬O鈧�:\[\Delta T_f = i \cdot m = 2 \cdot m\]- For (NH鈧�)鈧係O鈧�:\[\Delta T_f = i \cdot m = 3 \cdot m\]The larger the product \(i \cdot m\), the larger the freezing point depression.
05

Conclusion

Since the van't Hoff factor \(i\) for (NH鈧�)鈧係O鈧� is 3, compared to 2 for the other compounds, it will result in the largest freezing point depression among the options provided.

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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 is a crucial concept in understanding how solutes affect the freezing point of a solution. It represents the number of particles that the solute dissociates into when dissolved in a solvent.

For ionic compounds like NaCl, NH鈧凬O鈧�, and (NH鈧�)鈧係O鈧�, this factor describes how many ions are formed when they dissolve in water.
  • NaCl dissociates into 2 ions: Na鈦� and Cl鈦�. Hence, it has a van't Hoff factor of 2.
  • NH鈧凬O鈧� dissociates into 2 ions: NH鈧勨伜 and NO鈧冣伝, resulting in a van't Hoff factor of 2.
  • (NH鈧�)鈧係O鈧� breaks into 3 ions: 2 NH鈧勨伜 and 1 SO鈧劼测伝, making its van't Hoff factor 3.
The significance of the van't Hoff factor is that it allows us to calculate the effect a solute has on the freezing point depression of the solution. The higher the factor, the more significant the impact.
molality
Molality is a measure that represents the concentration of a solution in terms of the amount of solute per kilogram of solvent. It is crucial for calculating changes in physical properties that depend on temperature, such as boiling point elevation and freezing point depression.

To determine molality, use the formula:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]Molality is particularly useful because it does not change with temperature, unlike molarity, which is dependent on volume.
  • In our exercise, 100 grams of each compound are dissolved in 1 kg of water.
  • We first convert the mass of each solute to moles using its molar mass.
  • Then, divide by the mass of the water (1 kg) to find the molality.
Molality is used in tandem with the van't Hoff factor to calculate the freezing point depression.
ion dissociation
Ion dissociation is the process through which ionic compounds dissolve in water, separating into individual ions. This process is vital for understanding how the addition of a solute can affect a solution's colligative properties, including freezing point depression.

When an ionic compound like NaCl is added to water, it dissociates into its constituent ions, Na鈦� and Cl鈦�. This dissociation increases the number of particles in the solution:
  • NaCl yields 2 separate ions.
  • NH鈧凬O鈧� also yields 2 ions: NH鈧勨伜 and NO鈧冣伝.
  • (NH鈧�)鈧係O鈧� dissociates further into 3 ions: 2 NH鈧勨伜 ions and 1 SO鈧劼测伝 ion.
The dissociation process and the resulting increase in the number of particles directly contribute to properties like freezing point depression, as more particles interfere with the formation of a solid structure.
solubility effects
Solubility effects refer to how the solubility of a compound in a solvent affects its ability to cause phenomena like freezing point depression. The degree to which a solute dissolves can determine how efficient it is in influencing the physical properties of the solvent.

In the context of freezing point depression, a more soluble compound can introduce more ions into a solution. This increase in the number of particles can enhance the solution's freezing point lowering effect.
  • Highly soluble compounds tend to exhibit significant freezing point depression because they dissolve completely, releasing more ions into the solution.
  • Even if a compound has a high van't Hoff factor, if it is not very soluble, it might not lower the freezing point effectively.
Understanding solubility and its impact on freezing point depression helps in choosing the most effective solute for applications like melting ice without damaging surrounding plant life.

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Most popular questions from this chapter

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