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Both \(\mathrm{NaCl}\) and \(\mathrm{CaCl}_{2}\) are used to melt ice on roads and sidewalks in winter. What advantages do these substances have over sucrose or urea in lowering the freezing point of water?

Short Answer

Expert verified
NaCl and CaCl_2 dissociate into more particles, providing a greater freezing point depression compared to sucrose and urea.

Step by step solution

01

Understanding Freezing Point Depression

Freezing point depression is a colligative property, which means its effect depends on the number of solute particles in a solution, not the identity of the solute. When a solute is added to a solvent, it disrupts the crystallization process, lowering the freezing point.
02

Determine Ionic vs. Molecular Compounds

Sodium chloride (NaCl) and calcium chloride (CaCl_2) are ionic compounds, which dissociate into multiple ions in solution. NaCl dissociates into two ions (Na⁺ and Cl⁻), whereas CaCl_2 dissociates into three ions (Ca²⁺ and two Cl⁻). In contrast, sucrose and urea are molecular compounds that do not dissociate into ions.
03

Calculate Van't Hoff Factor

The Van't Hoff factor (i) represents the number of particles the solute dissociates into. For NaCl, i = 2, while for CaCl_2, i = 3. Sucrose and urea have i = 1 as they don't dissociate into ions.
04

Impact on Freezing Point Depression

The effectiveness of a solute in lowering freezing point is proportional to its Van't Hoff factor. Since NaCl and CaCl_2 dissociate into more ions, they have a greater impact on lowering the freezing point than sucrose or urea, which remain as single particles.
05

Identify Practical Advantages

NaCl and CaCl_2 are more effective in melting ice due to their higher ability to lower the freezing point. Additionally, they are generally more cost-effective and easier to spread and handle compared to organic compounds like sucrose.

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Key Concepts

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

Colligative Properties
Colligative properties are fascinating characteristics of solutions that depend entirely on the number of solute particles present, regardless of what those particles are. One key aspect of these properties is that they result in changes to the physical behavior of the solvent.
  • Freezing point depression
  • Boiling point elevation
  • Vapour pressure lowering
  • Osmotic pressure
In the context of lowering the freezing point, when solutes are added to a solvent, they interfere with the solvent's usual freezing process. This is why salt is used on roads during winter. The salt solutions formed lower the freezing point of water, helping to melt ice. Since this property is colligative, it's the number of particles the solute breaks into that matters, not the type of solute. Therefore, substances like \(\text{NaCl}\text{ and } \text{CaCl}_{2}\) are more effective, as they dissociate into more particles.Understanding colligative properties is critical for predicting how solute-solvent mixtures will behave in different conditions.
Van't Hoff Factor
The Van't Hoff factor \( (i) \) is an important concept in chemistry, especially when discussing colligative properties such as freezing point depression. It indicates the number of particles a solute splits into when dissolved.
  • For example, \(\text{NaCl}\) splits into two particles: \( \text{Na}^+ \) and \( \text{Cl}^- \).
  • \(\text{CaCl}_{2}\) splits into three particles: \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \) ions.
The larger the Van't Hoff factor, the more particles are present in solution, enhancing the colligative effects. This means substances like \(\text{CaCl}_{2}\) with a Van't Hoff factor of 3 will lower the freezing point more than \(\text{NaCl}\) with a Van't Hoff factor of 2.Because the freezing point depression depends on the number of particles, the Van't Hoff factor provides a measuring stick to estimate the effect on solutions. The higher the \( i \)-value, the greater the degree of freezing point depression, making ionic compounds highly effective in applications like de-icing roads.
Ionic vs. Molecular Compounds
The distinction between ionic and molecular compounds is crucial in understanding why some substances are more effective in freezing point depression.
  • Ionic Compounds: These consist of metals combined with non-metals and dissociate into ions in a solution. For instance, \(\text{NaCl}\) breaks into \( \text{Na}^+ \) and \( \text{Cl}^- \) ions, while \(\text{CaCl}_{2}\) dissociates into \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \) ions.
  • Molecular Compounds: Composed of covalently bonded non-metal molecules, these do not break into ions in solution. A good example is sucrose, which stays intact as a whole molecule when dissolved in water.
Ionic compounds are more effective for freezing point depression due to their dissociation. Breaking down into multiple ions provides more solute particles in the solution compared to one whole molecular entity. So, the use of ionic compounds like \(\text{NaCl}\) and \(\text{CaCl}_{2}\) translates into a greater effect on lowering the freezing point of water, making them preferred for winter ice melting.

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Most popular questions from this chapter

Hemoglobin, the oxygen-transport protein, binds about \(1.35 \mathrm{~mL}\) of oxygen per gram of the protein. The concentration of hemoglobin in normal blood is \(150 \mathrm{~g} / \mathrm{L}\) blood. Hemoglobin is about 95 percent saturated with \(\mathrm{O}_{2}\) in the lungs and only 74 percent saturated with \(\mathrm{O}_{2}\) in the capillaries. Calculate the volume of \(\mathrm{O}_{2}\) released by hemoglobin when \(100 \mathrm{~mL}\) of blood flows from the lungs to the capillaries.

The osmotic pressure of \(0.010-M\) solutions of \(\mathrm{CaCl}_{2}\) and urea at \(25^{\circ} \mathrm{C}\) are 0.605 and 0.245 atm, respectively. Calculate the van't Hoff factor for the \(\mathrm{CaCl}_{2}\) solution.

What are ion pairs? What effect does ion-pair formation have on the colligative properties of a solution? How does the ease of ion-pair formation depend on (a) charges on the ions, \((\mathrm{b})\) size of the ions, \((\mathrm{c})\) nature of the solvent (polar versus nonpolar), (d) concentration?

Calculate the molalities of the following aqueous solutions: (a) \(1.22 M\) sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution (b) \(0.87 \mathrm{M} \mathrm{NaOH}\) (density of solution \(=1.12 \mathrm{~g} / \mathrm{mL}\) ), solution (density of solution \(=1.04 \mathrm{~g} / \mathrm{mL}),\) (c) \(5.24 \mathrm{M}\) \(\mathrm{NaHCO}_{3}\) solution (density of solution \(=1.19 \mathrm{~g} / \mathrm{mL}\) ).

Arrange the following solutions in order of decreasing freezing point: \(0.10 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.35 \mathrm{~m} \mathrm{NaCl}, 0.20 \mathrm{~m}\) \(\mathrm{MgCl}_{2}, 0.15 \mathrm{~m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}, 0.15 \mathrm{~m} \mathrm{CH}_{3} \mathrm{COOH}\)

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