Question

If each substance listed here costs the same amount per kilogram, which would be most cost-effective as a way to lower the freezing point of water? (Assume complete dissociation for all ionic compounds.) Explain. a. \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) b. \(\mathrm{NaCl}\) c. KCl d. \(\mathrm{MgCl}_{2}\) e. \(\mathrm{SrCl}_{2}\)

Short answer

MgCl2 and SrCl2 would be the most cost-effective as they produce the most particles upon dissociation, which is 3 ions per formula unit, leading to a greater freezing point depression.

Step-by-step solution

Understand Freezing Point Depression

Freezing point depression is a colligative property, which depends on the number of particles dissolved in a solution, not their identity. When ionic compounds dissolve, they dissociate into individual ions, thus increasing the number of particles in solution. The greater the number of dissolved particles, the greater the freezing point depression.

Determine the Dissociation of Each Substance

For the given substances, determine how many particles they would produce in solution. (a) HOCH2CH2OH is a covalent compound and does not dissociate, (b) NaCl dissociates into 2 ions, (c) KCl dissociates into 2 ions, (d) MgCl2 dissociates into 3 ions, (e) SrCl2 dissociates into 3 ions.

Compare the Number of Particles Produced

Since the amount per kilogram is the same for all substances, the most cost-effective one will be the substance that produces the most particles per kilogram. Between NaCl, KCl, MgCl2, and SrCl2 - MgCl2 and SrCl2 produce the most with 3 particles after dissociation.

Evaluate Other Factors

If the molar mass of the substances is not the same, then the substance that produces more particles per unit mass would be more cost-effective. However, since the cost is per kilogram and all substances are assumed to dissociate completely, molar mass does not affect the cost-effectiveness in this case.

Key concepts

Colligative Properties

Colligative properties are intrinsic physical properties of solutions that depend on the concentration of dissolved particles, rather than the specific type of particle. One of the key colligative properties often examined in chemistry is freezing point depression.

This phenomenon occurs because the presence of solute particles in a solvent interferes with the formation of a solid structure, which effectively lowers the temperature at which the solution will freeze. It's important to note that this property only depends on the total number of solute particles in the solution, not on their individual characteristics. Therefore, in freezing point depression exercises, it's crucial to consider both the number of particles that disperse in the solution and the effect each particle has on the overall freezing point.

Dissociation of Ionic Compounds

Understanding the dissociation of ionic compounds is crucial when discussing freezing point depression. Ionic compounds are made up of positively and negatively charged ions held together by the strong electrostatic forces of attraction in a solid crystal lattice. When dissolved in water, these compounds dissociate into their respective ions.

For example, common table salt, NaCl, separates into two ions - sodium (Na+) and chloride (Cl-). The extent of dissociation is vital because it directly affects the number of particles in the solution and thus the degree of freezing point depression. Hence, if a substance fully dissociates into more ions, it will be more efficient in lowering the freezing point. This knowledge is applied when comparing the cost-effectiveness of various substances used to depress the freezing point of water.

Cost-Effectiveness in Chemistry

When considering cost-effectiveness in chemistry, especially for processes involving colligative properties like freezing point depression, we examine the efficiency and economic impact of different chemicals or compounds. Cost-effectiveness is not just about the price per kilogram; it also involves how the substance’s chemical properties contribute to achieving the desired outcome in the most resource-efficient manner.

For instance, if a substance can lower the freezing point of water by producing a greater number of particles per unit mass, it can be considered more cost-effective for this purpose, assuming equal pricing. In industrial applications, such as de-icing roads or creating anti-freeze solutions, selecting the right substance can significantly impact both performance and budget, making cost-effectiveness a critical aspect to understand in applied chemistry.