Arrange these aqueous solutions in order of decreasing freezing point and explain your reasons: (a) \(0.50 \mathrm{~m}\) \(\mathrm{HCl},\) (b) \(0.50 \mathrm{~m}\) glucose, (c) \(0.50 \mathrm{~m}\) acetic acid.

Short Answer

Expert verified
The arrangement from lowest to highest freezing point is HCl, Acetic acid, Glucose.

Step by step solution

01

Recognizing how solutes affect freezing point

It's essential to understand that for solutions with a similar molality, the one that yields the most particles will have the lowest freezing point. This happens because these particles interfere with the formation of an orderly pattern of particles during crystallization, which is required for the phase transition in freezing.
02

Identifying the particle contribution of each solute

The 0.50 m HCl solution breaks up into two particles because HCl is a strong acid which ionizes completely, a hydrogen ion (H+) and a chloride ion (Cl-). The 0.50 m glucose does not break up and so contributes a single particle. Acetic acid is a weak acid, so only a small portion of it ionizes in water. Thus, it contributes slightly more than one particle, but less than two.
03

Arranging the solutions

A solution with a higher molality will have more particles, therefore delaying the formation of the solid structure and thus having a lower freezing point. So, the arrangement from lowest to highest freezing point is HCl (most particles), Acetic acid (middle) and Glucose (fewest particles).

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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 characteristics of solutions that depend on the number of dissolved particles in solution, but not on the nature of those particles. This is a central aspect of chemistry that impacts boiling points, freezing points, vapor pressures, and osmotic pressures of solutions.

One of the most commonly studied colligative properties is freezing point depression, which is the process where the freezing point of a liquid is lowered by adding another compound to it. Why does this happen? It's because when a solute is added to a solvent, it disrupts the formation of the crystal lattice necessary for freezing. The presence of solute particles makes it harder for the solvent molecules to get into the right position to form a solid, hence a lower temperature is needed to achieve solidification.

Freezing point depression is directly proportional to the molality of the solution and the number of particles the solute provides once dissolved. This is where the concept of van't Hoff factor comes into play – it represents the number of particles a compound splits into in solution. For example, an ionic compound like HCl will dissociate completely in water into its ions, thus it has a higher van't Hoff factor compared to covalent compounds that do not dissociate (like glucose).
Molality
Molality is a measure of the concentration of a solution defined as the number of moles of solute per kilogram of solvent. It is represented with the unit 'm' and it is particularly useful because, unlike molarity, molality doesn't change with temperature since mass remains constant regardless of temperature fluctuations.

When discussing freezing point depression, molality is the preferred unit of concentration because it directly relates to the number of solute particles in the solution, which is essential for determining the extent of the freezing point change. If a solution has a higher molality, that means it contains more solute particles in the same amount of solvent, which generally leads to a greater freezing point depression.

In the exercise mentioned, all solutions have the same molality (\(0.50 \text{ m}\)). However, the freezing points differ because of their disposition to release different numbers of particles into the solution, which in turn affects the freezing point depression experienced by each solution.
Ionic and Covalent Solutes
Understanding the behavior of ionic and covalent solutes in solutions is vital when examining their effect on colligative properties. Ionic compounds, such as HCl, dissociate into ions when dissolved in water. This complete dissociation increases the amount of particles in solution, greatly affecting the freezing point when compared to covalent compounds.

Covalent compounds, like glucose and acetic acid, do not dissociate into ions in a solution. Glucose, which is a non-electrolyte, does not dissociate at all and therefore contributes to the solution it's one molecule per formula unit. Acetic acid, on the other hand, being a weak electrolyte, dissociates only partially, resulting in fewer particles compared to HCl but more than glucose.

In the given exercise, the ranking of freezing points from lowest to highest would indeed be determined by the number and type of particles each solute contributes. HCl, being an ionic solute, has the strongest effect on freezing point depression, followed by the weakly dissociating acetic acid, and finally glucose which causes the least freezing point depression because it does not dissociate into ions at all.

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Most popular questions from this chapter

A \(3.20-\mathrm{g}\) sample of a salt dissolves in \(9.10 \mathrm{~g}\) of water to give a saturated solution at \(25^{\circ} \mathrm{C}\). What is the solubility (in \(\mathrm{g}\) salt/ \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ) of the salt?

The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(20^{\circ} \mathrm{C}\) is \(44 \mathrm{mmHg},\) and the vapor pressure of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) at the same temperature is \(94 \mathrm{mmHg}\). A mixture of \(30.0 \mathrm{~g}\) of methanol and \(45.0 \mathrm{~g}\) of ethanol is prepared (and may be assumed to behave as an ideal solution). (a) Calculate the vapor pressure of methanol and ethanol above this solution at \(20^{\circ} \mathrm{C}\). (b) Calculate the mole fraction of methanol and ethanol in the vapor above this solution at \(20^{\circ} \mathrm{C}\).

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

Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains \(80.78 \%\) C, \(13.56 \% \mathrm{H}\), and \(5.66 \%\) O. A solution of \(1.00 \mathrm{~g}\) of this pheromone in \(8.50 \mathrm{~g}\) of benzene freezes at \(3.37^{\circ} \mathrm{C}\). What are the molecular formula and molar mass of the compound? (The normal freezing point of pure benzene is \(5.50^{\circ} \mathrm{C}\) ).

Liquids A (molar mass \(100 \mathrm{~g} / \mathrm{mol}\) ) and \(\mathrm{B}\) (molar mass \(110 \mathrm{~g} / \mathrm{mol}\) ) form an ideal solution. At \(55^{\circ} \mathrm{C}\), A has a vapor pressure of \(95 \mathrm{mmHg}\) and \(\mathrm{B}\) has a vapor pressure of \(42 \mathrm{mmHg}\). A solution is prepared by mixing equal masses of \(\mathrm{A}\) and \(\mathrm{B}\). (a) Calculate the mole fraction of each component in the solution. (b) Calculate the partial pressures of \(\mathrm{A}\) and \(\mathrm{B}\) over the solution at \(55^{\circ} \mathrm{C}\). (c) Suppose that some of the vapor described in (b) is condensed to a liquid. Calculate the mole fraction of each component in this liquid and the vapor pressure of each component above this liquid at \(55^{\circ} \mathrm{C}\).

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