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When a solution containing non-volatile solute freezes, which equilibrium would exist? (a) solid solvent \(\rightleftharpoons\) liquid solvent (b) solid solute \(\rightleftharpoons\) liquid solution (c) solid solute \(\rightleftharpoons\) liquid solvent (d) solid solvent \rightleftharpoons liquid solution

Short Answer

Expert verified
The correct equilibrium existing when a solution containing non-volatile solute freezes is (d) solid solvent \(\rightleftharpoons\) liquid solution.

Step by step solution

01

Understanding the Concepts Involved

Consider the process when a solution containing a non-volatile solute starts to freeze. The freezing process involves the solvent transitioning from a liquid phase to a solid phase, while the solute remains dissolved in the remaining liquid solvent or becomes part of the solid phase without changing its state. Non-volatile solute means it will not readily evaporate and, as such, does not influence the equilibrium of the phases of the solvent.
02

Identifying the Correct Equilibrium

During freezing, the solvent molecules start to arrange into a solid structure, such as ice in the case of water. The non-volatile solute does not participate in this phase change directly. So, the correct equilibrium is between the pure solid solvent and the liquid solution (which still contains the solute).
03

Determining the Correct Answer

With the understanding of the freezing process, we can conclude that equilibrium exists between the solid solvent and the liquid solution, which still contains the non-volatile solute dispersed throughout. Therefore, the correct answer is option (d) solid solvent \(\rightleftharpoons\) liquid solution.

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Key Concepts

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

Physical Chemistry
Understanding the concepts related to the freezing point depression in the context of physical chemistry requires us to delve into the nature of solutions and how they behave under certain conditions. Physical chemistry is the branch that deals with the way matter behaves on a molecular and atomic level and how chemical reactions occur. When it comes to solutions, which are homogeneous mixtures of two or more substances, physical chemistry looks at how the components interact and affect properties like boiling points, freezing points, and vapor pressures.

Specifically, freezing point depression is a colligative property, meaning it depends on the number of solute particles in a solution, not their chemical identity. This is a vital concept in physical chemistry because it allows us to predict how the addition of a solute will affect a solvent's freezing point. By observing these changes, we gain insights into the molecular nature of solutions.
Solution Equilibrium
Solution equilibrium refers to the state in which the processes of dissolution and crystallization of a solute occur at an equal rate, resulting in a stable mixture where the concentration of solute in the solution remains constant over time. Upon reaching equilibrium, the solution becomes saturated, meaning it cannot dissolve more solute at a given temperature and pressure.

In the context of freezing, when a solution starts to solidify, the solvent begins to form a pure solid phase, and the solute gets excluded from this newly forming solid. This equilibrium is not static but dynamic; it undergoes continuous microscopic changes even though the macroscopic state appears stable. It's essential for students to understand that the dissolved, non-volatile solute particles disrupt the crystalline structure of the solidifying solvent, lowering its freezing point.
Phase Transition
What we're discussing here is a phase transition, which is the transformation of a substance from one state of matter to another, such as from liquid to solid during freezing. During this process, energy is either absorbed or released, and the physical properties of the substance change.

During the phase transition of freezing, the arrangement of solvent molecules changes from a less organized liquid state to a more ordered solid state. The presence of a non-volatile solute interferes with the organization of solvent molecules, preventing them from reaching their solid state at the usual temperature. This leads to a lowering of the freezing point of the solution compared to the pure solvent. The equilibrium in a frozen solution is between the solid solvent and the liquid solution since the liquid continues to possess the non-volatile solute particles.
Non-Volatile Solute
A non-volatile solute is a substance that has little to no tendency to vaporize under existing conditions and thus does not significantly contribute to the vapor pressure of a solution. In the context of freezing, non-volatile solutes are critical because they alter the physical properties of the solvent by disrupting its normal freezing process.

When a non-volatile solute is added to a solvent, the solute particles obstruct the solvent molecules from organizing into a solid lattice structure. This obstruction requires a lower temperature to overcome, leading to a depression of the freezing point. For students, it is crucial to understand that the solute itself does not go through a phase change during this process; it stays dissolved within the liquid fraction of the solution until the solvent has wholly solidified.

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Most popular questions from this chapter

If two liquids \(A\left(P_{A}^{\circ}=100\right.\) torr \()\) and \(B\left(P_{B}^{\circ}=200\right.\) torr \()\) are completely immiscible with each other (each one will behave independently of the other) are present in a closed vessel. The total vapour pressure of the system will be: (a) less than 100 torr (b) greater than 200 torr (c) between 100 to 200 torr (d) 300 torr

When a liquid that is immiscible with water was steam distilled at \(95.2^{\circ} \mathrm{C}\) at a total pressure of 748 torr, the distillate contained \(1.25 \mathrm{~g}\) of the liquid per gram of water. The vapour pressure of water is 648 torr at \(95.2^{\circ} \mathrm{C}\), what is the molar mass of liquid? (a) \(7.975 \mathrm{~g} / \mathrm{mol}\) (b) \(166 \mathrm{~g} / \mathrm{mol}\) (c) \(145.8 \mathrm{~g} / \mathrm{mol}\) (d) None of these

The freezing point of a solution of \(2.40 \mathrm{~g}\) of biphenyl \(\left(\mathrm{C}_{12} \mathrm{H}_{10}\right)\) in \(75.0 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(4.40^{\circ} \mathrm{C}\). The normal freezing point of benzene is \(5.50^{\circ} \mathrm{C}\). What is the molal freezing point constant \(\left({ }^{\circ} \mathrm{C} / \mathrm{m}\right)\) for benzene? (a) \(-5.3\) (b) \(-5.1\) (c) \(-4.6\) (d) \(-4.8\)

A complex is represented as \(\mathrm{CoCl}_{3} x \mathrm{NH}_{3}\). Its \(0.1\) molal solution in water shows \(\Delta T_{f}=0.558 \mathrm{~K}\). \(K_{f}\) for \(\mathrm{H}_{2} \mathrm{O}\) is \(1.86 \mathrm{~K}\) molality \(^{-1}\). Assuming \(100 \%\) ionisation of complex and co-ordination number of Co is six, calculate formula of complex: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) none of these

Chloroform, \(\mathrm{CHCl}_{3}\), boils at \(61.7^{\circ} \mathrm{C}\). If the \(K_{b}\) for chloroform is \(3.63^{\circ} \mathrm{C} / \mathrm{molal}\), what is the boiling point of a solution of \(15.0 \mathrm{~kg}\) of \(\mathrm{CHCl}_{3}\) and \(0.616 \mathrm{~kg}\) of acenaphthalene, \(\mathrm{C}_{12} \mathrm{H}_{10} ?\) (a) \(61.9\) (b) \(62.0\) (c) \(52.2\) (d) \(62.67\)

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