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Phenol associates in benzene to a certain extent in dimerisation reaction. A solution containing \(0.02 \mathrm{~kg}\) of phenol in \(1.0 \mathrm{~kg}\) of benzene has its freezing point depressed \(0.69 \mathrm{~K}\). Hence, degree of association of phenol dimerized will be : \(\left[K_{f}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]\) (a) \(0.63\) (b) \(0.73\) (c) \(0.83\) (d) \(0.93\)

Short Answer

Expert verified
The degree of association of phenol dimerized is 0.73.

Step by step solution

01

Understand the Concept of Freezing Point Depression

The freezing point depression is the process where the freezing point of a solvent decreases due to the addition of a solute. In this case, phenol is the solute and benzene is the solvent. The extent to which the freezing point is depressed is related to the number of solute particles in the solution. If phenol dimerises (associates in pairs), the freezing point depression will be less than if it remained as individual phenol molecules because there will be fewer solute particles.
02

Calculate the Molality of the Solution

Use the freezing point depression to calculate the molality of the solution. The formula for freezing point depression is \(\[\Delta T_{f} = K_{f} \times m\]\), where \(\Delta T_{f}\) is the freezing point depression, \(K_{f}\) is the cryoscopic constant, and \(m\) is the molality. Rearranging this formula to solve for molality, we get \(m = \frac{\Delta T_{f}}{K_{f}}\). Plug in the values to get the molality of the solution.
03

Calculate the Molar Mass of the Associated Phenol

Use the weight of phenol and the molality to determine the molar mass of the associated solute. The formula is \(m = \frac{\mathrm{moles~of~solute}}{\mathrm{kg~of~solvent}}\), and since the weight of the solute is given, we can calculate the moles of solute (which in this case is the dimerised phenol) and then the molar mass.
04

Calculate the Degree of Association

The degree of association (\(\alpha\)) can be calculated using the formula \(\alpha = \frac{\mathrm{observed~molar~mass}}{\mathrm{theoretical~molar~mass}}\). The theoretical molar mass is the molar mass of a single molecule of phenol, while the observed molar mass is from the previous step considering dimerisation. A ratio of these gives the degree of association.
05

Determine the Degree of Association

The theoretical molar mass of phenol is the molar mass of a phenol molecule, C6H5OH, which is known. The observed molar mass is twice this value in case of complete dimerisation. Use the degree of association formula and the calculated molar mass to determine the degree of association and match it with the given options.

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Key Concepts

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

Cryoscopic Constant
The cryoscopic constant, denoted as \(K_f\), is crucial in understanding freezing point depression. It represents the freezing point lowering of the solvent when a one molal solution of a non-volatile solute is added. For benzene, \(K_f\) is \(5.12 \mathrm{~K \; kg \; mol}^{-1}\). This value is unique for each solvent and is based on its properties, such as the entropy and enthalpy of fusion.

Essentially, \(K_f\) helps us quantify how much a solute affects the freezing point of a solvent. It tells you how sensitive a solvent's freezing point is to the presence of solute particles. In the given problem, the depression in the freezing point of benzene helps determine the degree of phenol association - highlighting the intimate relationship between \(K_f\) and the behavior of solutions.
Molality Calculation
Molality is a measure of solute concentration in a solution, defined as the moles of solute per kilogram of solvent. It is represented by \(m\). Unlike molarity, molality is not affected by temperature changes because mass remains constant with temperature.

To calculate the molality, we use the formula \(m = \frac{\Delta T_f}{K_f}\). Here, \(\Delta T_f\) is the change in freezing point, and \(K_f\) is the cryoscopic constant. In the context of the given exercise, the molality calculation helps us find the molar mass of the associated phenol, allowing us to further understand the degree of association in the dimerization reaction.
Degree of Association
The degree of association is the fraction of solute molecules that associate (or dissociate) in a particular medium. It's denoted by \(\alpha\) and is calculated using the formula \(\alpha = \frac{\text{observed molar mass}}{\text{theoretical molar mass}}\). For molecules that dimerize, the observed molar mass will be twice that of the individual molecule's molar mass if the association is complete.

In our exercise, the determination of \(\alpha\) helps to reveal whether phenol molecules in benzene remain as individual entities or combine to form dimers. The degree of association affects the properties of the solution, including its boiling point, freezing point, and vapor pressure.
Colligative Properties
Colligative properties are properties of solutions that depend on the ratio of solute particles to solvent molecules, regardless of the nature of the solute particles. Freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure are all colligative properties.

These properties can provide insights into the molecular nature of solutions. For instance, when a solute such as phenol associates or dissociates in a solvent, the number of solute particles changes, which in turn affects the colligative properties. Understanding colligative properties, therefore, not only allows us to predict how a solution will behave but also to infer details about the solute's molecular interactions, such as the degree of association we investigated in the exercise related to phenol in benzene.

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Most popular questions from this chapter

Consider \(0.1 \mathrm{M}\) solutions of two solutes \(X\) and \(Y\). The solute \(X\) behaves as a univalent electrolyte while the solute \(Y\) dimerises in solution. Which of the following statements are correct regarding these solutions? (1) The boiling point of the solution of \(X\) will be higher than that of \(Y\) (2) The osmotic pressure of the solution of \(Y\) will be lower than that of \(X\) (3) The freezing point of the solution of \(X\) will be lower than that of \(Y\) (4) The relative lowering of vapour pressure of both the solutions will be the same Select the correct answer from the option given below : (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,2 and 4 (d) 1,3 and 4

The osmotic pressures of equimolar solutions of urea, \(\mathrm{BaCl}_{2}\) and \(\mathrm{AlCl}_{3}\) will be in the order : (a) \(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\) urea (b) \(\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}>\) urea (c) urea \(>\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}\) (d) \(\mathrm{BaCl}_{2}>\) urea \(>\mathrm{AlCl}_{3}\)

Select correct statement? (a) Heats of vaporisation for a pure solvent and for a solution are similar because similar intermolecular forces between solvent molecules must be overcome in both cases (b) Entropy change between solution and vapoür is smaller than the entropy change between pure solvent and vapour (c) Boiling point of the solution is larger than that of the pure solvent (d) All are correct statements

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

The vapour pressure of a given liquid will decrease if : (a) surface area of liquid is decreased (b) the volume of liquid in the container is decreased (c) the volume of the vapour phase is increased (d) the temperature is decreased

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