Question

\(\mathrm{C}_{6} \mathrm{H}_{6}\) freezes at \(5.5^{\circ} \mathrm{C}\). At what temperature will a solution of \(10.44 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{10}\) in \(200 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}\) freeze? \(K_{f}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=5.12^{\circ} \mathrm{C} / \mathrm{m}\) (a) \(4.608^{\circ} \mathrm{C}\) (b) \(0.892^{\circ} \mathrm{C}\) (c) \(5.5^{\circ} \mathrm{C}\) (d) none of these

Short answer

The new freezing point of the solution is approximately \(0.892^\circ C\).

Step-by-step solution

Calculate the molality of the solution

Use the formula for molality, which is the number of moles of solute per kilogram of solvent. First, calculate the number of moles of butane \(\mathrm{C}_{4}\mathrm{H}_{10}\) by dividing its mass in grams by its molar mass. Then, divide the number of moles by the mass of the benzene solvent in kilograms to find the molality.

Find the change in freezing point

Use the freezing point depression formula, \(\Delta T_f = K_f \times m\), where \(K_f\) is the cryoscopic constant and \(m\) is the molality of the solution. This will give you the change in freezing point.

Calculate the new freezing point

Subtract the change in freezing point from the original freezing point of benzene to find the new freezing point of the solution.

Key concepts

Cryoscopic Constant

The cryoscopic constant, denoted as \(K_f\), is a fundamental parameter in physical chemistry that helps us comprehend how the freezing point of a solvent is affected when a non-volatile solute is dissolved in it. The principle underlying this constant is simple: adding a solute to a pure solvent will generally lower the solvent's freezing point, a phenomenon known as freezing point depression. In the context of freezing point depression, the cryoscopic constant represents the freezing point lowering per molal concentration of the solute.

The value of \(K_f\) is unique for each solvent and is dependent on both the physical properties of the solvent and the nature of its interactions with the solute. Knowing the cryoscopic constant enables us to predict how much the freezing point of the solvent will drop for every mole of solute particles added per kilogram of solvent. In the given exercise, the cryoscopic constant for benzene is \(5.12^{\text{\textdegree}}C/m\).

Molality

Molality, often symbolized as \(m\), is a measure of the concentration of a solution, specifically defined as the number of moles of a solute dissolved per kilogram of solvent. Unlike molarity, which is moles per liter of solution, molality is dependent only on the masses of the solute and solvent, making it independent of temperature and pressure changes.

To calculate molality, you need the mass of the solute, its molar mass, and the mass of the solvent. Convert the mass of the solvent to kilograms to ensure proper units. For example, in the exercise where we dissolve butane in benzene, the calculation begins with finding the moles of butane by its given mass and molar mass, then dividing by the mass of benzene in kilograms, to yield the molality of the solution.

Solution Colligative Properties

Colligative properties are properties of solutions that rely on the ratio of solute particles to solvent molecules, rather than the identity of the solute itself. These properties include vapor pressure lowering, boiling point elevation, osmotic pressure, and freezing point depression. The impact on these properties is a function of the number of particles the solute contributes to the solution. It's important to note that for non-electrolytes, these particles are typically the molecules of the solute, whereas for electrolytes, they are the ions the solute dissociates into.

Freezing point depression is a particularly fundamental colligative property closely related to the exercise at hand. It is directly proportional to the molality of the solute in the solution and can be calculated using the cryoscopic constant, as demonstrated in the steps of the given problem. This property forms a core part of the curriculum in JEE Physical Chemistry due to its importance in both conceptual understanding and practical applications.

JEE Physical Chemistry

The Joint Entrance Examination (JEE) in India is a highly competitive test for aspirants aiming to enter engineering institutions. Physical Chemistry is a significant segment of the JEE Chemistry syllabus, combining principles of physics and chemistry to explain the physical structure of molecules, the forces that act upon them, and the chemical reactions they undergo.

Understanding the colligative properties of solutions, such as freezing point depression, is crucial for JEE aspirants because these concepts are frequently tested both conceptually and through problem-solving exercises. JEE Physical Chemistry questions often require a deep understanding of the principles and the ability to apply them to complex numerical problems, as can be seen in the step-by-step solution provided for determining the freezing temperature of a solution.