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The osmotic pressures of \(0.010 \mathrm{M}\) solutions of \(\mathrm{KI}\) and of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) are \(0.432\) atm and \(0.24\) atm respectively. The van't Hoff factor for KI is : (a) \(1.80\) (b) \(0.80\) (c) \(1.2\) (d) \(1.0\)

Short Answer

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The van't Hoff factor for KI is (a) 1.80.

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01

Understand Osmotic Pressure and Van't Hoff Factor

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Key Concepts

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

Osmotic Pressure
Osmotic pressure is a pivotal concept in physical chemistry that involves the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to one of higher concentration. It's a colligative property, meaning it depends solely on the number of solute particles present in a solution, rather than their identity.

This movement of solvent molecules leads to a pressure build-up on the side with the higher solute concentration, which we define as osmotic pressure. It can be calculated using the formula: \( \Pi = iCRT \) where \( \Pi \) is the osmotic pressure, \( i \) is the van't Hoff factor (which indicates the number of particles the solute splits into or forms when dissolved), \( C \) is the molar concentration of the solution, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

In the given exercise, the different osmotic pressures of KI and sucrose solutions can be attributed to the dissimilar van't Hoff factors for the solutes, which dictate the number of particles contributing to the osmotic pressure.
Physical Chemistry
Physical chemistry is a branch of chemistry concerned with the study of the physical properties and behavior of matter, including the understanding of how chemical structures impact their properties and reactions. It plays an essential role in developing a thorough comprehension of osmotic pressure as it combines principles of physics and chemistry to analyze the changes in matter at a molecular level.

Understanding osmotic pressure involves an in-depth look at the molecular interactions, especially when considering solutions and their behaviors under various conditions. This kind of interdisciplinary approach that physical chemistry offers is crucial for students, particularly when preparing for competitive examinations like JEE Chemistry, where a conceptual grasp of such topics is tested.
Colligative Properties
Colligative properties are those properties of solutions that depend on the ratio of the number of solute particles to the solvent molecules in a solution, regardless of the type of solute. Examples include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

The significance of colligative properties lies in their utility; they allow chemists to determine molecular masses and to understand solution behavior in biological and chemical processes. The van't Hoff factor is a crucial element in studying these properties as it helps quantify the effect of solutes on the properties of the solvent. A higher van't Hoff factor means a greater effect on the colligative properties, as more particles are interacting with the solvent.
JEE Chemistry Preparation
Effective JEE Chemistry preparation involves a balanced focus on conceptual clarity and problem-solving skills. Understanding concepts like osmotic pressure and the van't Hoff factor is fundamental for students aiming to excel in the Physical Chemistry section of JEE.

To prepare for such topics, students are encouraged to practice a plethora of problems, understand the underlying principles, and connect theoretical knowledge with practical applications. Tools like textbook solutions provide step-by-step guidance on how to approach complex problems, helping students to gradually build up their knowledge and confidence. Remember, it's not just about memorizing formulas but also about knowing when and how to apply them.

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Most popular questions from this chapter

The cryoscopic constant of water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). A \(0.01\) molal acetic acid solution produces a depression of \(0.0194^{\circ} \mathrm{C}\) in the freezing point. The degree of dissociation of acetic acid is: (a) zero (b) \(0.043\) (c) \(0.43\) (d) 1

Freezing point of a solution is smaller than that point of a solvent. It is due to : (a) \(\Delta H\) of solution and solvent is almost identical since intermolecular forces between solvent molecules are involved (b) \(\Delta S\) of solution (between solution and solid) is larger than that of the \(\Delta S\) of solvent (between solvent and solid) (c) \(\Delta S\) of the solution is smaller than that of the solvent (d) \(\Delta H\) of the solution is much higher than of solvent but \(\Delta S\) of solution is smaller than that of the solvent

Which of the following solutions will have maximum osmotic pressure? Assume \(90 \%\) dissociation of each salt: (a) decinormal aluminium sulphate (b) decinormal barium chloride solution (c) decinormal sodium sulphate solution (d) solution of equal volumes of decinormal barium chloride and decinormal sodium sulphate solutions

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

Bromoform has a normal freezing point of \(7.734^{\circ} \mathrm{C} / \mathrm{m}\) and it's \(K_{f}=14.4^{\circ} \mathrm{C} / \mathrm{m}\). A solution of \(2.60 \mathrm{~g}\) of an unknown in \(100 \mathrm{~g}\) of bromoform freezes at \(5.43^{\circ} \mathrm{C}\). What is the molecular weight of the unknown? (a) \(16.25\) (b) \(162.5\) (c) 100 (d) none of these

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