Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For each of the following dilute solutions, Van't Hoff's factor is equal of 3 , except (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{CaF}_{2}\) (c) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\)

Short Answer

Expert verified
The compound with a Van't Hoff factor not equal to 3 is (c) K3PO4.

Step by step solution

01

- Understanding Van't Hoff's Factor

The Van't Hoff factor, denoted by 'i', is the number of particles a compound dissociates into when it is dissolved in solution. For a compound that completely dissociates into three ions in solution (one cation and two anions, or one anion and two cations), the Van't Hoff factor is 3.
02

- Analyzing each compound

Examine each compound and predict the number of ions each will produce when dissolved in water. (a) Na2SO4 will dissociate into 2 Na+ cations and 1 SO4^2- anion, for a total of 3 ions. (b) CaF2 will dissociate into 1 Ca^2+ cation and 2 F- anions, for a total of 3 ions. (c) K3PO4 will dissociate into 3 K+ cations and 1 PO4^3- anion, for a total of 4 ions. (d) (NH4)2CO3 will dissociate into 2 NH4+ cations and 1 CO3^2- anion, for a total of 3 ions.
03

- Identifying the Exception

From the analysis in Step 2, it's clear that all compounds except for K3PO4 will produce 3 ions in solution. K3PO4 will produce 4 ions in solution, making it the exception with a Van't Hoff factor different from 3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Physical Chemistry and the Van't Hoff Factor
Physical chemistry is the branch that deals with the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. Understanding the Van't Hoff factor is integral to several concepts in physical chemistry, particularly those involving solutions. The Van’t Hoff factor (i) is significant because it provides insight into the number of particles formed when a substance is dissolved in a solution.

This factor affects the physical properties of solutions, such as freezing point depression, boiling point elevation, and osmotic pressure, all of which are fundamental when studying colligative properties. The step-by-step textbook solution explicitly illustrates the Van't Hoff factor by breaking down ionic compounds into their constituent ions. Evidently, physical chemistry requires a grasp of both theoretical and practical applications, where exercises like these highlight the practical aspects of these chemical principles in a straightforward manner.
Ionic Dissociation in Solutions
Ionic dissociation is a process where an ionic compound separates into its ions when dissolved in a solvent, typically water. This process is essential for understanding the behavior of salts and other ionic compounds in solution.

In the educational example provided, comprehension of ionic dissociation helps students identify the Van't Hoff factor for various compounds. For instance, Na2SO4 dissociates into three ions (2 Na+ and 1 SO42-), which correlates to the Van’t Hoff factor of 3. Conversely, compounds that produce a different number of ions, such as K3PO4, which dissociates into four ions, have a different factor. Highlighting the dissociation process emphasizes the predictive power of these concepts in determining the properties of a solution.
Colligative Properties and Their Reliance on the Van't Hoff Factor
Colligative properties are properties of solutions that depend on the number of dissolved particles in the solution, regardless of their nature. These include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. The Van't Hoff factor plays a crucial role in calculating these properties.

For a given solute, the higher the Van’t Hoff factor, the more significant the effect on the colligative properties. For example, K3PO4 having a Van’t Hoff factor of 4 means it would have a greater impact on the solution's colligative properties compared to a substance with a lower factor of 3, such as CaF2.

It is imperative for students to recognize that colligative properties are influenced by the amount of the solute rather than its identity, and exercises elucidating the Van't Hoff factor offer a practical approach to understanding these solution characteristics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The normal freezing point of nitrobenzene is \(278.82 \mathrm{~K}\). A non- volatile solute is dissolved in it and a solution of molality \(0.25 \mathrm{~m}\) is prepared. If the observed freezing point of the solution is \(276.82 \mathrm{~K}, K_{\ell}\) of nitrobenzene is (a) \(6 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\) (b) \(8 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\) (c) \(7 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\) (d) \(5 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\)

A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equal (a) potential energy (b) intermolecular force (c) kinetic energy (d) total energy

Which serves best as a semipermeable membrane? (a) copper ferrocyanide (b) vegetable membrane (c) animal membrane (d) cellophane

Pure water boils at \(373 \mathrm{~K}\) and nitric acid at \(359 \mathrm{~K}\). The azeotropic mixture of water and nitric acid boils at \(393.5 \mathrm{~K}\). On distillation of the azeotropic mixture, (a) pure nitric acid will distil over first. (b) pure water will distil over first. (c) one of them will distil over with small amount of the other. (d) both of them will distil over in the same composition as they are in the mixture.

The vapour pressure of an aqueous solution is found to be 750 torr at a temperature, \(T\), and the same solution shows an elevation in boiling point equal to \(1.04 \mathrm{~K}\). If \(T\) is the boiling point of pure water, then the atmospheric pressure should be \(\left(K_{\mathrm{b}}\right.\) of water \(\left.=0.52 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\right)\) (a) 760 torr (b) 777 torr (c) 746 torr (d) 750 torr

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free