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 \(>\mathrm{AlCl}_{3}\)

Short Answer

Expert verified
\(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\) urea

Step by step solution

01

Understand the Concept of Osmotic Pressure and Van't Hoff Factor

Osmotic pressure is a property related to the number of particles in a solution. It is higher for solutions with more particles. The Van't Hoff factor (i) represents the number of particles into which a solute dissociates in solution. For non-electrolytes like urea, the Van't Hoff factor is 1, as it does not dissociate. For electrolytes like \(\mathrm{BaCl}_{2}\) and \(\mathrm{AlCl}_{3}\), the factors are 3 and 4, respectively, as \(\mathrm{BaCl}_{2}\) dissociates into 3 particles (1 \(\mathrm{Ba}^{2+}\) and 2 \(\mathrm{Cl}^{-}\)) and \(\mathrm{AlCl}_{3}\) dissociates into 4 particles (1 \(\mathrm{Al}^{3+}\) and 3 \(\mathrm{Cl}^{-}\)).
02

Apply the Relationship between Van't Hoff Factor and Osmotic Pressure

Given that the solutions are equimolar, their osmotic pressures (\(\Pi\)) will be directly proportional to their Van't Hoff factors (\(i\)).
03

Determine the Order of Osmotic Pressures

Since \(\mathrm{AlCl}_{3}\) has the highest Van't Hoff factor (4), followed by \(\mathrm{BaCl}_{2}\) (3), and urea has the lowest (1), the order of osmotic pressures is \(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\) urea.

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.

Van't Hoff factor
Understanding the Van't Hoff factor is crucial for grasping the behavior of solutions in various conditions, such as osmotic pressure scenarios covered in the Joint Entrance Exam (JEE) Chemistry syllabus. The Van't Hoff factor, denoted by the symbol 'i', is a dimensionless quantity that indicates the number of particles a compound forms when it dissolves in a solution. For non-electrolytes, substances that do not produce ions in solution, the Van't Hoff factor is typically 1. This is because they do not dissociate upon dissolving.

For electrolytes, the factor differs depending on the degree of dissociation. Strong electrolytes fully dissociate into ions, resulting in a Van't Hoff factor equal to the total number of ions produced. For example, common table salt (NaCl) would have a factor of 2. Weak electrolytes only partially dissociate, and their Van't Hoff factors are usually between 1 and the total number of ions that could be formed.

Calculation of the Van't Hoff factor is essential when predicting properties such as boiling point elevation, freezing point depression, and osmotic pressure, all of which are influenced by the number of solute particles in the solution. In osmotic pressure, for instance, a solution with a higher Van't Hoff factor will have a higher osmotic pressure, assuming the solutions are equimolar.
Dissociation of Electrolytes
Dissociation is the process where ionic compounds separate into their constituent ions when dissolved in a solvent, like water. This process is vital for JEE Chemistry students to comprehend as it affects the properties of solutions, such as conductivity and reactivity, along with osmotic pressure. In the given example, \(\mathrm{BaCl}_{2}\) and \(\mathrm{AlCl}_{3}\) are both electrolytes, and they dissociate in water to a different extent because of their structure and the nature of their ionic bonds.

The dissociation of \(\mathrm{BaCl}_{2}\) results in one \(\mathrm{Ba}^{2+}\) ion and two \(\mathrm{Cl}^{-}\) ions, while \(\mathrm{AlCl}_{3}\) yields one \(\mathrm{Al}^{3+}\) ion and three \(\mathrm{Cl}^{-}\) ions. The total number of particles that result from this process determines the osmotic pressure of the solution, thus directly linking the concept of dissociation with osmotic pressure.
Properties of Solutions
The properties of solutions play a significant role in different analytical aspects of Chemistry, such as physical states, concentrations, and interactions between solutes and solvents. One important property is osmotic pressure, which can be influenced by factors like temperature, concentration, and the nature of the solute.

For a solution to exert osmotic pressure, it must be separated by a semipermeable membrane from pure solvent or a less concentrated solution. Osmotic pressure drives the solvent across the membrane into the more concentrated solution, seeking to equalize the concentrations on both sides of the membrane.

Colligative Properties

Additionally, osmotic pressure is a colligative property - a property that depends on the number of solute particles in a solvent, regardless of their identity. Hence, for equimolar solutions of different substances, the one with the higher Van't Hoff factor will exhibit a higher osmotic pressure because it releases more particles into the solution.

The study of these properties helps chemists to predict how substances will behave in various situations, enabling them to engineer solutions with desired properties for industrial, pharmaceutical, or environmental applications.

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

What is the correct sequence of osmotic pressure of \(0.01 \mathrm{M}\) aq. solution of : (1) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (2) \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) (3) \(\mathrm{BaCl}_{2}\) (4) Glucose (a) \(\pi_{4}>\pi_{2}>\pi_{3}>\pi_{1}\) (b) \(\pi_{3}>\pi_{4}>\pi_{2}>\pi_{1}\) (c) \(\pi_{3}>\pi_{4}>\pi_{1}>\pi_{2}\) (d) \(\pi_{1}>\pi_{2}>\pi_{3}>\pi_{4}\)

A semipermeable membrane used in the measurement of osmotic pressure of a solution allows the passage of: (a) solute molecules through it (b) solvent molecules through it (c) both solvent and solute molecules (d) either solvent or solute

An ideal solution is formed by mixing two volatile liquids \(A\) and \(B . X_{A}\) and \(X_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the solution and \(Y_{A}\) and \(Y_{B}\) are the mole fractions of \(A\) and \(B\) respectively in the vapour phase. A plot of \(\frac{1}{Y_{A}}\) along \(y\) -axis against \(\frac{1}{X_{1}}\) along \(x\) -axis gives a straight line. What is the slope of the straight line? (a) \(\frac{P_{B}^{\circ}}{P_{A}^{\circ}}\) (b) \(\frac{P_{A}^{\circ}}{P_{B}^{\circ}}\) (c) \(P_{B}^{\circ}-P_{A}^{\circ}\) (d) \(P_{A}^{\circ}-P_{B}^{\circ}\)

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) \(0.93\)

\(\mathrm{C}_{6} \mathrm{H}_{6}\) 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

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