Problem 42 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following salt will have the same volume of vant Hoff factor 'i'as that of \(\left.\mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})]\right\\}\) (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (b) \(\mathrm{NaCl}\) (c) \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

(a) \(\text{Al}_2(\text{SO}_4)_3\)

Step 1: Understand the Problem

We are asked to find a salt with the same Van't Hoff factor (i) as \(\text{K}_4[\text{Fe(CN)}_6]\). The Van't Hoff factor \(i\) indicates the number of particles the compound dissociates into in solution.

Step 2: Determine Van't Hoff Factor for \(\text{K}_4[\text{Fe(CN)}_6]\)

\(\text{K}_4[\text{Fe(CN)}_6]\) dissociates into 4 potassium ions (\(\text{K}^+\)) and one \([\text{Fe(CN)}_6]^{4-}\) ion. This results in a total of 5 particles. Therefore, the Van't Hoff factor \(i\) is 5.

Step 3: Calculate Van't Hoff Factor for Each Option

- (a) \(\text{Al}_2(\text{SO}_4)_3\) dissociates into 2 \(\text{Al}^{3+}\) and 3 \(\text{SO}_4^{2-}\), totaling 5 particles. - (b) \(\text{NaCl}\) dissociates into 1 \(\text{Na}^+\) and 1 \(\text{Cl}^-\), totaling 2 particles. - (c) \(\text{Al}(\text{NO}_3)_3\) dissociates into 1 \(\text{Al}^{3+}\) and 3 \(\text{NO}_3^-\), totaling 4 particles. - (d) \(\text{Na}_2\text{SO}_4\) dissociates into 2 \(\text{Na}^+\) and 1 \(\text{SO}_4^{2-}\), totaling 3 particles.

Step 4: Compare the Factors

The Van't Hoff factor for \(\text{K}_4[\text{Fe(CN)}_6]\) is 5, and the salt \(\text{Al}_2(\text{SO}_4)_3\) also yields a Van't Hoff factor of 5. Therefore, (a) \(\text{Al}_2(\text{SO}_4)_3\) is the correct option.

Key Concepts

Understanding Ionic Dissociation

Ionic dissociation is a fundamental part of chemistry that involves the separation of a compound into individual ions when it dissolves in water. This phenomenon is especially important when dealing with salt compounds, such as the ones mentioned in our exercise. When a substance like \( ext{K}_4[ ext{Fe(CN)}_6]\) dissolves, it separates into its constituent ions which are 4 potassium ions \( ext{K}^+\) and one \[\text{Fe(CN)}_6^{4-}\]. This results in five ions overall entering the solution.In the context of this problem, understanding ionic dissociation helps us calculate the Van't Hoff factor - a measure of how many particles the compound separates into. This makes ionic dissociation crucial because each particle in the solution can affect properties like freezing point and boiling point, as well as the so-called colligative properties of the solution.

Exploring Electrolyte Solutions

Electrolyte solutions are solutions that conduct electricity due to the presence of free ions. When salts like \( ext{NaCl}\) or \( ext{Al}_2( ext{SO}_4)_3\) dissociate in water, they produce ions, enabling the solution to carry an electric current. The concentration and nature of these ions determine the solution's conductivity.In an electrolyte solution, properties such as conductance and boiling point elevation are influenced by the presence of these free-moving ions. This is where the Van't Hoff factor comes into play. It helps determine the extent to which a compound will affect the electrolytic properties of a solution by indicating how many ions are formed per formula unit when the compound dissolves. If more ions are produced, the solution will show stronger electrolyte properties.

Chemical Compounds Analysis Using Van't Hoff Factor

Chemical compounds analysis often involves examining how a compound behaves in a solution, and calculating the Van't Hoff factor becomes an essential part of this analysis. The Van't Hoff factor, denoted as \(i\), is integral for understanding how a compound dissociates and helps predict the impact of this dissociation on the physical properties of solutions.Our exercise showcases this analysis clearly. We started with the compound \( ext{K}_4[ ext{Fe(CN)}_6]\), which we knew had a Van't Hoff factor of 5 because it dissociates into 5 particles. By comparing the Van't Hoff factors of several salts, we found that \(\text{Al}_2(\text{SO}_4)_3\) also dissociates into 5 ions, thus matching it with our original compound.Using the Van't Hoff factor in chemical analysis allows for deeper insights into a molecule or salt's effects on colligative properties of their solutions. This, in turn, aids in determining practical applications and reactions where these compounds might be used effectively.