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 \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] ?\) (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) \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) since it has the same van't Hoff factor of 5.

Step 1: Calculate van't Hoff factor for \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\)

For \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\), the dissociation in solution is: \[\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] \rightarrow 4\mathrm{K}^{+} + [\mathrm{Fe}(\mathrm{CN})_6]^{4-}\] This means it dissociates into 5 particles. Therefore, van't Hoff factor \(i = 5\).

Step 2: Calculate van't Hoff factor for \(\mathrm{Al}_2(\mathrm{SO}_4)_3\)

The dissociation reaction is: \[\mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow 2\mathrm{Al}^{3+} + 3\mathrm{SO}_4^{2-}\] This dissociates into 5 ions. Thus, for \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), van't Hoff factor \(i = 5\).

Step 3: Calculate van't Hoff factor for \(\mathrm{NaCl}\)

The dissociation reaction is: \[\mathrm{NaCl} \rightarrow \mathrm{Na}^+ + \mathrm{Cl}^-\] This dissociates into 2 ions. Thus, for \(\mathrm{NaCl}\), van't Hoff factor \(i = 2\).

Step 4: Calculate van't Hoff factor for \(\mathrm{Al}(\mathrm{NO}_3)_3\)

The dissociation reaction is: \[\mathrm{Al}(\mathrm{NO}_3)_3 \rightarrow \mathrm{Al}^{3+} + 3\mathrm{NO}_3^-\] This dissociates into 4 ions. Thus, for \(\mathrm{Al}(\mathrm{NO}_3)_3\), van't Hoff factor \(i = 4\).

Step 5: Calculate van't Hoff factor for \(\mathrm{Na}_2 \mathrm{SO}_4\)

The dissociation reaction is: \[\mathrm{Na}_2 \mathrm{SO}_4 \rightarrow 2\mathrm{Na}^+ + \mathrm{SO}_4^{2-}\] This dissociates into 3 ions. Thus, for \(\mathrm{Na}_2 \mathrm{SO}_4\), van't Hoff factor \(i = 3\).

Step 6: Compare and select

Both \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\) and \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) have a van't Hoff factor of 5. Therefore, \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) is the compound with the same van't Hoff factor as \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\).

Key Concepts

Dissociation of Salts

When salts are dissolved in water, they dissociate into their constituent ions. Dissociation is the process where a compound splits into two or more ions when added to a solvent. It's a critical concept in chemistry because it explains how salts conduct electricity in solution. This happens because ions are charged particles that carry electric current.

For example, when potassium ferricyanide, ext{K}_4[ ext{Fe}( ext{CN})_6], dissociates in water, it separates into four potassium ions and one ext{Fe}( ext{CN})_6^{4-} ion. Each salt has its own unique dissociation pattern depending on its composition. Understanding how different salts dissociate can help predict their behavior in solutions, including their reactivity and participation in further chemical reactions. This dissociation process is crucial for calculating the van't Hoff factor, which measures the effect of a solute on the colligative properties of a solution.

Ionic Compounds

Ionic compounds are made up of positively and negatively charged ions held together by electrostatic forces. These compounds typically form between metals and non-metals. Metals lose electrons easily and become positively charged ions, while non-metals gain these electrons and become negatively charged ions.

For example, sodium chloride (NaCl) is a classic ionic compound. Sodium (a metal) donates an electron to chlorine (a non-metal), resulting in the formation of ext{Na}^+ and ext{Cl}^- ions. These opposite charges attract each other, creating a strong ionic bond. Ionic compounds are generally solid at room temperature and have high melting and boiling points due to these strong interactions.

When dissolved in water, these compounds dissociate into their ions, aiding the process of conduction of electricity in the solution. Understanding ionic compounds and their dissociation is vital for applying the van't Hoff factor when calculating properties like boiling point elevation or freezing point depression.

Molecular Dissociation

Not only salts but even certain molecular compounds can dissociate in solution, albeit typically in water or other polar solvents. Dissociation increases the number of solute particles in a solution, directly impacting the colligative properties, such as boiling point, vapor pressure, and osmotic pressure.

For instance, when you dissolve acetic acid in water, it partially dissociates into hydrogen ions ( ext{H}^+) and acetate ions ( ext{CH}_3 ext{COO}^-). Though acetic acid is a weak acid and does not fully dissociate, understanding its behavior helps predict how it will affect the solution's properties.

In the case of salts, complete dissociation into ions occurs, which is essential for various applications, including in industrial and laboratory settings. Understanding the dynamics of molecular dissociation allows chemists to manipulate solution properties for desired outcomes.

Chemical Equilibrium

Chemical equilibrium is a state where the rate of a forward chemical reaction equals the rate of the reverse reaction. For solutions, this means that the rate of dissociation equals the rate of recombination of ions or molecules. At equilibrium, the concentrations of reactants and products remain constant over time.

Salt solutions often reach equilibrium during the dissociation process. This balance can be shifted by changing the conditions, such as the concentration of ions, temperature, and pressure. Le Chatelier's Principle explains how equilibria respond to such changes, asserting that the system will adjust to minimize the impact.

For example, if more of one ion is added to the solution, the system may respond by shifting the equilibrium position to reduce the concentration of the added ion. Understanding chemical equilibrium is essential for predicting how solutions behave in reality, especially when dealing with salts and calculating factors like the van't Hoff factor.