Problem 156 - A Complete Resource Book in Chemistry for JEE Main

Question

If a is the degree of dissociation of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) the van't Hoff factor (i) used for calculating the molecular mass is \([\mathbf{2 0 0 5}]\) (a) \(1+\alpha\) (b) \(1-\alpha\) (c) \(1+2 \alpha\) (d) \(1-2 \alpha\)

Step-by-Step Solution

Answer

The correct van't Hoff factor is option (c) \(1 + 2\alpha\).

Step 1: Understanding Degree of Dissociation

The degree of dissociation, \( \alpha \), indicates what fraction of a substance dissociates into its ions. In this case, for every molecule of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) that dissociates, it produces 2 \(\mathrm{Na}^+\) ions and 1 \(\mathrm{SO}_4^{2-}\) ion.

Step 2: Determining Total Number of Particles

Initially, we start with 1 mole of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \). If \( \alpha \) moles dissociate, it will produce \( \alpha \times 2 + \alpha = 3\alpha \) more moles of ions along with the 1 original mole if no dissociation happened.

Step 3: Calculating the Van't Hoff Factor, i

The van't Hoff factor, \( i \), is calculated by considering the total number of particles after dissociation. It is given by: \[ i = (1 - \alpha) + 3\alpha = 1 + 2\alpha \]Hence, for \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), \( i = 1 + 2\alpha \).

Step 4: Choosing the Correct Option

From the calculation, the van't Hoff factor \( i \) is \( 1 + 2\alpha \). This corresponds to option (c) in the given choices.

Key Concepts

Degree of Dissociation

The term 'degree of dissociation' refers to the proportion of a solute that splits into ions when dissolved in a solution. This is especially relevant for electrolytes like sodium sulfate (\(\mathrm{Na}_{2}\mathrm{SO}_{4}\)). Instead of remaining intact, \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) dissociates into its component ions: two \(\mathrm{Na}^+\) and one \(\mathrm{SO}_4^{2-}\). The degree of dissociation is symbolized by \(\alpha\), representing how much of the original compound separates into ions in the solution. For example, if \(\alpha = 0.5\), then 50% of the substance breaks down into ions, while the other 50% remains as the undissociated compound. Understanding this concept is crucial for predicting how substances will behave in different environments, affecting properties like boiling and freezing points.

Molecular Mass Calculation

Molecular mass calculations allow us to determine the mass of a molecule based on its composition. This is particularly significant when calculating colligative properties, which rely on the number of particles in a solution. When a substance like \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) dissociates, its effective number of particles in solution changes, altering properties dependent on molecular mass. The van't Hoff factor (\(i\)) helps adjust for this by accounting for the increase in particle count due to dissociation.To calculate the van't Hoff factor, you use the equation \(i = 1 + 2\alpha\). Here, \(\alpha\) determines how many of the original formula units separate into ions, affecting the overall count of particles and the calculated molecular mass.

Dissociation Equation for Na2SO4

The dissociation of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) in water can be represented by a chemical equation. When sodium sulfate dissolves, it separates into ions according to the equation:\[ \mathrm{Na}_{2}\mathrm{SO}_{4}(s) \rightarrow 2\mathrm{Na}^+(aq) + \mathrm{SO}_4^{2-}(aq) \]This equation shows that one formula unit of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) yields three ions in total: two sodium ions and one sulfate ion. It highlights the stoichiometry of the dissociation, where each unit of solid compound produces multiple ions, which is vital for calculating the van't Hoff factor and understanding electrolyte solutions.

Electrolyte Solutions

Electrolyte solutions are those that contain free ions and can conduct electricity. Substances like \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) are classified as electrolytes because they dissociate into ions, facilitating the flow of electric current.Understanding electrolytes is essential for various scientific applications, including physiology, where they influence nerve and muscle functions, and in industrial applications, such as battery technology. In solutions, electrolytes modify physical properties, influencing how solvents like water behave. This is why the degree of dissociation and the van't Hoff factor are critical; they help predict the extent to which a solute will impart electrical conductivity and modify colligative properties, including boiling and freezing points.