Step 1: Understanding the Concept:
The van’t Hoff factor ($i$) accounts for the effect of solute dissociation or association on colligative properties (like boiling point elevation or freezing point depression).
When a salt like $Na_2SO_4$ dissolves in water, it breaks apart into ions, increasing the total number of particles in the solution.
The degree of dissociation ($\alpha$) represents the fraction of the total moles of solute that has dissociated into ions.
Step 2: Key Formula or Approach:
For a general electrolyte $A_x B_y$ that dissociates into $n$ ions ($n = x+y$):
The total number of particles at equilibrium, starting from 1 mole of solute, is:
\[ i = (1 - \alpha) + n\alpha = 1 + (n - 1)\alpha \]
Where:
- $(1-\alpha)$ is the mole fraction of undissociated solute.
- $n\alpha$ is the total mole fraction of the ions produced.
Step 3: Detailed Explanation:
Let's consider the dissociation of sodium sulfate ($Na_2SO_4$) in an aqueous medium.
Sodium sulfate is a strong electrolyte that dissociates according to the following equilibrium equation:
\[ Na_2SO_4 (s) \rightleftharpoons 2Na^+ (aq) + SO_4^{2-} (aq) \]
From the balanced chemical equation, we can see that one formula unit of $Na_2SO_4$ dissociates to yield two sodium ions and one sulfate ion.
Therefore, the total number of ions produced per formula unit ($n$) is $2 + 1 = 3$.
Now, let's assume we start with 1 mole of $Na_2SO_4$ and it has a degree of dissociation $\alpha$.
At equilibrium:
- Moles of undissociated $Na_2SO_4 = 1 - \alpha$
- Moles of $Na^+$ ions $= 2\alpha$
- Moles of $SO_4^{2-}$ ions $= \alpha$
The total moles of particles present in the solution at equilibrium is the sum of these parts:
\[ \text{Total Moles} = (1 - \alpha) + 2\alpha + \alpha = 1 + 2\alpha \]
Since the van't Hoff factor $i$ is the ratio of total moles of particles to the initial moles of solute:
\[ i = \frac{1 + 2\alpha}{1} = 1 + 2\alpha \]
This matches the expression derived using the general formula: $i = 1 + (3-1)\alpha = 1 + 2\alpha$.
Step 4: Final Answer:
Substituting $n=3$ into the dissociation formula gives $i = 1 + 2\alpha$. This corresponds to option (D).