Animated Solution for Chemistry - Solutions: The degree of dissociation ($\alpha$) of a weak electrolyte, $A_x{B}_y$ is related to van't Hoff factor ($i$) by the expression

Imagine you drop a pinch of salt into a beaker of water. It vanishes, right? But at the microscopic level, a chaotic dance is happening. The solid crystal lattice breaks apart, and individual ions are surrounded by water molecules. This is the essence of dissolution.

However, not all substances behave the same way. Strong electrolytes, like sodium chloride, shatter completely into their constituent ions. But weak electrolytes are a bit more stubborn. They only partially dissociate, establishing a delicate equilibrium between the intact molecules and the free-floating ions. To quantify this partial breakup, chemists use a term called the degree of dissociation, denoted by the Greek letter $\alpha$.

When a solute dissociates (or associates) in a solvent, it changes the total number of particles in the solution. This change directly affects the colligative properties—like boiling point elevation or freezing point depression. To account for this, we introduce the van't Hoff factor, $i$. It is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved.

There is a beautiful, standard mathematical bridge that connects the degree of dissociation $\alpha$ with the van't Hoff factor $i$:

Here, $n$ is the crucial piece of the puzzle. It represents the total number of moles of ions produced when exactly one mole of the electrolyte undergoes complete dissociation.

Let's look at the specific weak electrolyte given in our problem: $A_x{B}_y$.

When one molecule of $A_x{B}_y$ decides to break apart, how many pieces does it form? The subscript $x$ tells us it will yield $x$ number of cations (let's call them $A^{y+}$), and the subscript $y$ tells us it will yield $y$ number of anions ($B^{x-}$).

So, the dissociation reaction looks like this:

Therefore, the total number of particles produced from one single molecule is simply the sum of these ions.

Now, we have all the ingredients. We take our master equation and substitute the specific value of $n$ we just found for our electrolyte $A_x{B}_y$.

Removing the parentheses for a cleaner look, we get:

And there we have it! A clean, elegant relationship between the degree of dissociation and the van't Hoff factor for any generic electrolyte. This perfectly matches option (a). Understanding this derivation is incredibly powerful, as it allows you to tackle any variation of this problem, whether the electrolyte is $MgC{l}_2$, $A{l}_2(S{O}_4{)}_3$, or an unknown compound $A_x{B}_y$.