Calculating van't Hoff Factor for $\text{Na}_{2}\text{SO}_{4}$
The van't Hoff factor ($i$) quantifies the effect of solute behavior in solution relative to non-electrolytes. For electrolytes like sodium sulfate ($\text{Na}_{2}\text{SO}_{4}$), it depends on the number of ions formed and the degree of dissociation ($\alpha$).
Sodium Sulfate Dissociation
In a dilute aqueous solution, sodium sulfate ($\text{Na}_{2}\text{SO}_{4}$) dissociates into ions:
$ \text{Na}_{2}\text{SO}_{4} \rightarrow 2\text{Na}^{+} + \text{SO}_{4}^{2-} $
From the dissociation equation, one formula unit of $\text{Na}_{2}\text{SO}_{4}$ produces 3 ions (two $\text{Na}^{+}$ and one $\text{SO}_{4}^{2-}$). Therefore, the number of ions, $n$, is 3.
Van't Hoff Factor Formula
The van't Hoff factor ($i$) is calculated using the formula:
$ i = 1 + (n-1)\alpha $
Where $n$ is the number of ions produced upon dissociation and $\alpha$ is the degree of dissociation.
Calculating $i$ for $\text{Na}_{2}\text{SO}_{4}$
Substituting $n=3$ for $\text{Na}_{2}\text{SO}_{4}$ into the formula:
$ i = 1 + (3-1)\alpha $
$ i = 1 + 2\alpha $
Thus, the van't Hoff factor ($i$) for a dilute aqueous solution of $\text{Na}_{2}\text{SO}_{4}$ is $1 + 2\alpha$. This corresponds to Option D.
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