The van't Hoff Factor (i) for a dilute aqueous solution of $Na_2SO_4$ is

The dissociation of $Na_2SO_4$ in an aqueous solution is given by:

$Na_2SO_4 \rightleftharpoons 2Na^+ + SO_4^{2-}$

Let the initial number of moles of $Na_2SO_4$ be $1$ and the degree of dissociation be $\alpha$.

Moles at equilibrium:
$Na_2SO_4 = 1 - \alpha$
$Na^+ = 2\alpha$
$SO_4^{2-} = \alpha$

Total number of moles at equilibrium $= 1 - \alpha + 2\alpha + \alpha = 1 + 2\alpha$

The van't Hoff factor $i$ is the ratio of the total number of moles of particles at equilibrium to the initial number of moles.

$i = \dfrac{1 + 2\alpha}{1} = 1 + 2\alpha$

Answer: $1 + 2\alpha$