The degree of dissociation of weak acid is directly proportional to

According to Ostwald's dilution law for a weak acid $HA$, the dissociation constant $K_a$ is given by:

$K_a = \dfrac{C\alpha^2}{1-\alpha}$

For a weak acid, the degree of dissociation $\alpha$ is very small ($\alpha \ll 1$), so $1-\alpha \approx 1$.

$K_a \approx C\alpha^2 \Rightarrow \alpha = \sqrt{\dfrac{K_a}{C}}$

The concentration $C$ is the number of moles per unit volume. For $1$ mole of the weak acid, $C = \dfrac{1}{V}$, where $V$ is the volume of the solution.

Substituting $C = \dfrac{1}{V}$ into the expression for $\alpha$:

$\alpha = \sqrt{K_a \times V}$

$\alpha \propto \sqrt{V}$

Thus, the degree of dissociation is directly proportional to the square root of the volume of the solution containing $1$ mole of the weak acid.

Answer: square root of the volume containing $1$ mole of the weak acid