Dissociation constant of 0.01 M weak acid is $10^{-4}$. What is percent dissociation of acid?

Dissociation of a weak acid follows the equilibrium: $HA \rightleftharpoons H^+ + A^-$. Given initial concentration $C = 0.01\text{ M}$ and dissociation constant $K_a = 10^{-4}$, the degree of dissociation $\alpha$ is determined from the expression:
$K_a = \frac{C\alpha^2}{1-\alpha}$.

For weak acids, $\alpha$ is typically small, allowing the approximation $1 - \alpha \approx 1$, simplifying to:
$K_a \approx C\alpha^2$.

Solving for $\alpha$:
$\alpha^2 = \frac{K_a}{C} = \frac{10^{-4}}{0.01} = 10^{-2}$
$\alpha = \sqrt{10^{-2}} = 0.1$

Multiplying by 100 gives the percentage dissociation:
Percent dissociation $= 0.1 \times 100\% = 10\%$

While $\alpha = 0.1$ slightly challenges the approximation, exact calculation yields $\alpha \approx 0.095$, still closest to $10\%$ among standard options. The correct choice is thus $\boxed{\text{C}}$.