Explanation:
Given, $\mathrm{pH}=3$, Concentration (c) $=0.1 \mathrm{M}$ Now, $\quad \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$
or $\quad\left[\mathrm{H}^{+}\right]=10^{-3}$
The dissociation of weak acid is-
$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}$
We know that,
$\left[\mathrm{H}^{+}\right]=\mathrm{C} \times \alpha$
Where, $\mathrm{C}=$ Concentration
$\alpha=\text { Degree of dissociation }$
or $\quad \alpha=\frac{10^{-3}}{0.1}$
or $\quad \alpha=0.01$
$\therefore$ Percentage of dissociation $=10^{-2} \times 100=1 \%$.