In a solvent 50 % of an acid HA dimerizes and the rest dissociates....

Let initial moles of $HA$ is $a$

$2HA \to {H}_{2}{A}_{2}0 . 5a0 . 25a$

$HA ⇌ {H}^{+} + {A}^{-}0 . 5a0 . 5a0 . 5a$

Now, $i = \dfrac{\text{final moles}}{\text{initial moles}} = \dfrac{0 . 25a + 0 . 5a + 0 . 5a}{0 . 5a + 0 . 5a}$

$= 1 . 25 = 125 \times {10}^{- 2}$