An aqueous solution of acetic acid has a molality of 0.1 m. If the...

Acetic acid dissociates as $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$.

The number of ions produced per molecule is $n = 2$.

The degree of dissociation is $\alpha = 1.35\% = 0.0135$.

The van't Hoff factor $i$ is given by $i = 1 + (n - 1)\alpha$.

$i = 1 + (2 - 1) \times 0.0135 = 1.0135$

The depression in freezing point is $\Delta T_f = i \times K_f \times m$.

Substituting the given values:

$\Delta T_f = 1.0135 \times 1.86 \times 0.1$

\Delta T_f = 0.188511 ^{}^{\circ}\text{C} \approx 0.189 ^{}^{\circ}\text{C}

Answer: 0.189^{}^{\circ} C