To solve this, we must account for the dissociation of the weak acid using the van't Hoff factor ($i$).

Step 1: Calculate the van't Hoff factor ($i$). The dissociation of a weak monobasic acid HA is given by: $HA \rightleftharpoons H^+ + A^-$ Here, the number of particles produced per molecule ($n$) is 2. The degree of ionization ($\alpha$) is given as 30% or 0.3. The formula for $i$ in case of dissociation is $i = 1 + (n - 1)\alpha$ . $i = 1 + (2 - 1)(0.3) = 1 + 0.3 = 1.3$

Step 2: Calculate the Depression of Freezing Point ($\Delta T_f$). Using the formula $\Delta T_f = i \cdot K_f \cdot m$ : $i = 1.3$ $K_f = 1.86 \text{ °C/m}$

• $m = 0.1 \text{ m}$

$\Delta T_f = 1.3 \times 1.86 \times 0.1 = 0.2418 \text{ °C}$

Step 3: Calculate the Freezing Point of the solution ($T_f$). The freezing point of pure water ($T_f^0$) is 0 °C. $T_f = T_f^0 - \Delta T_f = 0 - 0.2418 \approx -0.24 \text{ °C}$ .