If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)

<p>The degree of dissociation, often represented as $\alpha$, is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing point.</p> <p>In this case, a weak monobasic acid, when it dissociates, produces two particles: one $H^{+}$ ion and one anion. So if the degree of dissociation is 0.3 ($\alpha = 0.3$), the average number of particles per molecule of the acid ($i$), also known as the van&#39;t Hoff factor, will be $1 + \alpha = 1 + 0.3 = 1.3$.</p> <p>The decrease in freezing point ($\Delta T_{f}$) is given by the formula $\Delta T_{f} = i \cdot m \cdot K_{f}$, where $m$ is the molality of the solution and $K_{f}$ is the cryoscopic constant of the solvent. Therefore, the observed freezing point depression will be 1.3 times the theoretical freezing point depression for a non-dissociating solute (where $i = 1$).</p> <p>So, the observed freezing point will be 30% higher than the theoretical freezing point, given the degree of dissociation of 0.3.</p>