A 0.001 molal solution of a complex in water has the freezing point of . Assuming 100% ionization of — Solutions and Colligative Properties Chemistry Question
Question
A 0.001 molal solution of a complex $\text{MA}_8$ in water has the freezing point of $-0.0054^\circ\text{C}$. Assuming 100% ionization of the complex in water, which of the following is the correct representation of the complex? ($K_f$ of water = $1.86 \text{ Km}^{-1}$)
💡 Solution & Explanation
The observed freezing point depression $\Delta T_f = 0.0054^\circ\text{C}$. Using $\Delta T_f = i \times K_f \times m$, we have $0.0054 = i \times 1.86 \times 0.001$. Solving for the van't Hoff factor gives $i = 0.0054 / 0.00186 = 2.903 \approx 3$. An $i$ value of 3 implies that the complex yields 3 ions upon dissociation. The coordination complex $[\text{MA}_6]\text{A}_2$ dissociates into 1 complex cation $[\text{MA}_6]^{2+}$ and 2 $\text{A}^-$ anions, giving exactly 3 ions. Therefore, correct answer is C.