A quantity of 1.04 g of (mol. wt. = 267) was dissolved in 100 g of . The freezing point of the solut — Solutions and Colligative Properties Chemistry Question
Question
A quantity of 1.04 g of $\text{CoCl}_3 \cdot 6\text{NH}_3$ (mol. wt. = 267) was dissolved in 100 g of $\text{H}_2\text{O}$. The freezing point of the solution was $-0.29^\circ\text{C}$. How many moles of solute particle exist in solution for each mole of solute introduced? $K_f$ for water = $1.86^\circ\text{C m}^{-1}$.
Answer: 4
💡 Solution & Explanation
Molality of the complex $m = (1.04 / 267) / 0.100 = 0.003895 / 0.100 = 0.03895 \text{ m}$. Using $\Delta T_f = i \times K_f \times m$, we have $0.29 = i \times 1.86 \times 0.03895$. Solving for van't Hoff factor $i = 0.29 / 0.072447 \approx 4.0$. This means the complex dissociates into 4 particles per molecule (likely $[\text{Co(NH}_3\text{)}_6]^{3+}$ and $3\text{Cl}^-$). Therefore, correct answer is 4.
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