In the above problem, if the solution is cooled to , how many gram of ice would separate? — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

At $-0.200^\circ\text{C}$, the remaining liquid must have $\Delta T_f = 0.200^\circ\text{C}$. Using $\Delta T_f = K_f \times (\text{moles} / W_{\text{kg\_remaining}})$, we have $0.200 = 1.86 \times (0.0833 / W_{\text{kg}})$. This gives $W_{\text{kg}} = (1.86 \times 0.0833) / 0.200 = 0.775 \text{ kg} = 775 \text{ g}$ of liquid water remaining. Ice separated = $1000 \text{ g} - 775 \text{ g} = 225 \text{ g}$. Therefore, correct answer is B.